Unit 3: STATES OF MATTER

Transcription

1 Unit 3: STATES OF MATTER Chapter 5: Gases 5.: Pressure Pressure: - the amount of force per unit of area [ N/m Pascal (Pa)], measured in kilopascal (kpa), mm g, torr, atmospheric pressure (atm). - in a pressurized container, a pressure can be felt as the particles pushed on the inside wall of the container. Vacuum Barometer: - an instrument (invented by Evangelista Torricelli, hence, the old unit of pressure torr was named after him) to measure atmospheric pressure using a dish filled with mercury and an inverted glass tube filled with mercury. - when the glass tube is placed into the dish filled with mercury, the level of mercury in the tube will drop. owever, the atmospheric pressure of the surface of mercury dish will keep the mercury column at a certain level. - the higher the atmospheric pressure, the higher the mercury is in the column. - at normal atmospheric pressure at sea level, the mercury column is at 760 mm. Manometer: - a device consists of a tube connected to a container of gas to measure the pressure in the gas container. - there are two kinds of manometer: an open-ended manometer and a close-ended manometer. Mercury filled Dish Column of Mercury Atmospheric Pressure A Torricellian Barometer. Open-Ended Manometer: - the tube is open ended to allow atmospheric pressure to come in. 760 mm P gas P atm h h P gas P atm + h h Gas Pressure Atmospheric Pressure Gas Pressure < Atmospheric Pressure Gas Pressure > Atmospheric Pressure Page 6.

2 Unit 3: States of Matter. Close-Ended Manometer: - the tube is a vacuum that is closed ended. The difference in height of the U-tube is the pressure of the gas container. Close-Ended Vacuum Tube P gas h h (Check out animation at under Part : Reading a Manometer) Units of Pressure standard atmosphere (atm) 760 mm g 760 torr 0.35 kpa Note: mm g torr Example : Convert 55 mm g to torr, atm and kpa. 55 mm g 55 torr ( mm g torr) 55 mm g atm 760 mm g 0.69 atm 55 mm g 0.35 kpa 760 mm g 70.0 kpa Example : Convert 350 kpa to atm, torr and mm g. 350 kpa atm 0.35 kpa 3.45 atm 350 kpa 760 torr 0.35 kpa 65. torr torr torr mm g ( mm g torr) Page 7.

3 5.: The Gas Laws of Boyle, Charles, Guy-Lussac, Avogadro & the Combined Gas Law Variables to Describe a Gas:. Pressure (P): - the amount of force per unit of area, measures in kilopascal (kpa) or standard atmosphere (atm).. Volume (V): - the amount of space the gas is occupied; measures in Litre (L). 3. Temperature (T): - the average of kinetic energy of the gas; measures in Kelvin (K). 4. Moles (n): - the amount of gas particle in a closed system; measures in moles (mol). Laws that Relate Gas Variables:. Boyle s Law: - at a constant temperature and moles, pressure is inversely proportional to the volume. Boyle's Law (Constant Temperature) P V k P or PV k V k constant of proportion Volume Pressure Pressure (kpa) (P, V ) 50 kpa L P V (P, V ) Volume (L) Boyle s Law P V P V P Pressure at Initial Condition V Volume at Initial Condition P Pressure at Final Condition V Volume at Final Condition Example : A gas cylinder changed its volume from.50 L to 6.5 L. If it were at 0.35 kpa initially, what would be its final pressure? P 0.35 kpa P? V.50 L V 6.5 L P V P V PV P V 0.35 kpa.50 L ( 6.5 L) As Volume, Pressure P ( )( ) P 40.5 kpa Page 8.

4 Unit 3: States of Matter. Charles s Law: - at constant pressure and moles, volume is directly proportional to the temperature. V T V kt or V k T k constant of proportion Temperature Volume Volume (L) Charles's Law (Constant Pressure) Absolute Zero 0 Kelvin ( 73.5 C) (T, V ) Temperature (degree Celsius) (T, V ) Charles s Law T Temperature at Initial Condition T Temperature at Final Condition V T V T V Volume at Initial Condition V Volume at Final Condition Example : A balloon is has a volume of 3.5 L at 5.0 C. Determine the volume of the same balloon when the temperature is dropped to 5.00 C. V 3.5 L T 5.0 C 98.5 K V? T 5.00 C 78.5 K (Change C to K) V T V T T V T V As Temp, Volume ( 3.5 L )( 78.5 K ) V ( 98.5 K) V 3.03 L 3. Gay-Lussac s Law: - at constant moles and volume, pressure is directly proportional to the temperature. Gay-Lussac's Law (Constant Volume) P T P kt or P k T k constant of proportion Temperature Pressure Pressure (kpa) Absolute Zero 0 Kelvin ( 73.5 C) (T, P ) Page Temperature (degree Celsius) (T, P )

5 Gay-Lussac s Law T Temperature at Initial Condition T Temperature at Final Condition P T P T P Pressure at Initial Condition P Pressure at Final Condition Example 3: A canister is has a pressure of 8.00 atm at 5.0 C. Calculate its pressure if the temperature was to increase to 00.0 C. (Don t do this at home!) P 8.00 atm T 5.0 C 88.5 K P? T 00.0 C K (Change C to K) (P can be in atm) P T P T T P T P As Temp, Pressure ( 8.00 atm )( K ) P ( 88.5 K) P 0.4 atm 4. Avogadro s Law: - at constant pressure and temperature, volume is directly proportional to amount of moles of gas present. V n V kn or V k n k constant of proportion Avogadro' Law (Constant Temperature and Pressure) Constant P at.000 atm Constant T at 73.5 K Moles Volume Volume (L) (n, V ) (n, V ) Amount (mols) Avogadro s Law V n Amount of Moles at Initial Condition n Amount of Moles at Final Condition n V n V Volume at Initial Condition V Volume at Final Condition Page 0.

6 Unit 3: States of Matter Example 4: An excess amount of solid phosphorus is reacted with 9.00 mol of chlorine gas at 3. L to produce phosphorus trichloride gas under constant temperature and pressure. a. Write a balanced equation for this reaction. b. Assuming a complete reaction, what is the volume of phosphorus trichloride produced? a. P 4 (s) + 6 Cl (g) 4 PCl 3 (g) b. n 9.00 mol (Cl ) V 3. L n 9.00 mol Cl 4 mol PCl 6 mol Cl n 6.00 mol (PCl 3 ) V? 3 V n n n V V n V As Moles, Volume ( 6.00 mol )( 3. L ) V ( 9.00 mol) V 49 L 5. Combined Gas Law: - a formula that summarizes Boyle s Charles s and Guy-Lussac s Gas Laws. - allows the user of the formula to determine the change in conditions of the same amount of gas. Combined Gas Law P V P Pressure at Initial Condition V Volume at Initial Condition T Temperature at Initial Condition T P V T P Pressure at Final Condition V Volume at Final Condition T Temperature at Final Condition Example 5: A high altitude weather balloon has a volume of 57. L at 700 mm g and 5.0 C. Determine its volume at its maximum height where the pressure is atm and the temperature is at 45.0 C. V 57. L atm P 700 mm g 760 mm g P atm T 5.0 C 98.5 K V? P atm T 45.0 C 8.5 K (Change C to K) (P can be in atm) P V T PV T T P PV T V V ( atm)( 57. L)( 8.5 K) ( 98.5 K)( atm) V 0 L Because P much more than T, the final Volume. Assignment 5. pg #5 to 7, 9 5. pg. 33 #3 to 34 Page.

7 5.3: The Ideal Gas Law Ideal Gas Law: - a formula that relates pressure, volume, amount, and temperature of an ideal gas (gaseous volume does not account for total particles volumes) at one specific condition. Ideal Gas Law PV nrt P Pressure (kpa) n Amount of Gas (mol) V Volume (L) T Temperature (K) R Gas Constant (L atm) / (K mol) or 8.3 J / (K mol) or 8.3 (L kpa) / (K mol) Example : Determine the mass of propane if it is in a 00 L container at 5.0 C and at 3.0 atm. V 00 L T 5.0 C 88.5 K P 3.0 atm R (L atm)/(k mol) m? n? (need to find n first) PV nrt PV n RT n ( 3.0 atm)( 00 L) L atm K mol n mol ( 88.5 K) For propane, C 3 8, M 44. g/mol m nm m ( mol)(44. g/mol) m g m g.9 kg Example : A 0.85 mol of xenon sample is heated from 3.0 C to 48.0 C. The pressure the gas has also changed from 75 mm g to 854 mm g. What is the change in volume the sample of xenon gas is experienced? n 0.85 mol T 3.0 C 96.5 K T 48.0 C 4.5 K atm P 75 mm g 760 mm g P atm atm P 854 mm g 760 mm g P atm R (L atm) / (K mol) V? V? This might seem to be a combined gas law problem, but we don t have either volume to find both V and V. P V nrt V (.85 mol) ( 96.5 K) nrt K mol P ( atm) V L V L atm 0 L atm 0 (.85 mol) ( 4.5 K) nrt K mol P ( atm) V L V V V L L V 3. L (Volume has Decreased by 3. L) Page.

8 Unit 3: States of Matter Using the Ideal Gas Law formula, we can find the Density (D) or the Molar Mass (M) of any gas. PV nrt PV m m RT (Substitute for n) M M m m Solve for for Density (D ) Solve for Molar Mass (M) V V Denisty (g/l) Example 3: Calculate the molar mass of a gaseous compound containing carbon and hydrogen if its density is g/l at 00.0 kpa and at 5.0 C. Example 4: Calculate the density in g/l of a sulfur dioxide gas at 74 torr and at 8.47 C. 5.4: Gas Stoichiometry Mole-Volume Relationships: PM m mrt M RT V PV PM D RT. Standard Temperature and Pressure (STP): - the amount of any gas at 0 C and 0.35 kpa (Earth s atmospheric pressure at sea level).. Standard Ambient Temperature and Pressure (SATP): - the amount of any gas at 5 C and 00 kpa. STP C and 0.35 kpa ( atm) SATP C and 00.0 kpa Note: The amount of gas is determined by temperature, pressure and volume. The type of gas particles has no effect on these variables. (Avogadro s ypothesis) Page 3. M m RT V P Note: Do NOT memorize these formulas but learn the derivations! D g/l P 00.0 kpa T 5.0 C 98.5 K R 8.3 (L kpa) / (K mol) (We use this R because we are given P in kpa) M? atm P 74 torr 760 torr P atm T 8.47 C K R (L atm) / (K mol) For sulfur dioxide, SO, M g/mol D? M DRT P L kpa g/l K K mol M P ( 00.0 kpa) M g/mol DRT ( ) ( ) D RT M 6.0 g/mol PM ( atm)( g/mol) D g/l L atm K mol D.8 g/l ( K)

9 Example : Determine the amount of oxygen gas in a 5.00 L container under STP and SATP. a. STP b. SATP STP.4 L/mol mol n 5.00 L.4 L n 0.3 mol SATP 4.8 L/mol mol n 5.00 L 4.8 L n 0.0 mol Example : Determine the volume of 3.50 g of nitrogen gas under STP and SATP. a. STP b. SATP n M m 3.50 g 8.04 g/mol mol n M m 3.50 g 8.04 g/mol mol STP.4 L/mol V ( mol)(.4 L/mol) V.80 L SATP 4.8 L/mol V ( mol)(4.8 L/mol) V 3.0 L Gaseous Stoichiometry Procedure (Ideal Gas, STP, or SATP). Predict the products and balance the chemical equation.. Put all the information given under the appropriate chemicals. m PV 3. Find the moles of the given chemical: n or (n CV) or n or M RT mol mol (at STP: n Volume ; at SATP: n Volume )..4 L 4.8 L 4. Check for limiting reagent if necessary (if you are given enough information to find the moles of two chemicals). Use the Limiting Reagent for further calculation. Ignore the Excess Reagent. 5. Find the mole of the required chemical using mole ratio: require coefficient mol of require mol of given given coefficient 6. Convert mole of the required chemical to its mass, concentration or volume equivalence: (m nm) or n C or (PV nrt) or (at STP: V n.4 L/mol; at SATP n 4.8 L/mol). V Page 4.

