CHEMISTRY 109 #25 - REVIEW

Transcription

1 CHEMISTRY 109 Help Sheet #25 - REVIEW Chapter 4 (Part I); Sections ; Ch. 9, Section 9.4a-9.4c (pg 387) ** Review the appropriate topics for your lecture section ** Prepared by Dr. Tony Jacob (Resource page) Nuggets: Terms; Work; Sign conventions; H - enthalpy; Specific Heat Capacity; q = C s mδt; Heat Transfer: Heat Lost + Heat Gained = 0; Phase Changes (Δ fusion H and Δ vaporization H) TERMS Energy: Capacity to do work or to produce heat Kinetic energy: energy due to motion; E k = 1 / 2 mv 2 (units: m = mass (kg); v = velocity (m/s)); thermal (microscopic, e.g., atoms); mechanical (macroscopic objects); acoustic (sound) Molecules absorb/store energy via vibration, rotation, or translation Potential energy: energy from relative position; gravitational; PE = mgh (m = mass (kg); h = height above Earth (m); g = Earth s gravitational constant = 9.8m/s 2 ); composition - chemical or electrostatic System: the part of the universe that is under observation; Surroundings: everything that is not the system Heat: amount of energy flow from one region to another; denoted by the letter q Temperature: the average kinetic energy of the molecules Thermal Energy: sum of all the kinetic energies of all the molecules State function: a property that is independent of the pathway chosen (all variables that are capitalized are state functions, e.g., E, H, V, etc.; all variables that are lower case letters are not state functions, e.g., q) Exothermic: energy is released from system surroundings (i.e., heat given off/produced/yielded/released; the beaker feels warm) Endothermic: energy is absorbed from surroundings system (i.e., heat is absorbed; the beaker feels cool) UNITS of energy, work, or heat = J (joules) (J = kg (m 2 /s 2 ); Other less common energy units: 1 (L)(atm) = 101.3J; 1 (L)(bar) = 100.0J WORK (if covered) 1 st Law of Thermodynamics: energy is conserved: ΔE = q + w q = heat; w = work; ΔE = internal energy; units: ΔE, q, and w J HEAT: q < 0 ( ) exothermic (gives off heat); q > 0 (+) endothermic (absorbs heat) WORK: defined as PV work; w = PΔV (change in gas volume with constant pressure); ΔV = V f V i VOLUME: ΔV > 0 (+) expansion of gases (V f > V i ); ΔV < 0 ( ) compression of gases (V i > V f ) From w = PΔV: When ΔV > 0 (+) w < 0 ( ) work done by the system (work is done for us; yea!); When ΔV < 0 ( ) w > 0 (+) work done on the system (work is done by us; boo!) SIGN CONVENTIONS (sign of ΔE and w are determined by: w = PΔV and ΔE = q + w) q ΔV w ΔE + (endo) 0 0 (+) internal E increasing (exo) 0 0 ( ) internal E decreasing 0 (compression) + (work done on the system) (+) internal E increasing 0 + (expansion) (work done by the system) ( ) internal E decreasing q w ΔE + (endo) + (work done on the system) (+) internal E increasing (exo) (work done by the system) ( ) internal E decreasing + (endo) (work done by the system) depends on the size of q and w (exo) + (work done on the system) depends on the size of q and w

