Preparation of the standard solution Exp 5: Copyright Houghton Mifflin Company.All 1 1
Mass of KHP: 5.2 5.5 g Volume of volumetric flask: 250.0 cm Molarity of standard (KHP) solution: M = n/v Copyright Houghton Mifflin Company.All 1 2
n M V 5.27 g /204.2 g 0.250 dm mol 1 Copyright Houghton Mifflin Company.All 1
KOOCC 6 H 4 COOH + NaOH KOOCC 6 H 4 COONa + H 2 O Molarity of NaOH : n NaOH/1 = nkhp/1 (MNaOH x VNaOH)/1 = (MKHP x VKHP)/1 MNaOH = (MKHP x 0.025 ) / VNaOH*** ***VNaOH(average vol. in dm from titration) Copyright Houghton Mifflin Company.All 1 4
M NaOH = -------- mol/dm concentration of NaOH (g/dm ): = (------ mol/dm / x molar mass of NaOH (g/mol) = ----------g/dm 1 5
Gases: Chapter 9 Characteristics: Unlike liquids and solids, they: have large spaces between the gas molecules not tightly packed expand to fill the container take the shape of the container diffuse into one another and mix 4 properties that affect its physical behaviour: amount of gas (in moles) volume temperature pressure 6
7
What is pressure? Pressure Pressure is the amount of force applied to an area. it is the force per unit area (force divided by the area over which the force is applied) The unit is N.m -2 which is one pascal (Pa) F( N) P( Pa) A( m 2 ) 8
Units of Pressure Pascals Bar 1 Pa = 1 N/m 2 1 bar = 10 5 Pa = 100 kpa mm Hg or torr The difference in the heights in mm (h) of two connected columns of mercury Atmosphere 1.00 atm = 760 torr 9
Barometric Pressure The simplest device used to measure pressure is the Toricelli barometer Standard atmosphere (atm) is the pressure exerted by a mercury column of 760 mm (density of Hg = 1.5951 g.cm - (0 0 C and gravity = 9.80665 m.s -2 ) 10
1 atm = 760 mmhg 1 atm = 760 Torr Example: What is the height of a column of water that exerts the same pressure as a column of mercury 76.0 cm (760 mm) high? P H2 O = P Hg Using equation (g x h x d) g x h H2 O x 1.00 g/cm = g x 76.0 cm x 1.6 g/cm h H O 2 10.m 76.0cm 1.6 g / cm 1.010 1.00 g / cm cm 11
Common Units of Pressure Atmosphere atm Millimeter of mercury mmhg 1 atm = 760 mmhg Torr Torr = 760 Torr Pascal Pa = 101 25 Pa Kilopascal kpa = 101.25 kpa Bar bar = 1.0125 bar Millibar mb = 101.25 mb 12
Manometer Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel. Figure 9.2
Standard Pressure Normal atmospheric pressure at sea level. It is equal to: 1.00 atm 760 torr (760 mm Hg) 101.25 kpa
Manometers 15
Example: When a manometer is filled with liquid mercury (d = 1.6 g/cm ), the barometric pressure is 748.2 mmhg, and the difference in mercury levels is 8.6 mmhg. What is the gas pressure P gas? P gas = P bar + P = 748.2 mmhg - 8.6 mmhg = 79.6 mmhg 16
Gas Laws Boyle s Law: The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure. P 1 V 17
Charles s Law: - The volume of a fixed amount of gas at constant pressure is directly proportional to its temperature. V T A plot of V versus T will be a straight line. 18
Standard Temperature and Pressure STP conditions: standard temperature for gases = 0 0 C = 27.15 K standard pressure, 1 atm = 760 mmhg Avogadro s Law: - at a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas n V 1 mol gas 22.4 L gas ( at STP 19 )
Example: What is the mass of 1.00 L of cyclopropane gas, C H 6, when measured at STP? 1L =1dm 1 mol x x 1.00 1.00 22.4 L L 1 mol 22.4 L L gas ( at STP) 0.0446 mol mass of C H 6 n m. m 0.0446 mol 42.08 g. mol 1.88 g 1 20