10 Unit 3: States of Matter Example 3: A piece aluminium metal is placed in an excess amount of sulfuric acid until all the metal is reacted. Calculate the mass of the aluminium used if 5.76 L of hydrogen gas is evolved at STP. Al (s) + 3 SO 4 (aq) 3 (g) + Al (SO 4 ) 3 (aq) M 6.98 g/mol 5.76 L? g STP.4 L/mol n 5.76 L mol.4 L mol mol Al n Al mol mol Al 3 mol m Al nm ( mol Al) (6.98 g/mol) m Al 4.63 g Example 4: 35.4 kg of liquid octane is burned under excess oxygen. Determine the volume of carbon dioxide gas produced at SATP. n C 8 8 C 8 8 (l) + 5 O (g) 6 CO (g) + 8 O (g) 35.4 kg SATP 4.8 L/mol M 4.6 g/mol? L 35.4 kg kmol 4.6 g/mol n CO kmol C mol CO mol C kmol CO V ( kmol CO ) (4.8 L/mol) V CO 6. kl CO Example 5: Ammonia is reacted with oxygen to form nitrogen monoxide and water vapour. n N 3 4 N 3 (g) + 5 O (g) 4 NO (g) + 6 O (g) If 50.0 L of ammonia at 90.0 kpa at 5.0 C were allowed to react with excess oxygen, what would be the pressure of nitrogen monoxide in a collector vessel measuring 30.0 L at a temperature of 0.0 C? 4 N 3 (g) + 5 O (g) 4 NO (g) + 6 O (g) 50.0 L 30.0 L 5.0 C 98.5 K 0.0 C 83.5 K 90.0 kpa? kpa PV (90.0 kpa) (50.0 L) mol RT kpa L ( 8.3 )(98.5K) mol K n NO mol N 3 4 mol NO 4 mol N mol NO P NO nrt V kpa L ( 7 mol)( 8.3 )( 83.5 K) mol K 30.0 L kpa P NO 4 kpa Page 5.

11 Example 6: If 5.5 L of hydrogen at STP is reacted with 7.85 L of nitrogen at 98.0 kpa and at 8.6 C, determine the volume of ammonia formed at SATP. 3 (g) + N (g) N 3 (g) 5.5 L 7.85 L; 98.0 kpa? L STP.4 L/mol 8.6 C K SATP 4.8 L/mol R 8.3 (L kpa) / (K mol) Since there is enough information to determine the moles of two reactants, we need to determine which one is the limiting reagent. mol n 5.5 L mol.4 L n N RT PV ( 98.0 kpa)( 7.85 L) L kpa 8.3 K mol ( K) mol N Let s assume N is the limiting reagent. Calculate the mol actually needed. n mol N 3 mol mol N mol needed But we don t have mol of, we only have mol of. Therefore, is the limiting reagent. (Note: the limiting reagent is NOT always the chemical with the smaller number of moles. You have to always compare like we did above.) Now, we calculate the moles of N 3 formed by using moles of limiting reagent,. n N mol 3 mol N 3 mol mol N 3 Finally, we determine the volume of N 3 produced. V ( mol N 3 )(4.8 L/mol).5595 L V N.3 L 3 N 3 Assignment 5.3 pg #35 to 45, pg #49 to 6 Page 6.

12 Unit 3: States of Matter 5.5: Dalton s Law of Partial Pressures Partial Pressure: - the pressure exerted by one component of a gas mixture. Dalton s Law of Partial Pressures: - the total pressure in a gas mixture is equal to the sum of the pressures exerted by individual gaseous components. n Total ( ) Example : Write the equations for partial pressures and mole components for air, which is composed of N, O, Ar, CO, and other trace gases ( P total P + + n total n + no + nar + nco + trace gases N PO + PAr + P CO P trace gases Dalton s Law of Partial Pressures P Total P + P + P 3 + (Using Ideal Gas Law P nrt n Total RT n RT nrt nrt ) V V V V V (R, T, and V are common to all components in gas mixture) RT RT RT n + n + n (Take out Common Factor and cancel both sides) V V V n Total n + n + n 3 + (Mole Components of Gas Mixtures) N n Mole Fraction: - a unit-less ratio of the mole of any one component to the mole of the entire gas mixture. - since pressure is directly proportional to mole (PV nrt), mole fraction is also the ratio between the partial pressure of one gaseous component to the total pressure of the entire gas mixture. Mole of a Gas Component Partial Pressure of a Gas Component Mole Fraction Total Moles in Gas Mixture Total Pressure in Gas Mixture n χ P n Total PTotal χ Mole Fraction (χ is pronounced chi like chi-tea, not tai-chi) has NO UNITS. Example : Air is composed of of oxygen. (We would spontaneously combust if air has over 3% oxygen.) Determine the amount, in moles, and the pressure of oxygen if a compressed air tank at 4.50 atm has a total volume of 6.50 L and it is at 8.0 C. χ P Total 4.50 atm V 6.50 L T 8.0 C 9.5 K n oxygen? P oxygen? χ oxygen PO P Total P Total V n Total RT V n Total RT P oxygen χ oxygen P Total (0.0947)(4.50 atm) P Total ( 4.50 atm)( 6.50 L) L atm K mol ( 9.5 K) P oxygen atm mol χ oxygen no n Total n oxygen χ oxygen n Total (0.0947)( mol) n oxygen 0.56 mol Page 7.

13 Example 3: A 36.0 g of sample of oxygen gas and 8.5 g of hydrogen gas is mixed in a 3.00 L container at 0.0 C before reacting them to make water. What is the partial pressure of each gas and total pressure of the entire gas mixture? n oxygen m M O O n oxygen.5 mol n hydrogen m M 36.0 g 3.00 g/mol 8.5 g.0 g/mol n hydrogen mol V 3.00 L T 0.0 C 93.5 K P oxygen? P hydrogen? P Total? P oxygen V n oxygen RT P oxygen no V RT P hydrogen V n hydrogen RT P hydrogen n V RT L atm.5 K mol ( mol) ( 93.5 K) ( 3.00 L) L atm K mol P oxygen 9.0 atm ( 6 mol) ( 93.5 K) ( 3.00 L) P Total P oxygen + P hydrogen 9.0 atm atm P hydrogen 3.7 atm P Total 4.8 atm Example 4: A.50 L flask of hydrogen gas at 850 torr is connected to a.50 L flask of ethene by a valve. When the valve is opened, the final pressure of the mixture is atm, what was the initial pressure of ethene in the.50 L flask? V hydrogen.50 L atm P hydrogen 850 torr 760 torr P hydrogen atm (This is the initial partial pressure, not the final partial pressure of. We cannot subtract P Total with. atm for P ethene ) V ethene.50 L P ethene? V Total.50 L +.50 L 4.00 L P Total atm (Note that P Total is already SMALLER than P hydrogen. Total pressure should always be greater than partial pressure. This is telling you the partial pressure given was at initial conditions). Page 8. Since we do not have T, we have to solve for nt. P total V Total n total RT P V ( Total Total atm )( 4.00 L ) n total T mol K R L atm K mol The number of moles for both and C 4 is the same before and after they mix together. P hydrogen V hydrogen n hydrogen RT P V ( n hydrogen T atm )(.50 L ) mol K R L atm K mol n total T n hydrogen T + n ethene T n ethene T n total T n hydrogen T n ethene T (4.6 mol K) ( mol K) (mol K) Finally, find initial partial pressure of ethene before the mixing. P ethene V ethene n ethene RT ( ) L atm ( mol K) nc T R K mol P ethene V.50 L C ( ) P ethene atm

14 Unit 3: States of Matter Example 5: Sodium azide, NaN 3, can be electronically ignited to produced nitrogen used in automobile airbags. The reaction is as follows. NaN 3 (s) heat Na (s) + 3 N (g) An experiment is set up where the nitrogen produced is bubbled through water for collection and has become saturated with water vapour. Suppose 800 ml of nitrogen gas is collected at 0.0 C and has a total pressure of 05 kpa. (Vapour Pressure of water at 0.0 C is torr the concept of vapour pressure will be covered in section 0.8.) a. Determine the partial pressure of the nitrogen gas in this experiment. b. Calculate the mass of sodium azide needed to react to produce this amount of nitrogen. NaN 3 (s) N (g) with O (g) Liquid O 0.35 kpa P water torr 760 torr P hydrogen kpa P nitrogen? V Total 800 ml L P Total 05 kpa T 0.0 C 93.5 K m NaN 3? a. P Total P water + P nitrogen P nitrogen P Total P water P nitrogen 05 kpa kpa P nitrogen kpa b. Solving for n N allows us to use Stoichiometry to find m of NaN 3 P nitrogen V n nitrogen RT P n nitrogen V ( N kpa )( L ) mol RT L kpa 8.3 ( 93.5 K) K mol NaN 3 (s) heat Na (s) + 3 N (g) 65.0 g/mol mol? g n NaN mol N 3 mol NaN 3 mol N mol NaN 3 m nm ( mol NaN 3 )(65.0 g/mol) m NaN.46 g 3 NaN 3 Page 9.