2 Example 1: The following process occurs in a piston: H 2 (g) 2H(g) What are the signs for q, ΔV, w, and ΔE? Answer 1: q > 0 (+); ΔV > 0 (+); w < 0 (-); ΔE can t be determined unless the values of q and w are known q: it requires energy to break a bond: H H 2H (think about a pencil representing the bond it takes energy to break that bond/pencil); putting energy into a system is endothermic and q > 0 (+) ΔV: the number of moles of gases increases from 1mol to 2mol as the reaction proceeds; hence, the V is increasing (more gas mol) and therefore ΔV > 0 (+) w: w = -PΔV; the value of P is always a positive (+) value; plug in the signs of the equation: w = (-)(+)(+) = (-), hence, w < 0 (-) ΔE: ΔE = q + w; since q > 0 (+) and w < 0 (-), the sign of ΔE will be determined by the magnitudes of q and w; for example, if q = 5 and w = -2 ΔE = (+5) + (-2) = +3 and ΔE > 0 (+); however, if q = 2 and w = -5 ΔE = (+2) + (-5) = -3 and ΔE < 0 (-); hence, the sign of ΔE will depend on the magnitudes of q and w ENTHALPY Change in Enthalpy: ΔH; ΔH = H f - H i with i = initial conditions and f = final conditions Change in enthalpy for a reaction, ΔH is expressed as: ΔH = H products - H reactants (ΔH is often written ΔH rxn ) Enthalpy: H is enthalpy defined as H = E + PV where E is internal energy and PV is work Change in enthalpy, ΔH, at constant P is: ΔH = ΔE + PΔV substituting ΔE = q + w and w = PΔV ΔH = (q p + w) + ( w) which yields: ΔH = q p where q p is heat at constant pressure (i.e., ΔH = the heat when P is constant; e.g., a piston) Now Start with ΔE = q + w and substitute w = PΔV to yield: ΔE = q + PΔV ; at constant V ΔV = 0 and q q v ; to yield: ΔE = q v where q v is heat at constant volume (i.e., ΔE = the heat when V is constant; e.g., a rigid box) Now Start with ΔH = ΔE + PΔV and substitute ΔH = q p and ΔE = q v to yield: q p = q v w Summary: ΔE = q + w ; ΔH = q p ; ΔE = q v ; q p = q v w ; w = PΔV Example 2 (skip if q v and q p are not covered!): The vaporization of methanol was performed in a piston: CH 3 OH(l) CH 3 OH(g) The q p of this process is 35.3kJ/mol at 64.7 C. a. Is this reaction endothermic or exothermic? b. If the work from this process was -2.50kJ, what is the value of q v? c. What is the value of ΔE in this reaction? d. What would the value of the work be in the following reaction run under the same conditions? 3CH 3 OH(g) 3CH 3 OH(l) e. What is the enthalpy change for the reaction shown in part d? Answer 2: a. endothermic {since q p > 0 endothermic} b. 32.8kJ {q p = q v w; q v = q p + w = (-2.50) = 32.8kJ} c. 32.8kJ {ΔE = q v } d. 7.50kJ {reverse rxn w -w; triple rxn triple work; (-2.50)(-1)(3) = 7.50kJ} e kJ {ΔH = q p = 35.3kJ for the original rxn; for the new rxn it is reversed and x3 (35.3)(-1)(3) = kJ} HEAT CAPACITY - used within one phase (s, l, or g) - not between phases Specific heat capacity, C s or just C (or sometimes called just specific heat): units = J/g C; the energy required to raise 1g of material 1 C Heat = Specific heat capacity x grams x ΔT or q = C s mδt where ΔT is T f - T i ; note how units work out: J (heat) = (J/g C) x g x C Molar heat capacity, C m : units = J/mol C; the energy required to raise 1mol of material 1 C Heat = Molar heat capacity x moles x ΔT or q = C m (mol)δt (some instructors skip C m ) where ΔT is T f - T i ; note how units work out: J (heat) = (J/mol C) x mol x C Example 3: You wish to heat water to make coffee. How much heat (in kj) must be used to raise the temperature of 0.180kg water (~1 cup) from 15.0 C (cold tap water) to 96.0 C (brewing temperature)? C = J/g C Answer 3: q = CmΔT; q = (4.184J/g C)(0.180kg)(1000g/1kg)(96 15); q = 61,003J(1kJ/1000J); q = 61.0kJ

3 HEAT TRANSFER: Heat lost + Heat gained = 0 C s1 x mass 1 x (T f T i1 ) + C s2 x mass 2 x (T f T i2 ) = 0; (T f1 = T f2 = T f in thermal equilibrium) Example 4: 10.0 gram metal (C = 9.75 J/g C) at 120. C is dropped into 100. ml water (C = J/g C; D = 1.00g/ml) at 25.0 C. What is T final? Answer 4: Heat gained + heat lost = 0; C metal m metal (T f T i,metal ) + C water m water (T f T i,water ) = 0; (9.75J/g C)(10g)(T f 120) + (4.184J/g C)(100ml)(1g/ml)(T f 25) = 0; 97.5T f 11, T f 10,460 = 0; 515.9T f 22,160 = 0; 515.9T f = 22,160; T f = 43.0 C PHASE CHANGE (page 387 in text): There is no temperature change during a phase change; see Heating Curve below Solid Liquid: Δ fusion H = Δ fus H; Liquid Solid: -Δ fus H = Δ freezing H; Liquid Gas: Δ vaporization H = Δ vap H; Gas Liquid: -Δ vap H = Δ condensation H s l: Heat = Δ fus H x mol (if Δ fus H = J/mol) or Heat = Δ fus H x grams (if Δ fus H = J/g) l g: Heat = Δ vap H x mol (if Δ vap H = J/mol) or Heat = Δ vap H x grams (if Δ vap H = J/g) 150 Heating ice from -50 C to water gas at 150 C Steps 1, 3, and 5 use: q = Cm!T Steps 2 and 4 use: q =!H x mass Total heat required = Steps temperature heating ice melting ice s " l heating water boiling water l " g heat water gas heat added Example 5: How much heat is required to heat 25.0g H 2 O(l) from C to 125 C? (CH 2 O(s) = 2.11J/g C; CH 2 O(l) = 4.184J/g C; CH 2 O(g) = 2.00J/g C; Δ fus H = 333J/g; Δ vap H = 2256J/g) Answer 5: q total = (q heat ice to mp ) + (q melt ice ) + (q heat water to bp ) + (q boil water ) + (q heat H 2 O(g) ) Step 1: q = (2.11J/g C)(25.0g)(0-(-75.0 C)) = 3956J Step 2: q melt = (333J/g)(25.0g) = 8325J Step 3: q = (4.184J/g C)(25.0g)( C) = 10460J Step 4: q boil = (2256J/g)(25.0g) = 56400J Step 5: q = (2.00J/g C)(25.0)( ) = 1250J q total = q 1 + q 2 + q 3 + q 4 + q 5 = = 80391J = 80.4kJ (Note: all the signs of q should be the same; in this case, heating is endothermic so positive (+); if the system is cooled: all the signs of q should be negative (-); when cooling a system remember to change the signs of Δ fus H and Δ vap H because you have reversed the process) 1. When a molecule absorbs energy, how is the energy manifested in molecular motion? 2. a. A piston is the system of interest. If the only change that occurs is the volume increases what is the sign of work? b. If the only change to a system is that heat goes into the system what is the sign of q? c. If the changes above in parts a and b occur, what can be said about the sign of the internal energy? Is internal energy increasing or decreasing? 3. If a piston s volume decreases from 4.50L to 1.70L and the pressure around the piston remains constant at 0.950atm, what is the value of the work? Is the work done on the system or by the system? 4. Describe the following processes as exothermic or endothermic.