Ideal-Gas Equation So far we ve seen that: V 1/P (Boyle s law) V T (Charles s law) V n (Avogadro s law) Combining these, we get, if we call the proportionality constant, R: V = R nt P i.e. PV = nrt 21
The Ideal Gas Equation PV nrt Where R = gas constant R = 8.145 m.pa.mol -1.K -1 (SI unit-use this) = 8.145 J.mol -1.K -1 (SI unit)(1m Pa = 1J) = 0.082057 L.atm.mol -1.K -1 = 8.14 kpa. dm mol -1.K -1 22
Units (1m Pa =1J) R = 8.145 m Pa mol -1 K -1 R= 8.145 J mol -1 K -1 P m Pa F ( N) A ( m 2 ) kg m 1 N s 2 kg m s 2 Pa m 2 kgm 1 Pa s 2 m 2 Pa kg m 1 s 2 kg m 2 s 2 1J 2
24
Volume conversion (from milliliters to litres and then to cubic meters) 1 L 1 m V 05 ml 1000 ml 1000 L.05 10 4 m 25
Pressure conversion 1 atm 10125 Pa 1.7 atm x Pa X = (1.7 atm x 10125 Pa) / 1 atm = 18815 Pa = 1.8 x 10 5 Pa 26
Example: What is the pressure, in kpa, exerted by 1.00 x 10 20 molecules of N 2 in a 05 ml flask at 175 0 C? 2 1 mol 6.0210 molecules x x 1.0010 0.000166 20 mol molecules 20 1.0010 molecules 1 mol 2 6.02210 molecules N 2 P nrt V 2.010 0.000166 Pa mol 8.145 m.0510. Pa. mol 4 m 1. K 1 448 K 2.0 kpa 27
General Gas Equation PV i n T i i i P V n f f T f f For a fixed mass of gas, PV i T i i P V f T f f 28
Example: If a fixed amount of gas held at constant volume goes from STP conditions to boiling at 100 0 C, what is the new pressure? Pi T i P T f f PV i n T i i i P f P V n f f 1.00 1.7 P T i f f T T i atm 27 atm f K 7 K 29
Applications of the Ideal Gas Equation Molar mass Molar mass determination PV = nrt m n M PV mrt M 0
Example: Propylene is used in the production of plastics. A glass vessel weighs 40.105 g when clean, dry and evacuated; it weighs 18.2410 g when filled with water at 25 0 C(density of water = 0.9970 g/ml) and 40.2959 g when filled with propylene gas at 740. mmhg and 24.0 0 C. What is the molar mass of propylene? Mass of water to fill vessel = 18.2410 g 40.105 g = 98.1105 g Density = mass/volume Volume = mass/density Volume of water (vol. of vessel) = 98.1105 g/0.9970 g/ml = 98.41 ml = 0.09841 L 1
Mass of gas = 40.2959 g 40.105 g = 0.1654 g Temperature (K) = (24 0 C + 27.15 0 C) x 1K/1 0 C = 297.15 K R = 8.145 m Pa mol -1 K -1 Pressure = 740. mmhg/760 mmhg = 0.9741 atm (convert to Pa) = 98700 Pa mrt M PV 1 0.1654 g 8.145 m Pa mol K 98700 Pa 9.84110 42.08 g / mol 1 5 297.15 K m 2
Gas Densities and Molar Mass If we divide both sides of the ideal-gas equation by V and by RT, we get: PV = nrt n V = P RT
Applications of the Ideal Gas Equation - Gas Densities d m V n M V n V M We can then replace n/v with P/RT m MP d V RT The density of a gas at STP is calculated by dividing its molar mass by the molar volume 4 (22.4 L/mol)
Example: What is the density of O 2 at 298 K and 0.947 atm? (Convert 0.947 atm to Pa) R = 8.145 m Pa mol -1 K -1 d m V MP RT g 2.00 8.145 m 1 mol Pa mol 95954 Pa 1 1 K 298 K 129.5g m 1 m 1000 L 1.24 g / L 5
Gases in Chemical Reactions Air bags use the decomposition of sodium azide, NaN to produce N 2 (g) that inflates the air bag Example: From the following equation determine what volume of N 2 is produced when 70.0 g of NaN 2 is decomposed at 75 mmhg and 26 0 C? 2NaN (s) 2Na(l) + N 2 (g) n( NaN n( N 2 70.0 g ) 1.08 mol 65.01 g / mol 1.08 mol ) 1.62 mol 2 6
P = 75 mmhg/760 mmhg = 0.967 atm (convert to Pa) n = 1.62 mol R = 8.145 m Pa mol -1 K -1 T = (26 0 C + 27 0 C) x 1K/1 0 C = 299 K V nrt P 1.62 mol 8.145 m 97981 Pa Pa mol 1 K 1 299 K 0.0411 m 1000 1 m L 41.1 L 7