15 5.6: The Kinetic Molecular Theory of Gas In order to explain the behaviour of an ideal gas as we have related with pressure, volume, moles and temperature, a model is needed to give us a picture of how gas particles behave in the molecular level. This model is called the Kinetic Molecular Theory (KMT) of Gas. The Kinetic Molecular Theory of Gas. All gaseous particles are so small that their volumes are essentially zero especially compared to the amount of space between them.. All gaseous particles are constantly moving (hence the word kinetic ). Pressure exists because of the collision of these gaseous particles against the wall of the container. 3. There are no attracting or repelling forces between the particles (again due to the large distances between them). 4. Temperature, express in Kelvin, is the average kinetic energy of the gas particles. (This is also the definition of temperature.) Root Mean Square Velocity ( u or u rms ): - the average velocity of a single gaseous molecule or atom in a closed system. (E k ) avg per Atom / Molecule Derivation of Pressure as it relates to Average Kinetic Energy This derivation is a complicated procedure (see Appendix of page A4 to A7 in the textbook if you have background in senior physics!). In any case, the total pressure of a gas relates to kinetic energy by the formula below. P nn 3 A ( E ) Relating Temperature and Average Kinetic Energy V k ( E ) avg N A (E k ) avg Average Kinetic Energy per Mole nn A k avg Starting with P 3 V PV NA (E k ) avg (Manipulating to almost isolate (E k ) avg ) n 3 RT 3 NA (E k ) avg Average Kinetic Energy m u (E k ) avg per Mole N A (E k ) avg N A (E k ) avg Average Kinetic Energy u Root Mean Square Velocity m mass of one atom or molecule in kg N A Avogadro s Number (Ideal Gas Law: PV nrt PV RT ) n m u Page 30.

16 Unit 3: States of Matter 3 3 RT NA (E k ) avg (In the AP Info Sheet: KE per mol RT ) 3 RT NA m u (Substitute m u for (E k ) avg ) 3RT N m A u OR u u (Solving for u ) 3RT N m A 3kT m [N A m M (Molar Mass in kg/mol) because m is in kg for Physics] ( R N A is replaced by k Boltzmann s Constant) Root Mean Square Velocity u rms u R 8.3 J / (mol K) m mass of one atom / molecule in kg 3RT M 3kT m M Molar Mass in kg/mol k Boltzmann s Constant J/K Example : A sample of argon gas is at STP in a halogen tube. a. Determine the average kinetic energy per mole of the gas. b. Calculate the root mean square velocity of an argon atom. c. Will the average kinetic energy be decreased, increased or remain the same for the following conditions? i. the temperature is increased to 400 C as electricity is applied into the tube. ii. the volume is allowed to double as the argon gas is evacuated into a larger tube. iii. the number of moles is halved as a vacuum is used to draw out some of the argon gas. d. If the argon gas is replaced by the same amount of neon gas under STP, describe the change in the average kinetic energy and the average velocity. At STP: T 73.5 K P.000 atm M g/mol M kg/mol R 8.3 J / (mol K) KE per mol? u rms? a. KE per mol 3 RT 3 (8.3 J / (mol K))(73.5 K) J/mol KE per mol J/mol 3.40 kj/mol b. Since we do not have the mass of Ar, we have the use the first variation of the u rms formula. u rms 3RT M J mol K ( 73.5 K) ( kg / mol) c. (i) As T, KE (ii) As V, KE remain unchanged (KE depends on T only) (iii) As n, KE remain unchanged (KE depends on T only) d. As Ar is replaced by Ne, T remains unchanged, therefore KE remains unchanged. owever, M, u rms J / kg u rms 4 m/s Assignment 5.5 pg. 35 #63 to pg #73 to 80 Page 3.

17 5.7: Effusion and Diffusion Effusion: - when gas passes through from one container to the next through a small hole. Rate of Effusion: - the speed of which a gas pass through a small hole between two containers. - it is the same as the root mean square velocity, u rms. This is because the small hole on the separting wall of the two containers allow one gas particle to pass through at a time. Therefore, the rate of effusion u rms - rate of effusion is generally measures in ml/min Graham s Law of Effusion: - the ratio between the effusion rate of two gases. Rate of Effusion (Gas) u Rate of Effusion (Gas ) u r r rms rms (Gas) (Gas ) 3RT M 3RT 3RT M M Graham s Law of Effusion M 3RT Rate of Effusion for Gas Rate of Effusion for Gas r r M M M Molar Mass in g/mol or kg/mol M (g/mol), Effusion Rate (ml/min) Example : Calculate the ratio of effusion rates between N 3 (g) and Cl (g) under the same temperature. M Cl g/mol M 7.04 g/mol r r N 3 N 3 Cl? r r N 3 Cl M M Cl N g/mol 7.04 g/mol r r N 3 Cl N 3 effuse about.5 times faster than Cl..463 Example : Determine the effusion rate F through a porous barrier if the effusion rate of N through the same barrier is 74 ml/min. r N 74 ml/min M g/mol F M 8.0 g/mol F N r? r r F N M M N F r M ( ) N N r F M F Since F is bigger than N, rf is less than N ( 74 min/l ) 8.0 g/mol g/mol r F r 35 ml/min Page 3.

18 Unit 3: States of Matter Example 3: Using the Effusion Rate Animation by the Chemistry Department of Iowa State University ( ffusion_macro.html), the volumes and times for a selected gas are determined. Complete the table below and find the molar mass of unknown gas X. Gas Molar Mass (g/mol) Volume Time Effusion Rate (ml/min) 00 ml 5 sec O 00 ml 0 sec Xe 00 ml 40.6 sec Kr 00 ml 3.4 sec Unknown X? 00 ml 3.5 sec Gas Molar Mass (g/mol) Volume Time Effusion Rate (ml/min).0 00 ml 5.00 sec 00 ml 00 ( 5.00 / 60)min O ml 0.0 sec 00 ml ( 0.0 / 60)min Xe ml 40.6 sec 00 ml ( 40.6 / 60)min 47.8 Kr ml 3.4 sec 00 ml 85. ( 3.4 / 60)min Unknown X? 00 ml 3.5 sec 00 ml ( 3.5 / 60)min 55.3 We can use any gas to set up a ratio with r X. We choose O because its rate is an even number. r O 300 ml/min M O 3.00 g/mol r X 55.3 ml/min M X? ( r ) ( r ) X M r r O X O O M M X O M X ( r ) O O ( r ) X M ( ) r ( ) O M O M X M X ( r ) X ( ) 300 ml/min ( 3.00 g/mol ) ( 55.3 ml/min) Since r X is less than r O, X is bigger than O. M X 44. g/mol Page 33.

19 Diffusion: - the natural tendency of a gas from an area of high concentration travels to an area where it has a lower concentration without any barriers (we sometimes called it moving down the concentration gradient ) until an equilibrium state is reached. - in general, the Rate of Diffusion is SLOWER than the Rate of Effusion. This is because even without the barrier, there are air particles the gas has to fight through to reach the area of low concentration. This is also true where both areas have roughly equal pressures. - the only time where the Rate of Diffusion is Greater than the Rate of Effusion is when the gas is traveled FROM an are of igh Pressure to very Low Pressure (near vacuum) like a breach on the hull of a starship or a plane. (See Diffusion Animation at 5.8: Real Gases Real Gas: - a gas where its properties deviate from an ideal gas. - when we account for the fact that gaseous particles have volume, and they actually have attraction between particles (intermolecular bonds), the ideal gas law become less accurate. - this happens mainly when pressure is extremely high, volume is small and temperature is low. At such conditions, the small volume will make the particle size an important matter. This is the same for attraction forces between particles because temperature is low and they are moving a lot slower. Ideal Gas Low Pressure igh Temperature Real Gas igh Pressure Low Temperature Various Gases under igh Pressures at 00 K Strong e Repulsion between molecules due to Very Small Volume Nitrogen Gas at various Temperatures Note that at Low T (00 K), nitrogen deviates greatly at high P Strong attractions between molecules due to Stronger Intermolecular forces (section 0.) Note that at igh T (000 K), nitrogen behaves closely to Ideal Gas even at high P Commonly use Low Pressures (0 to 0 atm) and conditions that conform to an Ideal Gas (PV nrt) Page 34.

20 Unit 3: States of Matter van der Waals Equation: - Johannes van der Waals first proposed correctional factors (volume and pressure due to intermolecular attractions) to correct for real gases under low volume and high pressure. Real Gas Pressure V actual V container nb Container Volume P real n number of moles b van der Waals Volume Constant V actual Volume of Spaces between Molecules nrt n a V nb V Volume Correction Pressure Correction n P real P a V n number of moles a van der Waals Pressure Constant V Volume of Container P Pressure with volume correction but prior to Pressure Correction At low pressure, there are a lot of spaces between molecules. Therefore, V actual V container and (nb 0). At high pressure, there is a lot less space between molecules and V actual < V container and (nb is significant). van der Waals Constants for Various Gases Intermolecular Forces or Electron Repulsions can significantly affect the u rms and the collision rate against the wall of the container. This causes pressure to behave differently than ideal gas. Gas M a b M a b (g/mol) (atm L ) / mol Gas (L/mol) (g/mol) (atm L ) / mol (L/mol) e Ne N Ar O Kr Cl Xe C N CO O CCl Note: - In general, a >> b (>> much bigger). This is because intermolecular forces has a greater effect on pressure than the amount of volume need to be corrected due to increase pressure. - As M, a and b. This is due to more electrons available for intermolecular forces (section 0.) - Polar Molecules especially the ones with hydrogen bonds (N 3 and O) have bigger a and b values than non-polar molecules with the similar molar mass due to stronger intermolecular forces. Page 35.

21 n a P + V ( V nb) nrt van der Waals Equation for Real Gases P real nrt n a V nb V (rearranged to solve for nrt as it appears in the AP Information Sheet) Example : 50 kg of oxygen gas is being transported in a 00 L truck under high pressure at 50 C (boiling point of oxygen is 83 C). Determine the pressure in the tank of the truck using ideal gas law and van der Waals equation of real gas. Contrast the differences in the results g n mol 3.00 g/mol V 00 L T 50 C 3.5 K R 0.08 (atm L) / (mol K) a.36 (atm L ) / mol b L/mol P ideal? P real? P ideal nrt V atm L ( mol)( 0.08 )( 3.5 K) mol K 00 L nrt n P real a V nb V mol 0.08 P real 00 L mol atm L ( )( )( 3.5 K) mol K L [ ( )( )] mol atm L (.36 ) P ideal 37 atm mol mol 00 L P real 83 atm (P ideal is like Theoretical result and P real is like Experimental result) Pideal Preal % Pressure Difference 00% P ideal 37 atm 83 atm % Pressure Difference 00% 37 atm % Pressure Difference.8% P real < P ideal by.8%, which is a significant margin. The drop in pressure is due to the low temperature and the high pressure (pressure correction, an /V, alone is 747 atm). These two conditions combined with a relatively small volume for the large number of moles (nb 49 L of molecules volume out of 00 L of container space) caused the intermolecular force to attract the molecules much more than otherwise. This result is a significant drop in the overall pressure. Assignment 5.7 pg. 36 #8 to pg. 36 #85 and 86 Page 36.