4 a. O 2 (g) O 2 (l) b. O 2 (g) 2O(g) c. CO 2 (s) CO 2 (g) d. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) 5. Which of the following processes is endothermic? a. Aluminum is burned in air to produce Al 2 O 3. b. Liquid oxygen is boiled into gaseous oxygen. c. Water gas is condensed into liquid water. d. A lake freezes in Wisconsin. e. None of these are endothermic. 6. An iron skillet contains 1510g iron and is at 178 C, and the skillet is cooled to room temperature, 21.0 C. How much heat was given off if the specific heat of iron is J/g C? a b c d x 10 4 e x J will raise T from 25.0 C to 60.0 C of 10.0g metal. What is the metal s specific heat capacity? a b c d x 10 4 e x A 5.00g piece of Cu is at 100. C, and is dropped into 10.0g of water at 20.0 C with a specific heat capacity of 4.184J/g C. If the final temperature of the water is 23.5 C, what is the specific heat capacity of the copper? 9. When 25.0ml H 2 O(l) at 75.0 C with a heat capacity of 4.184J/g C is mixed with 15.0ml H 2 O(l) at 94.0 C, what is the final temperature of the mixture? (Assume the density of water is 1.00g/ml) 10. Consider the following specific heat capacities of metals. Metal Specific Heat (J/g C) Cu Co Cr Au Ag If the same amount of heat is added to the same mass of each of the metals all at the same temperature, which metal will reach the highest temperature? a. Cu b. Co c. Cr d. Au e. Ag

5 A 11. If heating curves A and B represent the same chemical, then a. the mass of A is twice the mass of B b. the specific heat capacity of B is double the specific heat capacity of A c. the temperature recorded during the experiment must be incorrect since both A and B should have the same heating curves since they are the same chemical d. the mass of A is half the mass of B e. the mass of A equals the mass of B T q B 12. (Do if phase changes are covered.) How much heat is required to convert 10.0g ice at 25.0 C into water at 50.0 C? The specific heat capacity of ice is 2.11J/g C, the specific heat capacity of water is 4.18J/g C, and the heat of fusion for ice is 333J/g. ANSWERS 1. The molecule moves (translates), rotates, and vibrates. 2. a. w < 0 ( ); since w = PΔV and ΔV > 0 (expansion +) then w = (+)(+) < 0 ( ) b. q > 0; since heat is going into the system endothermic and q > 0 c. The ΔΕ sign depends on the magnitudes of q and w; if q > w ΔE > 0 (+); if w > q ΔE < 0 ( )} 3. w = 269J {w = PΔV; w = (0.95atm)( L) = ( 2.66Latm) x (101.3J/1L atm) = 269.4J; since w > 0 work is done on the system} 4. a. exothermic b. endothermic c. endothermic d. exothermic 5. b 6. e {q = C s m(t f T i ) = (0.45)(1510)(21-178) = J = J} 7. a {314 = C s (10)(60-25); C = 0.897J/g C} J/g C {C s1 m 1 (T f -T 1 ) + C s2 m 2 (T f -T 2 ) = 0; C s (5)( ) + (4.184)(10)( ) = 0; solve for C s } C {(4.184)(25)(T f 75) + (4.184)(15)(T f 94) = 0; T f = 0; T f = C} 10. d {the smaller the heat capacity the greater the ΔT given the same amount of heat being applied} 11. d {Choose a specific q and draw a vertical line; note that A has risen twice as much in temperature as B at this chosen q value; A and B have the same specific heat capacity since they are the same chemical; using q = CmΔT, if q and C are the same and DT for one is twice as large then m must be half as much} J {heat = q heat ice + q melt ice + q heat water = (2.11)(10.0)(0.00 (-25.0)) + (333)(10.0) + (4.18)(10.0)( ) = J}