Law of Combining Volumes Example: For the following reaction, determine what volume of SO 2 is produced per liter of O 2 consumed? Both gases are measured at 25 0 C and 745 mmhg. 2ZnS ( s) O2 2ZnO ( s) 2SO2 ( g) 1 L mol x 2 mol 2 mol 1 L x mol 0.667 L SO SO O 2 2 2 ( g) ( g) ( g) 8
Mixtures of Gases 9
Dalton s law of partial pressures: The total pressure of a mixture of gases is the sum of the partial pressures of the components of the mixture P tot = P A + P B + V A n A P RT tot V B n B P RT tot V tot V A V B 40
P P A tot n n A tot ( RT ( RT / V / V tot tot ) ) n n A tot and V V A tot n n A tot ( RT ( RT / / P P tot tot ) ) n n A tot n n A tot P P A tot V V A tot x A Where X A is the mole fraction (fraction of all the molecules in a mixture contributed by that component) 41
Example: What are the partial pressures of H 2 and He in a gaseous mixture containing 1.0 g of H 2 and 5.00 g He that are in a confined volume of 5.0 L at 20 0 C? (convert volume) n tot 1.00 g 2.02 g / mol H 0.50 mol 1.25 mol 5.00 g 4.00 g / mol 1.75 mol 2 He ntotrt P V 1.75 mol 8.145 m Pa mol 0.0050 m 1 K 1 29 K 8.510 5 Pa 42
Partial pressure of H 2 and He n n H 2 tot P P H 2 tot P H 2 P tot n tot n H 2 0.50 8.4 1.75 atm 2.4 atm P He P tot n tot n He 1.258.4 1.75 atm 6.0 atm 4
Partial Pressures and Mole Fraction Figure 9.1 So, when one collects a gas over water, there is water vapour mixed in with the gas. To find only the pressure of the desired gas, one must subtract the vapour pressure of water from the total pressure.
Collecting a gas over water Gas collected over water is wet made up of the desired gas and water vapour P tot = P atm = P gas + P H2O or P gas = P atm P H2O 45
Kinetic-Molecular Theory 1.Gases consist of large numbers of molecules that are in continuous, random motion. 2.The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained..attractive and repulsive forces between gas molecules are negligible.
Kinetic-Molecular Theory 4.Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant. 5.The average kinetic energy of the molecules is proportional to the absolute temperature.
1. Gas molecules contain translational kinetic energy (energy of objects moving through space). The faster the molecules move, the greater is their translational energies greater forces exerted when they collide. 2. Frequency of molecular collisions no. of collisions per second. The higher the frequency, the greater the total force of the collisions (increases with no. of molecules per unit volume and with molecular speed).. When a molecule hits the walls of a container, momentum is transferred as the molecule reverses direction impulse. The magnitude of the impulse is directly proportional to the mass and velocity of a molecule. 48
The Kelvin temperature (T) of a gas is directly proportional to the average translational kinetic energy of its molecules. Absolute zero of temperature the temperature at which the translational molecular motion should cease. 49
Nonideal (Real) Gases Compressibility factor: PV nrt For an ideal gas, PV/nRT = 1 For nonideal gas behaviour, at high pressures the volume of a gas becomes very small 0 but this is not so as gas molecules occupy a volume themselves. 50
Real Gases Figure 9.21 In the real world, the behaviour of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.