22 Unit 3: States of Matter Chapter 0: Liquids and Solids 0.: Intermolecular Forces Intermolecular Forces: - attraction forces between molecules in a compound - the strengths of the intermolecular forces explain the physical properties of compounds (solubility, boiling and freezing points). a. van der Waals Forces: - named after Johannes van der Waals who studied real gases and interaction between molecules. - there are two kinds of van der Waals forces. - they are Dispersion Forces and Dipole-Dipole Interactions. i. Dispersion Forces: - also known as London Dispersion Forces (named after Fritz London who first proposed how this force works). - on average, the non-polar molecules do not have any permanent dipoles like polar molecules - the dispersion is the temporary dipole that forms within the molecules even in non-polar molecules due the constant motions of electrons. In one instance, they can move to one side of the molecule making it temporary polar. In another instance, electrons will move and the direction of this temporary dipole will switch. Electrons sloshing causes Temporary dipoles δ + δ δ δ + - This constant sloshing around of electrons causes non-polar molecules to have these temporary dipoles. These temporary induced dipoles are what cause the attractions between non-polar molecules. induced dipole δ + δ δ + Dispersion Forces δ Dispersion Forces δ + δ δ + Dispersion Forces δ δ δ + δ δ + δ δ + δ δ + Page 37.

23 - even monoatomic element like elium has London Forces. (Check out animation at - in general, the higher the molar mass or the more electrons there are in a molecule, the stronger the London Dispersion Force (attraction between molecules intermolecular force). This causes an increase in melting and boiling points of the chemical. - Note: All molecules have electrons. ence, ALL molecules have London Dispersion Force. # of e or molar mass in atom or molecule, London Dispersion Force, Melting and Boiling Point Example : Explain the boiling points and the melting points of the noble gases. Noble Gases # of e Molar Mass (g/mol) Melting Point Boiling Point e C ( K) 69 C (4 K) Ne C (4 K) 46 C (7 K) Ar C (84 K) 86 C (87 K) Kr C (6 K) 53 C (0 K) Xe C (6 K) 08 C (65 K) Rn C (0 K) 6 C ( K) All atoms of noble gases are monoatomic non-polar. The only intermolecular force that governs the melting and boiling points is the London Dispersion Force. As the number of electrons in the noble gases increase, London dispersion force makes the attraction between the atoms greater. This in turn has an effect of increasing the boiling and melting point of the noble gas as one goes down the column. ii. Dipole-Dipole Interactions: - also known as simply Dipole Interactions. - intermolecular forces resulted from polar molecules. - dipole interaction is much stronger than Dispersion Force. δ + 3. Cl O 3.4 δ Cl δ + 3. δ + δ Overall Dipole for OCl Dipole-Dipole Interactions Page 38.

24 Unit 3: States of Matter Example : Order the boiling points from the least to greatest for the following compounds with similar molar mass. P 3 (34.00 g/mol), C 3 F (34.04 g/mol), and Si 4 (3.3 g/mol) Since P 3, C 3 F and Si 4 have similar molar mass, any differences in boiling points cannot be due to London Dispersion forces. Since dipole-dipole interactions exist in polar molecules, we have to examine the molecular geometry and structure of each compound... P. δ Overall Dipole of P3 δ +. P 3 has a trigonal pyramid geometry (VSEPR) and is Polar. Even though the P bonds have no polarity (electronegativities of P and are the same), the lone pair on one end of the P atom causes an uneven distribution of electrons F C δ Overall Dipole of C3F.. δ +. C 3 F has a tetrahedral geometry and is very polar. The C F bond along with the C bonds have strong polarity. The overall dipole moment for the molecule has electrons around the F atom.. Si 4 has a tetrahedral geometry with equal dipoles of Si bonds cancels out all bond polarities. ence, Si 4 is Non-Polar...6 Si.. Since non-polar molecule have no dipole interactions, Si 4 should have the lowest boiling point. P 3 is less polar than C 3 F due to the difference in electronegativities between P bond and C F with C bonds. Therefore, C 3 F must have the highest boiling point. Boiling Point: Si 4 < P 3 < C 3 F b. ydrogen Bonds: - are intermolecular bonds that involve hydrogen atom with very electronegative atom that also consists of lone pairs. - these include O, N, and X (halogen atoms) bonds. - the resulting molecule is always polar. Therefore, all hydrogen bonding molecules also have dipole interactions. - hydrogen bond is the STRONGEST of the intermolecular bonds. Page 39.

25 δ+ 3δ N δ+ δ+ ydrogen Bonding (Check out the ydrogen Bond Animation at Example 3: Account for the differences in the boiling points of the compounds listed below. Molecule Molar Mass London Dipole ydrogen (g/mol) Dispersion Forces Interactions Bonds Boiling Point OF C (8 K) Ne C (7 K) F C (9 K) O C (373 K) N C (40 K) C C ( K) Again, we need to draw the structural formulas of these molecules and compare their polarities. F O F OF is polar with dipole interactions Ne F Ne is non-polar with dispersion forces only F is polar with hydrogen bonds O O is polar with hydrogen bonds Notice that the hydrogen bond molecules (F, O and N 3 ) have boiling points much higher than molecule with just dipole interactions (OF ) and the ones with only London Dispersion Forces (Ne and C 4 ). N C N 3 is polar with hydrogen C 4 is non-polar bonds with dispersion forces only Page 40.

26 Unit 3: States of Matter Example 4: Given the graph below on the boiling points of hydrogen compounds with different group series, explain the following using the concepts of chemical bonding. Boiling Point (degree Celsius) O F N 3 C 4 Boiling Point of ydrogen Compounds P 3 S Cl Si 4 As 3 a. The hydrogen compounds in the Group (VIA) series have higher boiling points than hydrogen compounds in the other series. b. The first hydrogen compounds in Groups (VA), (VIA) and (VIIA), namely N 3, O and F, have higher boiling points than most other hydrogen compounds in their respective series. On the other hand C 4 has a lowest boiling point in its own Group (IVA) series. a. All hydrogen compounds in the Group (VIA) series are very polar and have hydrogen bonds. The V-shape molecules characterized in Group (VIA) create a greater dipole moment than other series (Group (VA) with its trigonal pyramid shape and Group (VIIA) with its linear form). On the other hand, all hydrogen compounds in the Group (IVA) series are non-polar and only have London dispersion forces. Since hydrogen bonds are stronger intermolecular forces than London dispersion forces, the hydrogen compounds in the Group (IVA) series have the lowest boiling points than the counterparts in the other series. b. N 3, F and O have stronger hydrogen bonds than most other hydrogen compounds in their series. The difference between the electronegativities with is the greatest in row (Electronegativities increase from left to right and from bottom to top of the Table). This huge difference in electronegativities in N 3, F and O is what causes their boiling points to buckle the trend. After N 3, F and O the rest of the hydrogen compounds in the respective series follow the effect of London dispersion forces, the higher the molar mass, the stronger the dispersion forces, and the increase in boiling points is the result. C 4 in the Group (IVA) series do not buckle the trend because the entire series are non-polar. The only intermolecular force at work is the London dispersion force. ence, C 4 has a lower boiling point than Si 4. Page 4. Se Ge 4 Br I Te Sb 3 Sn Periods Group (IV) Bp Group (V) Bp Group (VIA) Bp Group (VIIA) Bp

27 Summary of Intermolecular Forces. Intermolecular Bonds involve in a compound explain its physical properties such as solubility ( like dissolves like ), boiling and melting points (energy involved in physical phase change).. van der Waals Forces consist of London Dispersion forces (apply to all molecules) and Dipole Interactions (apply to polar molecules). 3. ydrogen Bonding is the strongest of all intermolecular bonds. Strength of Intermolecular Forces van der Waals Forces ydrogen Bond > Dipole Interaction >> London Dispersion Force >> (much stronger than) 0.: The Liquid State Surface Tension: - the inward force of a liquid to minimize its surface area. - water s hydrogen bond cannot attract with molecules in the air. Therfore, the higher net force is pushed inward, leaving a spherical surface with a higher surface tension. Water Droplet ydrogen Bonds drawn inward and sideways only creating Surface Tension Surfactant: - a surface-active agent that decreases the surface tension (example: detergent). Water molecules inside the droplets have hydrogen bonds in ALL directions Capillary Action: - when liquid suddenly rises in a narrow tube. - there are two forces that causes capillary action to occur. They are cohesive forces and ahesive forces. a. Cohesive Forces: - forces between molecules (intermolecular forces like London Dispersion Force, Dipole interactions, and ydrogen Bonding). b. Adhesive Forces: - forces between the liquid and the inner surface of the container. - in the case of capillary action, it would be the inside of the narrow tube. - polar liquid molecules with inner surface of the container that are also polar will create a large adhesive force (example: water and glass). - non-polar liquid molecules with polar inner surface of the container will create a small adhesive force (example: mercury and glass). Page 4.

28 Unit 3: States of Matter Polar Glass Surface cohesion weight of mercury Both adhesion and cohesion forces are both strong. owever, adhesion > cohesion. Therefore, water meniscus is concave. Cohesion forces are weak (g is non-polar). There is no adhesion. owever, cohesion > adhesion. Therefore, mercury meniscus is convex. Viscosity: - the measure of a liquid resistence to flow. - polar liquids tend to have high viscosity because they have both adhesive foce (with the container s inner surface) and cohesive forces (with other molecules within the liquid). These attractions slows down the flow rate. Thus, viscosity increases. Assignment 0. pg #35 to pg. 50 #4 to 44 Page 43.

29 0.3: An Introduction to Structure and Types of Solids Structural Classification of Solids. Amorphous Solids: - solids where the arrangements of atoms are very disoragnized with no repeating patterns. Examples: glass (heated silica), quartz glass (quartz crystal has a crystalline structure) and opal. There is no regular molecular arrangement in a quartz glass Structure of Silica. Crystalline Solids: - solids where the arrangemnets of atoms are highly organized. - the 3-dimensional positions of the atoms in a crystalline solid is called a lattice. - the smallest repeating structure in a lattice is called a unit cell. X-Ray Diffraction: - a diffraction method using x-ray to analysis the crystalline structure of a solid. - basically the x-ray is scattered because the crystalline solid provides an opening in the unit cell to interfere the x-ray. Basic Concept of X-Ray Diffraction X-Ray Diffraction and the representation of a simple cubic unit cell (one out of many cubic crystalline structures) X-Ray Diffraction and the representation of a hexagonal closest packed unit cell (one out of many closest packing arrangement) Page 44.

30 Unit 3: States of Matter Intramolecular Forces: - attraction forces between atoms WITIN a molecule. - the strengths of the intramolecular forces explain the amount of energy involved in a chemical change. Types of Crystalline Solids a. Ionic Solids: - solids that involved a combination of metal and non-metal elements. - intramolecular bonds are ionic bonds. b. Covalemt Solids: - solids that involved a combination of different non-metals. c. Atomic Solids: - elemental solids that involve: i. physical mixtures of metals (metal alloys) and metal elements (gold, lead etc.). ii. metalloids (elements close to the staircase of the Table carbon, silicon etc.). iii. Group (VIIIA) Elements (e (s), Ne (s), Ar (s), Kr (s) and Xe (s)) at extremely low temeperature where the noble gases freeze. 0.4 & 0.5: Structure and Bonding in Metal & Network Atomic Solids Different Types of Atomic Solids. Metals and Alloys: - characterized by delocalized non-directional covalent bonds (can be pictured as a bunch of positive charged nuclei floating in a sea of valence electrons ). See animation ( - These valence electrons have not been ionized, but they are usually so far remove from the nucleus that they can be viewed as ionized (hence the word delocalized). There is no dipole interactions involved in metallic bonds (therfore the term non-directional), but they do exhibit small covalent properties due to their close proximity to each other. - since there is a sea of valence electrons, metals and alloys are very good conductors of electricity. - all metals and alloys are closest packing (atoms are packed in layers to minimize spaces in between). This explains the high densities, good heat conductivities. - have different ranges of hardness and melting points. Iron-Iridium Alloy where atoms Metallic Bonds in metals and Alloys with are organized in layers metallic nuclei in a sea of valence electrons Page 45.

31 . Metalloids: - consists of elements near the staircase of the Table (examples are carbon in a form of diamond and silicon dioxide in a form of quartz crystal). - commonly called network solids because they form giant molecules characteriezed by directional covalent bonding. - contains no discrete molecular units where an array or network of atoms are held together by conventional covelent bonds (which are directional with dipoles) of neighboring atoms. - due to a more organized crystalline structure, they are typically hard and have high melting points. - because there is no sea of valence electrons as in metallic solids, network solids tend to be poor electric conductors (good insulator). Exception is with silicon elements. Si has smaller networks than diamonds, allowing some electrons to pass through. Therefore, silicon is sometimes called a semiconductor. Carbon as Graphite has weak layered network with delocalized bonding network (only some carbon atoms are connected) Carbon as Diamond has strong tetrahedral network where all four bonding sites of each carbon atoms are connected Carbon Network as C 60, Buckminsterfullerene, was discovered in 985. Quartz crystal has a hexagonal network structure. For a 3-D look, check out the website at Exploring Earth ( 3. Solid Group (VIIIA) Elements: - when noble gases freeze, they form a cubic closest packing structures, but are non-directional and non-covalent. - the only intermolecular forces between them are London Dispersion forces. Solid Ar Crystal Structure Solid Ne Crystal Structure Page 46.

32 Unit 3: States of Matter 0.6: Molecular Solids Properties of Molecular (Covalent) Compounds. Molecular Compounds tend to have much Lower Boiling and Melting Points than ionic compounds. This is because solid covalent compounds uses weak intermolecular forces to form their lattice structures, which does not take much energy to break them. Their boiling points are lower than ionic compounds because there are no ion interactions in liquid state, only intermolecular forces.. Molecular Compounds are Soft. Again, covalent compounds have a weak lattice structure made of intermolecular bonds that makes them soft. 3. Molecular Compounds tend to be More Flammable than ionic compound. This is due to the some non-metals like carbon and sulfur, which combine readily with oxygen in combustion reactions. 4. Most Molecular Compounds are Insoluble in Water. Because water is very polar and has lots of hydrogen bonds, it can only dissolve covalent compounds that are polar as well Like Dissolves Like. Since most covalent compounds are fairly non-polar, they do not dissolve in polar water well. 5. Molecular Compounds do NOT Conduct Electricity in their Solid States due to a lack of delocalized electrons. 6. Soluble Molecular Compounds do NOT Conduct Electricity in Water. This is simply due to the fact that covalent compounds do not dissociate into ions or electrolytes like soluble ionic compounds do. Due to the hydrogen bonds in water, it forms a honeycomb shape and expands in volume when it crystallizes into ice Even though no two snowflakes are alike, all of them have a basic hexagonal shape as dictated by the bent shape of water molecule and its hydrogen bonds Dry Ice, CO (s), is a covalent compound Even a halogen like I (s) Phosphorus, P 4 (s) can form that has a crystalline structure has a crystalline structure crystalline structure Page 47.

33 0.7: Ionic Solids Properties of Ionic Compounds. Ionic Compounds have a definite Crystalline Structure and are Poor Conductors of Electricity and eat in their Solid Form. Conduction of electricity and heat requires ions to move freely within the solid. The lattice structures of the solid ionic compounds do not allow ions to move freely. + Na + Cl Cs + Cl Simple Cubic Unit Crystalline Structure of NaCl Body Centred Cubic Unit Crystalline Structure of CsCl Face Centred Cubic Unit Crystalline Structure of ZnS. Ionic solids are generally igh Melting Points (typically 300 C to 000 C). Since a strong force can only shatter the crystal but not bend it as in metals, the energy needed to completely break up the lattice structure (lattice energy) is very large and it is the same energy needed to melt the ionic compounds. 3. Ionic solids are ard and Brittle. The lattice structure of all ionic compounds holds the ions in definite positions. When the compound encountered a strong force, the close proximity of the ions stay close together. This causes the crystal to shatter, not bent like metal solid would. 4. Ionic solids can be Melted to form Liquids that are Electrical Conductors. Ionic solids melt when the ions gain enough energy to break the lattice structure. They are move freely and can carry electrical charge through the liquid. This explains why a molten ionic substance conducts electricity, but a solid ionic material doesn't. The ions move through the liquid can carry charge from one place to another. 5. Soluble ionic solids dissolve to form solutions that are Electrical Conductors. (Not all ionic substances are soluble in water.) Soluble ionic compounds form electrolytes (ions in aqueous from) that allow the conduction of electricity. General Strength of Intramolecular Bonds in Solids Metallic > *Ionic > Network > Molecular > Group (VIIIA) *Ionic Bonds can be Stronger than some Metallic Bonds (depending on the metal / alloy) Atomic Solids Intramolecular Bonds >> Intermolecular Bonds Assignment 0.7 pg #67, 68 Page 48.

34 Unit 3: States of Matter 0.8: Vapour Pressure Vapour Pressure: - the pressure existed above a liquid when its rate of evaporation is the same as the rate of its condensation. - sometimes refers to as equilibrium vapour pressure because equilibrium means the same rate of a two-way process. Vapour Pressure P atmosphere P vapour + P g column P vapour P atmosphere P g column - in general, for a particular compound, as temperature increases, the vapour pressure increases. - This is because the higher the temperature, kinetic energy increases and more lquid molecules will overcome the intermolcular forces to become a gas. Therby, increasing vapour presuure. Temperature, Vapour Pressure Number of Molecules with given energy Note that more molecules can overcome the intermolecular forces needed to be in gaseous state at a higher temperature. T T > T T Energy required to overcome intermolecular forces. KE Volatile: - when liquids has a high vapour pressure. - when comparing compounds at the same temperature, the compound that has a lower boiling point is considered more volatile because it will have a higher vapour pressure. - when intermolecular forces are weak, the compound is more volatile. This is simply because it takes less energy for the liquid to evaporate. eat of Vaporization ( vap ): - amount of heat needed to boil mole of a compound from lquid to gas at its boiling point. - the stronger the intermolecular force, the higher the heat of vaporization. Intermolecular Force, vap Page 49.

35 Calculating Vapour Pressure P vap Vapour Pressure (torr) vap eat of Vaporization (J/mol) T Temperature in K R Gas Constant (8.345 J/ (K mol)) A Constant Factor relating to Vapour Pressure of a Particular Liquid vap P vap A e RT (P vap is an exponential function of the reciprical of T) vap ln (P vap ) RT ln Ae (Natural Log [ln] both sides of the equation can bring the exponent of the natural number down) vap ln (P vap ) ln (A) + RT ln e [Using Logarithm Law: ln(mn) ln (M) + ln (N)] vap ln (P vap ) ln (A) + RT [ln (e x ) x] and let ln (A) C vap ln (P vap ) + C R T [Now, ln (Pvap ) has a linear relationship with (/T)] Note that at a certain temperature, let s say 300 K, the P vap in increasing order is Ethylene Glycol < Water < Ethanol < Diethyl Ether. This is because diethyl ether, unlike other compounds has no hydrogen bonding (least intermolecular forces), and therefore largest P vap. Ln(P) Ethanol Water Vapour Pressure (torr) Diethyl Ether Vapour Pressure of Various Substances Diethyl Ether Ethanol Water Ethylene Glycol 760 torr atm 300 K Temperature (K) Boiling Point for water Boiling Point for ethanol Note: For substances, when P vap atm 760 torr, the corresponding temperature represents the boiling point. When y ln(p vap ) is plotted against x (/T), the graphs turned linear vap ln (P vap ) + C R T -0-5 Ethylene Glycol Page 50. /T (K) vap with slope and y-intercept C ln(a) R (Check out Animation at portfolio/text_images/030_vaporpvstemp.mov)

36 Unit 3: States of Matter Example : The following data is collected for octane. Using your TI-83 Plus Calculator, a. graph P vap versus T. b. linearized the previous graph by graphing ln(p vap ) versus (/T). c. obtain the equation of the linearized graph. d. calculate the vap of octane. e. determine the normal boiling point at 760 torr. a. Graph P vap versus T. Entering Data using TI-83 Plus Calculator: Temperature (K) Vapour Pressure (torr) STAT ENTER Enter Values Check if you entered the Correct Number of Data Values 6 th Score entered To plot Scatter Plot on TI-83 Plus Calculator:. Because we are plotting from a set of data, not from an equation, we have to turn on STAT PLOT. nd STAT PLOT Y. Set Window by identifying the Minimum and Maximum Values of Both Variables ENTER Select Scatter Plot 3. Graph Scatter Plot 800 GRAP WINDOW x: [60, 460, 0] y: [0, 800, 00] Pvap (torr) 400 The reason we set Y max 800 is because we will have to solve for the boiling point at 760 torr in part d. T (K) 460 Page

37 b. Linearized the previous graph by graphing ln(p vap ) versus (/T).. Use L 3 /L to obtain (/T). Use L 4 ln(l ) to obtain ln(p vap ) Cursor must be on the column heading before entering formula Cursor must be on the column heading before entering formula must be used to enter formula ALPA MEM nd L must be used to enter formula ALPA MEM nd L 3. Set up Plot nd + ENTER + Select L 3 nd STAT PLOT Y Select Plot Select On L 3 3 Select L 4 4. Turn off Plot nd nd ENTER L 4 STAT PLOT Y Select Plot Select Off 4 5. Set Window by Using ZoomStat (Automatic Adjusted Window Settings) ENTER Graph will appear ZOOM Select Option 9 ln(pvap) /T (K) Page 5.