Deviations from Ideal Behaviour Figure Figure 9.2 9.22 The assumptions made in the kinetic-molecular model break down at high pressure and/or low temperature.
Corrections for Non-ideal Behaviour The ideal-gas equation can be adjusted to take these deviations from ideal behaviour into account. The corrected ideal-gas equation is known as the van der Waals equation.
The van der Waals Equation (P + n 2 a V 2 ) (V nb) = nrt Table 9.2
Since molecules are attracted to each other, the force of collisions of gas molecules with the walls of the container are less than expected for an ideal gas P n V 2 2 a V nb nrt Gases behave ideally at high temperatures and low pressures. Gases tend to behave nonideally at low 55 temperatures and high pressures.
Example: Use the van der Waals equation to calculate the pressure exerted by 1.00 mol Cl 2 (g) confined to a volume of 2.00 L at 27 K. The volume of a = 0.657 (m ) 2 Pa.mol -2, and that of b = 0.0000562 m mol -1. R = 8.145 m Pa mol -1 K -1 2 nrt n a P 2 V nb V 1 1.00 mol 8.145 m Pa mol. K 0.002 m 1mol x 0.0000562 m 1 27K 1 mol 1.00 mol 2 0.657( m 2 (0.002) ( m ) ) 2 2 Pa mol 2 1004 Pa 1.00 10 6 Pa 56
To simplify calculation : R = 8.145 m Pa mol -1 K -1 nrt = 1.00 1 1 mol 8.145 m Pa mol. K 27K n 2 a = 1.00 mol 2 0.657( m ) 2 Pa mol 2 V-nb = 0.002 m 1mol x 0.0000562 m mol 1 57
EX. A 1.65 g sample of an unknown metal M reacts with excess HCl and the liberated hydrogen is collected over water at 25 o C at a barometric pressure of 744 mmhg. The gas collected has a volume of 267 ml. The equation for the reaction is: 2M(s) + 6HCl (aq) 2MCl (aq) + H 2 (g) The gas constant: R = 8.145 m Pa mol -1 K -1 Vapour pressure of H 2 O at 25 o C is 2.76 mmhg 1 atm = 760 mmhg 1 atm = 10125 Pa Calculate the following: (i) partial pressure of dry hydrogen gas in SI units (ii) moles of dry hydrogen gas in SI units (iii) moles of M (iv) molar mass of M (v) density of hydrogen gas, in g L -1 at STP. 58
(i) partial pressure of dry hydrogen gas using SI units P H2 = (744 2.76) mmhg = 720 mmhg/760 mmhg atm -1 = 0.947 atm (convert to Pa) PH 2 = 0.947 atm x 10125 Pa atm -1 = 9.60 x 10 4 Pa (ii) moles of dry hydrogen gas using SI units n = PV/RT = (9.60 x 10 4 Pa x 2.67 x 10 - m )/8.145 m Pa mol -1 K -1 x 298K ) = 0.0917 mol 59
(iii) moles of M From the balanced equation: moles of M = 2/ mol of H 2 = 2/ x 0.0917 = 0.0611 mol of M (iv) molar mass of M molar mass of M = mass/mol = 1.65 g/0.0611 mol = 27.0 g mol -1 (M = Al) (v) density of hydrogen gas, in g L -1 at STP. Density = PM/RT = (10125 Pa x 2.02 g mol -1 )/8.145 m Pa mol -1 K -1 x 27 K. = 90.2 g m - ( convert volume from cubic metres to litres ) = 0.0902 g L -1 60