38 c. obtain the equation of the linearized graph. Turn Diagnostic On nd CATALOG 0 ENTER Again ENTER Select DiagnosticOn Unit 3: States of Matter Note: After DiagnosticOn is selected, it will remain ON even when the calculator is turned Off. owever, resetting the calculator will turn the Diagnostic Off (factory setting). Turning Diagnostic On gives correlation (a measure of how well the equation fits the data points). Obtain LinReg(ax + b), Copy Equation to Screen and Graph Best Fit Line on Scatter Plot To access Y, press ENTER STAT Select CALC, use VARS Select Option ENTER for Y Choose Option 4 Linear Regression (ax + b) LinReg on L 3, L 4 Select Option LinReg (a + bx), L 3, L 4, Y calculates the equation and copied into the Y Screen. ENTER When r ±, it is Y Be sure Plot is On and Plot is Off considered a good fit 3. Set Window by Using ZoomStat (Automatic Adjusted Window Settings) ENTER Graph will appear ZOOM Select Option 9 ln(pvap) Linear Equation: y x ln (P vap ) (/T) /T (K) Page 53.

39 d. Calculate the vap of octane ln (P vap ) (/T) vap ln (P vap ) + C vap J/mol 40. kj/mol R T slope vap K R vap K (8.345 J/ (K mol)) e. Determine the normal boiling point at 760 torr. P vap 760 torr vap J/mol R J / (K mol) vap K R C T? vap ln (P vap ) + C R T ln (760) (/T) ln (760) (/T) ln( T T ln ( ) 847 T 396 K 3 C Comparing P vap of the Same Liquid at Two Different T: ( ) ln P + vap, T vap ln( ) R T P + vap, T vap C (Equating each equation with C) R T ln ( P ) ln( ) vap, T P vap, T vap R T vap R T (Bring both P vap and T on either side) P ln P vap, T vap, T vap M [Using Logarimic Law ln(m) ln(n) ln ] R T T N Clausius-Clapeyron Equation of Vapour Pressures Pvap, T vap ln P vap, T R T T P P vap (torr) at T vap eat of Vaporization (J/mol) vap,t P vap,t P vap (torr) at T T and T Different Temperatures (K) Page 54.

40 Unit 3: States of Matter Example : At the top of Mount Everest with an altitude of 9,08 feet, the atmospheric pressure drops to 53 torr. Given that the heat of vaprization of water is kj/mol at its normal boiling point, which is 00 C at atm, determine the boiling point of water at that elevation. P vap,t atm 760 torr T 00 C K P 53 torr vap,t vap kj/mol J/mol R J/ (K mol) T? R vap P ln P P ln P vap, T vap, T vap, T R T vap vap, T T T T R Pvap, T ln + vap P vap, T T T T R Pvap, T ln + vap P vap, T T T - J K K mol 760 torr ln + ( J/mol) 53 torr K T 344 K 7. C Example 3: Liquid oxygen at 00 C has a vapour pressure of 0.7 kpa. It boils at 83 C. What pressure msut be exerted to gaseous oxygen to condense it to a liquid at 0 C if its heat of vaporization is 6.8 kj/mol? P vap,t 0.7 kpa 80.3 torr T 00 C 73.5 K T 0.0 C 53.5 K vap 6.8 kj/mol 680 J/mol R J/ (K mol) P vap,t? 760 torr 0.35 kpa An enormous pressure is needed to keep oxygen in liquid phase at a temperature above its regular boiling point. P ln P, vap T, vap T 80.3 torr ln P vap, T R vap T T ( 680 J/mol) ( J/ ( K mol) ) 80.3 torr ln P vap, T 80.3 torr ln P vap, T 53.5 K 73.5 K e e (e ln(x) x) 80.3 torr P vap, T P vap, T 80.3 torr P vap,t torr 37.0 atm Page 55.

41 eating Curve: - a graph of temeparture versus time as a substance is heated from a solid phase to a gaseous phase. - when a substance is undergoing a phase change, its temperature remains at a constant (the plateau on the heating curve) until all molecules aquired enough energy to overcome the intermoelcular forces nexessary. This is commonly refered to as the potential change of a substance. - when a substance is undergoing temperature change within a particular phase, it is refered to as kinetic change. eat of Fusion ( fus ): - the amount of heat needed to melt one mole of substance from solid to liquid at its melting point. eating Curve of Water 00 C Temperature 0 C Phase Change Solid / Liquid at Melting Point Water Phase Change Liquid / Gas at Boiling Point Steam Ice Time Normal Melting Point: - the temperature where vapour pressures of liquid is equaled to that of a solid under normal atmospheric condition of atm. - when temperature is at the normal melting point, both liquid and solid coexist. - when T < Melting Point, P vap Solid < P vap Liquid. ence, any liquid s vapour pressure will be used by the solid to equilibrate its own vapour pressure. Thus, liquid will slowly become a solid below its melting point. - when T > Melting Point, P vap Liquid < P vap Solid. As such, any solid s vapour pressure will be used by the liquid to equilibrate its own vapour pressure. Therfore, solid will slowly become a liquid above its melting point. Normal Boiling Point: - the temperature where vapour pressures of liquid is equaled to that of a gas under normal atmospheric condition of atm. - when temperature is at the normal boiling point, both liquid and gas can coexist. - when T < Boiling Point, P vap Liquid < P vap Gas. ence, any vapour pressure of a gas will be used by the liquid to equilibrate its own vapour pressure. Thus, the gas will slowly become a solid below its boiling point. - when T > Boiling Point, P vap Gas < P vap Liquid. As such, any liquid s vapour pressure will be used by the gas to equilibrate its own vapour pressure. Therfore, liquid will slowly become a gas above its boiling point. Page 56.

42 Unit 3: States of Matter Sublimation: - when a solid becomes a gas, and vice-versa, directly without undergoing a liquid phase in between. Sublimation Point: - the temperature where vapour pressures of solid is equaled to that of a gas. Normal Sublimation Point: - the temperature where vapour pressures of solid is equaled to that of a gas under standard atmospheric pressure of atm. Supercooled Liquid: - when a liquid is cooled too rapidly, its temperature is below the melting point. - happens because molecules need time rearrange themselves for reystallization to become a solid. Superheated Liquid: - when a liquid is heated too rapidly, its temperature is above the boiling point. - happens because molecules need time to completely break apart their intermolecular bonds to become a gas. - the bubbles of a superheated liquid tend to be large that burst violently (known as bumping) as it s vapour pressure of the liquid is greater than that of the atmospheric preessure. 0.9: Phase Diagrams Assignment 0.8 pg. 504 #75 to 8 Phase Diagrm: - a digaram with axes of Presuure versus Temperature to illustrate all three phases of a substance. - it is more useful than a heating curve because boiling and melting points are pressure dependent, and the user now can see where the melting and boiling point are at any pressure. The curve of the vapour pressure versus temperature of a liquid is only a line on the phase diagram. - it allows the user to identify the phase of a substance at a certain pressure and temperature. Solid-Liquid Line: - a line that outlines the temperatures and the corresponding pressures where solid and liquid phases coexist. - used to find the melting point of a substance at any given pressure. Liquid-Gas Line: - a line that outlines the temperatures and the corresponding pressures where liquid and gas phases coexist. - used to find the boiling point of a substance at any given pressure. Solid-Gas Line: - a line that outlines the temperatures and the corresponding pressures where solid and gas phases coexist. - used to find the sublimation point of a substance at any given pressure. Triple Point: - a point indicating the pressure and temperature where all solid-liquid, liquid solid, and solid-gas line meet. - this is the pressure and temepertaure conditions where all three phases can coexist. Page 57.

43 Critical Point: - a point at the end of the liquid-gas line which indicates an intermediate fluid region where liquid can be coverted to gas instantaneously without waiting for phase change. - the temperature at critical point is called critical temperature and the pressure at critical point is called critical pressure. - at this region (beyond the critical pressure and temperature), the liquid will become a gas withoutout going through a change of state. Phase Diagram of Water From the phase diagram of water: - at atm, water has a normal melting point of 0 C and a normal boiling point at 00 C. - there is no normal sublimation point because the solid-gas line ends well under atm. - the negative slope of the solid-liquid line indicates that as pressure increases, the melting point of water decreases. - the positive slope of the liquid-gas line illustates that as pressure increases, the boiling point of water also increases. - the positive slope of the solid-gas line shows that as pressure increases, the sublimation point of water increases as well. - the triple point, where all three phases can coexist, is at C and atm (4.56 torr). - the critical point of water is at C and 7.7 atm. Beyond this critical temperature and pressure, water will become gaseous instantenously. Page 58.

44 Unit 3: States of Matter Phase Diagram of Carbon Dioxide From the Phase Diagram of Carbon Dioxide: - at atm, carbon dioxide has a normal sublimation point of 78.5 C. - there is no normal melting and boiling points because carbon dioxide s solid-gas line is well above normal atmospheric condition of atm. - the positive slope of the solid-liquid line indicates that as pressure increases, the melting point of carbon dioxide increases. - the positive slope of the liquid-gas line illustates that as pressure increases, the boiling point of carbon dioxide also increases. - the positive slope of the solid-gas line shows that as pressure increases, the sublimation point of carbon dioxide increases as well. - the triple point, where all three phases can coexist, is at 56.4 C and 5. atm (58 kpa or 3884 torr). - the critical point of carbon dioxide is at 3. C and 73 atm. Beyond this critical temperature and pressure, liquid carbon dioxide will become gaseous instantaneously. Assignment 0.9 pg. 505 #87 to 89 Page 59.

45 Chapter : Properties of Solutions.: Solution Composition Solution: - a homogeneous mixture of two or more substances (solute(s) and solvent). Solvent: - the largest component of a solution. Solutes: - the smaller components of a solution. Molarity (Concentration): - moles of solute per Litre of solution (M mol/l). Molarity (Concentration) n C V C Concentration (M mol/l) n moles of Solute V Total Volume of Solution Molality: - moles of solute per kilogram of solvent (m mol/kg). Molality Molality Molality m (mol/kg) n moles of Solute m solvent Mass of Solvent in kg Mass Percent: - the ratio of mass of a solute (g) and mass of a solution (g) expressed in percent. Mass Percent n m solute solvent msolute Mass Percent 00% m solution m solute Mass of Solute (g) m solution Total Mass of Solution (g) Mole Fraction (χ): - the ratio of the moles of a solute and the moles of the entire solution (solute and solvent). n A moles of component A Mole Fraction χ A n n A Total n Total Total Moles of Solution n A + n B + n C + Page 60.

46 Unit 3: States of Matter Parts per Million (ppm): - the amount of solute (usually ion) in milligram per Litre of solution. - use for measuring trace amount of solutes. Parts per Billion (ppb): - the amount of solute (usually ion) in mircogram per Litre of solution. - use for measuring trace amount of solutes. Parts per Million ppm m V solute solution (mg) (L) Parts per Billion ppb m ( µ g) V solute solution (L) Example : 40.0 g of -propanol with a density of g/ml is dissolved in 50 ml of water. Describe the composition of the solution by a. molarity b. molality c. mass percent d. mole fraction a. Molarity 40.0 g n mol C 3 7 O g/mol 40.0 g V propanol ml g/ml V Total 50 ml ml V Total ml L C propanol [C 3 7 O] mol L [C 3 7 O] 3.33 mol/l c. Mass Percent m propanol 40.0 g m solution m propanol + m water 40.0 g + 50 g m solution 90 g msolute Mass Percent 00% msolution 40.0 g Mass Percent 00% 90 g Mass Percent.% b. Molality 40.0 g n mol C 3 7 O g/mol m solvent 50 g 0.50 kg (Assume g ml O) nsolute Molality m solvent mol Molality 0.50 kg Molality 4.44 m d. Mole Fraction 40.0 g n mol C 3 7 O g/mol 50 g n water mol O g/mol n Total n propanol + n water n Total mol mol n Total mol χ propanol n propanol n Total mol mol χ propanol Page 6.

47 Example : Alcoholic beaverages have labels indicating alcoholic content (C 5 O) using (%v/v), ml of solutes per ml of solution. Suppose a Canadian whiskey has a 35% alcohol content, and the density of pure ethanol is g/cm 3, determine the composition of the whiskey by a. molarity b. molality c. mass percent d. mole fraction a. Molarity 35 ml C 5O 35% (v/v) 00 ml Total m ethanol g/cm 3 35 ml 7.65 g 7.65 g n mol C 5 O g/mol V Total 00 ml 0.00 L C ethanol [C 5 O] mol 0.00 L [C 5 O] 5.99 mol/l b. Molality n mol C 5 O V solvent V Total V ethanol 00 ml 35 ml 65 ml m solvent 65 g kg (Assume g ml O) nsolute Molality m solvent mol Molality kg Molality 9. m c. Mass Percent m ethanol 7.65 g m solution m ethanol + m water 7.65 g + 65 g m solution 9.65 g msolute Mass Percent 00% msolution 7.65 g Mass Percent 00% 9.65 g Mass Percent 9.8% d. Mole Fraction n ethanol mol C 5 O 65 g n water mol O g/mol n Total n propanol + n water n Total mol mol n Total mol χ propanol n propanol n Total χ propanol mol mol Example 3: g of sodium sulfate is dissolved in 500 ml of water. Calculate the concentration of sodium ion in ppm. This is a Stoichiometry problem. We have to figure out the mass of Na + in mg. n Na SO 4 Na SO 4 (s) Na + (aq) + SO 4 (aq) g mol Na SO g/mol n mol Na Na SO 4 m + Na + Na mol Na mol Na SO mol Na +.99 g/mol m g Na mg Na mol Na + msolute (mg) ppm Vsolution (L) mg ppm L 7.5 ppm Page 6.

48 Unit 3: States of Matter Normality (N): - the number of equivalence (active chemical agent) per Litre of solution. - for Acid or Base Solutions, the equivalence number of moles of + or O per molecular formula. Normality N Equivalence Molarity (C) Example 4: The density of a 0.37 M of barium hydroxide solution is.8 g/ml. Determine the composition of the solution in terms of its a. mass percent b. molality c. normality a. Mass Percent [Ba(O) ] 0.37 mol/l In L of solution, m 0.37 mol g/mol m g Ba(O) In L of solution, m total.8 g/ml 000 ml m total 80 g msolute Mass Percent 00% m solution g Mass Percent 00% 80g b. Molality n 0.37 mol Ba(O) m solvent m Total m solute m solvent 80 g g m solvent g kg Molality Molality n solute Molality 0.54 m m solvent 0.37 mol kg Mass Percent.57% c. Normality Ba(O) (s) Ba + (aq) + O (aq) Since there are moles of O for every mole of Ba(O) dissolves, the equivalence. Normality 0.37 M Normality N Page 63.

49 .: The Energies of Solution Formation The Dissolving Process:. Solute Molecules needs to overcome their Intermolecular Forces or the breaking up of the Lattice ( E lattice Change in Lattice Energy Endothermic Input Energy). Solvent Molecules needs have energy input to overcome their Intermolecular Forces ( ). 3. The Molecules of Solute and Solvent Interact to release energy ( 3 is exothermic Energy is Released). Enthalpy of ydration ( hyd ): - the combined energy of solvent s intermolecular forces ( ) and the release of energy between solute and solvent molecules ( 3 ). hyd + 3 Enthalpy of Solution ( aol ): - the total energy of dissolving a solute in a solvent. sol lattice + hyd Like-Dissolves-Like : - a dissolving rule that states polar solutes will dissolve in polar solvents and nonpolar solutes will dissolve in non-polar solvents. - the reason why polar solutes cannot dissolve in non-polar solvents and vice versa is due to the relatively small 3 between solute and solvent molecules. Since non-polar and polar molecules do not interact, the remaining energies involve in the dissolving process ( lattice and ) are too large (endothermic) for solution to form spontaneously. Energy Solute and Solvent BEFORE Dissolving sol 0 Solute and Solvent with no intermolecular bonds / lattice lattice + 3 Solute and Solvent Dissolve Completel Energy Solute and Solvent BEFORE Dissolving Solute and Solvent with no intermolecular bonds / lattice lattice + 3 sol > 0 Solute and Solvent Dissolve Completel When Like Dissolves Like, sol is exothermic or slightly endothermic Page 64.

50 Unit 3: States of Matter Solute and Solvent with no intermolecular bonds / lattice Solute and Solvent Dissolve 3 Completely Energy lattice + Solute and Solvent BEFORE Dissolving sol >> 0 When Polar and Non-polar molecules try to dissolve each other, sol is highly endothermic because 3 is small. Example : The lattice energy of silver chloride is 905 kj/mol. If the hydration of silver chloride in water has an enthalpy of 757 kj/mol, determine the heat of solution of silver chloride. Comment of the solubility of silver chloride based on your result. lattice 905 kj/mol hyd 757 kj/mol sol? sol lattice + hyd sol 905 kj/mol + ( 757 kj/mol) sol 48 kj/mol AgCl is only slightly soluble in water. This is because its sol is quite endothermic, which indicates that 3 is every small compared to its lattice energy and. Example : The enthalpy of solution for magnesium chloride is 74 kj/mol. Calculate its heat of hydration if magnesium chloride has a lattice energy of 56 kj/mol. lattice 56 kj/mol sol 74 kj/mol hyd? sol lattice + hyd hyd sol lattice hyd ( 74 kj/mol) 56 kj/mol hyd 600 kj/mol Assignment. pg. 548 #5 to 3. pg. 548 #33 to 35, 37 to 39 Page 65.

51 .3: Factors Affecting Solubility Solubility: - the amount of solute that can dissolve in a given amount of solvent at a specific temperature and pressure. Factors Affecting Solubility. Structure Effect:- in general, polar solutes dissolve in polar solvents whereas non-polar solutes dissolve in non-polar solvents - Like Dissolves Like. - bigger molecules are harder to dissolve (agitation might leviate this problem). ydrophobic: - non-polar molecules water fearing. ydrophilic: -polar molecules water loving. Agitation Solid Solute Solubility Size of Solute Rate of Solubility Like Dissolves Like. Pressure: - as pressure increases, solubility of gas solutes generally increases. (More force is exerted down to force gas particles to dissolve in a denser liquid solvent). - pressure has very little effect on solubility of liquid and solid solutes. Pressure enry s Law: - states that the concentration of gas in a solution is directly proportional to the pressure above the solution. enry s Law Constant (k): - relates concentration of gas in a solution with the pressure above the solution. - dependant on temperature and the identity of the gas solute. Example : Some health food vendors claim that drinking oxygenated beverage will increase physical peformance. Suppose an oxygenated drink is under 7.50 atm and the enry s law constant for oxygen is mol / (L atm) at 5 C. a. Calculate the concentration of oxygen in this beverage. b. If the partial pressure of oxygen in the atmosphere atm, what will be the concentration of oxygen in the beverage if it was left open for a long time. a. k mol / (L atm) P 7.50 atm C? C kp (0.03 mol / (L atm)) (7.50 atm) C mol/l Gas Solute Solubility enry s Law C kp C Concentration of Gas in Solution (mol/l) P Pressure above the solution (atm) k enry s Law Constant [mol / (L atm)] b. k mol / (L atm) P atm C? C kp (0.03 mol / (L atm)) ( atm) C mol/l Page 66.

52 Unit 3: States of Matter 3. Temperature: - as temperature increases, solubility of some solid solutes generally increases (exceptions are sodium sulfate and cerium sulfate). -when solutes have sol > 0, the increase in temperature supplies the energy needed to facilitate the dissoliving process. Therby, solubility increases. (Examples are KNO 3 and NaNO 3 ) - when solutes have sol < 0, the increase in temperature hinders the energy needed to be released during the dissolving process. Thereby, solubility decreases. (Examples are Na SO 4 and Ce (SO 4 ) 3 ) - as temperature increases, the solubility of gas solutes decreases. (The increase temperature causes gas solutes to move faster, breaking the intermolecular bonds they established with the molecules of the liquid solvent.) Temperature Solid Solute Solubility Gas Solute Solubility.4: The Vapour Pressure of Solutions Vapour Pressure of Solution: - in general, vapour pressure of solution tends to decrease as more solutes is dissoved into the solvent. χ - this is because the presence of solutes molecules create more solute χ solvent P soln intermolecular bonds that hinder the vaporization of the solvent. Raoult s Law: - the vapour pressure of a solution is directly proportional to the mole fraction of the solvent - pure solvent (χ solvent and χ solute 0) has the maximum vapour pressure. Raoult s Law for Solid Solute in Liquid Solvent P soln χ solvent P 0 solvent χ solvent Mole Fraction of Solvent P 0 solvent Vapour Pressure of Pure Solvent P soln Vapour Pressure of Solution Note: Moles of Ionic Solutes Total Number of Ions Dissociated Page 67. Psoln 0 P vap of Pure Solid Solute P vap of Pure Liquid Solvent χ solvent

53 Example : At 0 C, the vapour pressure of pure water is 9.3 torr. The density of water at 0 C is g/cm 3. Determine the vapour pressure of the resulting solution if a g glucose is dissoved in 50 ml of water at 0 C. b g of magnesium chloride is dissolved in 50 m L of water at 0 C. a. P 0 water 9.3 torr 5.00 g n glucose mol C 6 O g/mol n water ( 50 ml )( g/ml ) mol O g/mol n total mol mol n total mol χ water n water mol mol n Total χ water P soln χ water P 0 water P soln ( )(9.3 torr) P soln 9.0 torr b. Since MgCl is an ionic solute, we must write the dissociation equation MgCl (s) Mg + (aq) + Cl (aq) P 0 water 9.3 torr 5.00 g n ions mol Mg + and Cl ions g/mol (need to multiply by 3 due to a total of 3 moles of ions produced) n water ( 50 ml )( g/ml ) mol O g/mol n total mol mol n total mol χ water n water mol mol n Total χ water P soln χ water P 0 water P soln ( )(9.3 torr) P soln 9.06 torr Note that the P vap of a Solution consists of a Solid Solute is LOWERED than that of a Pure Solvent. Ideal Solution: - when liquid-solute and liquid-solvent solution follows Raoult s Law. - occurs when there are silimiar intermolecular forces between solute-solute, solventsolvent, and solute-solvent molecules P vap of Pure B P vap of Ideal Solution P vap of Pure A Ideal Solution of Liquid-Solute and Liquid-Solvent P Total χ A P 0 A + χ B P 0 B (χ A + χ B ) Pvap χ A 0 χ B 0 χ A Mole Fraction of Component A P 0 A Vapour Pressure of Pure A χ B Mole Fraction of Component B P 0 B Vapour Pressure of Pure B P Total Total Vapour Pressure of Solution Page 68.

54 Unit 3: States of Matter Example : Carbon tetrachloride (CCl 4 ) is commonly used to dissolve many non-polar liquid solutes. At 0 C, pure CCl 4 has a vapour pressure of 9.0 kpa. An.50 g of unknown liquid solute is dissolved in 90.0 g of CCl 4. The resulting vapour pressure was determined to be 84. kpa. Assuming the solution is ideal, calculate the molar mass of this unknown liquid solute. 0 P CCl kpa P soln 84. kpa m unknown.50 g 90.0 g n CCl g/mol mol CCl 4 χ CCl? χ 4 unknown? n unknown? M unknown? P soln χ CCl 4 0 P CCl 4 CCl 4 P χ 0 P soln CCl4 84. kpa 9.0 kpa χ unknown χ CCl nunknown n χ unknown n n n unknown Total unknown + nccl 4 unknown mol Example 3: At 60 C, the vapour pressures of benzene and toluene are 384 torr and 33 torr respectively. Determine the composition of a benzene-toluene solution if its vapour pressure is 00 torr assuming it behaves ideally. n unknown (n unknown ) n unknown n inknown n unknwon n unknwon n unknown n unknwon n unknown M unknown m n unknown unknown.50 g mol n unknown mol M unknown 5. g/mol P 0 benzene 384 torr P 0 toluene 33 torr P Total 00 torr χ benzene? χ toluene? Since χ benzene + χ toluene, then ( χ benzene ) χ toluene P Total χ benzene P 0 benzene + χ toluene P 0 toluene P Total χ benzene P 0 benzene + ( χ benzene ) P 0 toluene 00 χ benzene (384) + ( χ benzene ) (33) χ benzene χ benzene χ benzene 33χ benzene 67 5χ benzene 67 χ benzene χ toluene χ benzene χ benzene 0.67 χ toluene Page 69.

55 Non-Ideal Solution: - when liquid-solute and liquid-solvent solution deviates from Raoult s Law. P vap of Actual Non-Ideal Solution P vap of Actual Non-Ideal Solution P vap of Predicted Ideal P vap of Predicted Ideal Pvap Pvap χ A 0 χ B 0 Positive Deviation Non-Ideal Solution - when sol > 0 (endomethermic), solute-solvent molecules interactions are weak. Thereby, vapours from both components escape easily causing P actual > P predicted χ A 0 χ B 0 Negative Deviation Non-Ideal Solution - when sol < 0 (exomethermic), solute-solvent molecules interactions are strong. Thereby, vapours from both components cannot escape easily causing P actual < P predicted Example 4: At 5 C, the vapour pressures of pure ethanol and hexane are 7.87 kpa and 30 kpa. A solution of 4.0 g of ethanol and 5.0 g of hexane was found to have a vapour pressure of 54.4 kpa. Is the solution behaving ideally. Justify you answer mathematically. P 0 ethanol 7.87 kpa P 0 hexane 30 kpa m ethanol 4.0 g m hexane 5.0 g P actual 54.4 kpa P Total (predicted)? 4.0 g n ethanol mol C 5 O g/mol 5.0 g n hexane mol C g/mol n Total mol mol n Total mol mol mol χ ethanol χ hexane mol mol χ ethanol χ hexane P Total χ ethanol P 0 ethanol + χ hexane P 0 hexane P Total ( )(7.87 kpa) + ( )(30 kpa) P predicted 38.6 kpa Because P actual > P predicted, 54.4 kpa > 38.6 kpa, this is a non-ideal solution. In fact, it deviates positively from the ideal solution due to the non-polar nature of hexane dissolving a polar solvent of ethanol. Page 70. Assignment.3 pg. 549 #4, 43 and 44.4 pg #45 to 49, 5, 53, 55 and 56

56 Unit 3: States of Matter.5: Boiling-Point Elevation and Freezing-Point Depression Colligative Properties: - changes in physically properties of a pure substance as it is mixed with a solute. - these include freezing-point depression, boiling-point elevation and osmotic pressure. Boiling-Point Elevation: - the presence of solute in a solution raises the boiling point of a pure solvent due to the extra intermolecular forces between them (more energy is needed to boil the solution as vapour pressure is lowered). - the amount of temperature elevation to boil is directly proportional to the molality of the solute in the solution. - can be used to determine molar mass of a solute. Molal Boiling-Point Elevation Constant (K b ): - a constant relating the change in boiling point temperature and the molality of the solute in the solution. Freezing-Point Depression: - the presence of solute decreases the vapour pressure of the solvent. Since ice has a higher vapour pressure than that of water in the solution, the freezing point has effectively been lowered. (In order for ice to form, vapour pressure of the ice has to be lowered than that of water). - the amount of temperature depression to freeze is directly proportional to the molality of the solute in the solution. - used most commonly to determine molar mass of a solute. Molal Freezing-Point Elevation Constant (K f ): - a constant relating the change in freezing point temperature and the molality of the solute in the solution. Solvent Boiling Point T b Solution Boiling Point Page 7.

57 Boiling Point Elevation and Freezing Point Depression of Non-Electrolytic Solutions T b K b Molality solute T f K f Molality solute T b Change in Boiling Point Elevation ( C) T f Change in Freezing Point Depression ( C) K b Molal Boiling-Point Constant ( C kg/mol) K f Molal Freezing-Point Constant ( C kg/mol) Molality solute Molality of Solute (mol/kg of solvent) Example : Antifreeze, ethylene glycol (C OC O), is commonly used to prevent water from freezing in the engine in cold temperature as well as overheating. The maximum temperature a radiater can reach is 0 C. Given the densities for ethylene glycol and water are. g/ml and.00 g/ml respectively, and K b and K f for water are 0.5 C kg/mol and.86 C kg/mol, a. determine the volume of antifreeze needed to add to 0.0 L of water to sustain the maximum radiater temperature. b. what is the freezing point of the solution? a. T b 0 C 00 C 0 C K b 0.5 C kg/mol m water ml.00 g/ml m water g 0.0 kg D antifreeze. g/ml M antifreeze g/mol n antifreeze? m antifreeze? V antifreeze? n antifreeze T b K b Molality antifreeze K b kg of water o T b ( kg of water) n antifreeze ( 0 C)( 0.0 kg) mol o K b ( 0.5 C kg/mol) m antifreeze ( mol)( g/mol) g mantifreeze g V antifreeze ml D.g/mL antifreeze V antifreeze.9 L b. K f.86 C kg/mol m water 0.0 kg n antifreeze mol T f? T f K f Molality antifreeze K f n antifreeze kg of water mol T f (.86 C kg/mol) C 0.0 kg New Freezing Point 0 C C T f (soln) 7.9 C Example : A 0.0 g of newly synethsized enzyme can lowered the freezing point of 00.0 g CCl 4 by 4.70 C. Given that K f for CCl 4 is 30.0 C kg/mol, calculate the molar mass of this new enzyme. T f 4.70 C n T K f 30.0 C kg/mol f K f Molality enzyme K f enzyme kg of CCl 4 m 00.0 g kg CCl 4 T f ( kg of CCl ) ( o C m enzyme 0.0 g n enzyme )( kg ) mol n enzyme? M enzyme? M enzyme m n enzyme enzyme K f 0.0 g mol o ( 30.0 C kg/mol) M enzyme g/mol g/mol Page 7.

58 Unit 3: States of Matter.6: Osmotic Pressure Semipermeable Membrane: - a fine flter that allows solvent molecules to pass through but solute molecules are left behind the filter. Osmosis: - the flow of solvent of a solution through a semipermeable membrane. - the pure solvent (high solvent concentration) will flow into the solution (low solvent concentration) until a state of equilibrium is reached. Osmotic Pressure (Π): - the resulting pressure as pure solvent flows into a solution through the semipermeable membrane. - varies with temperature and molarity. - the most easiest method to find molar mass of solute. Π Osmotic Pressure (atm) C Molarity (mol/l) Osmotic Pressure of Non-Electrolytic Solutions Π CRT Example : Calculate the molar mass of a newly synthezied progesterone (pregnancy hormone secreted by the ovaries) if its osmotic pressure at 37.0 C is 4. torr when.0 mg is dissolved in 50.0 ml of water. nrt V R Gas Constant (L atm)/(mol K) T Temperature (K) T 37.0 C 30.5 K atm Π 4. torr atm 760 torr m progesterone.0 mg g V 50.0 ml L R (L atm)/(mol K) M progesterone? Π nrt V M progesterone m M RT V mrt M progesterone ΠV L atm K mol K L 3 (.0 0 g) ( ) ( atm)( ) M progesterone 07 g/mol Page 73.

59 Isotonic Solutions: - when solutions have exactly the same osmotic pressures. Example : Calculate the concentration of an aqueous glucose solution needed to be isotonic with human blood at the osmotic pressure of 8.0 atm at 37.0 C. T 37.0 C 30.5 K Π 8.0 atm R (L atm)/(mol K) C? Π CRT C RT Π ( 8.0 atm) L atm mol K ( 30.5 K) [glucose] 0.35 mol/l mol/l ypertonic Solutions: - when solutions have greater osmotic pressure than pure solvent (reverse osmosis). - the main priciple used in desalination (removal salt from salt water). P applied > Π Assignment.5 pg. 550 #57 to 64.6 pg. 550 #65 to 67 Page 74.