(second-order kinetics)

Transcription

1 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 1 PART A QUESTION 1 Given: Reaction studied: NO 2 (g) NO(g) + O(g) Arrhenius equation: k e! 8900 T L mol!1 s!1 (a) E a k A e! E a R T where A = ! E a / R =!8900 mol K ˆ E a = (8900 mol K) (8.314 J mol!1 K!1 ) = J = 74 kj (b) the rate constant (k) at 298 K k A e! E a R T k e! ˆ k = L mol!1 s!1 = 0.28 L mol!1 s!1 (c) [NO 2 after 10 minutes Given: [A o = [NO 2 = !2 mol L!1 t = 10 min = s = 600 s units: L mol!1 s!1 indicating second-order kinetics To find: [A at t = 10 min 1 [A k t % 1 [A o (second-order kinetics) 1 [A ( L mol!1 s!1 ) (600 s) % L mol! !2 mol L!1 ˆ [A after 10 min = ( L mol!1 )!1 = !3 mol L!1 = !3 mol L!1

2 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 2 PART A QUESTION 2 Given: Reaction studied: S 2 O 8 2! (aq) + 3 I! (aq) I 3! (aq) + 2 SO 4 2! (aq) a plot of log k versus 1/T slope from a straight line =!2500 K R = !3 kj K!1 mol!1 To find: activation energy (E a ) for the reaction log k! E a R 1 T % log A Compare to a straight-line linear equation: y = m x + b where y = log k x = 1 / T m = slope =! E a / R b = y-intercept = log A l o g k l o g A s l o p e = - E a / R 1 / T slope! E a R! 2500 K! E a ( !3 kj K!1 mol!1 ) ˆ E a = kj mol!1 = 48 kj mol!1

3 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 3 PART A QUESTION 3 Given: Reaction studied: H 2 (g) + I 2 (g) 2 HI(g) = !5 M!1 s!1 T 1 = 393 o C = ( ) K = 666 K k 2 = !4 M!1 s!1 T 2 = 443 o C = ( ) K = 716 K To find: E a in kj mol!1 Rn ( k 2 ) E a R ( T 2! T 1 T 2 T 1 ) Rn ( !4 M!1 s!1 ) E a ( !5 M!1 s! !3 kj mol!1 K!1 716 K! 666 K 716 K 666 K ) ˆ E a = kj mol!1 = 182 kj mol!1 PART A QUESTION 4 Given: Elementary reaction: A + B C E a = 30.0 kj mol!1 )H =!5.0 kj mol!1 (ˆ exothermic reaction) To find: E a for the reverse reaction (E a ) Energy As shown in the diagram, The reverse reaction is: E a = 30.0 kj E a = 35.0 kj C A + B A + B )H = 5.0 kj Reaction Profile C E a = 35.0 kj mol!1

4 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 4 PART A QUESTION 5 Given: E a = 20 kj )H = +8 kj (ˆ endothermic reaction) E a (cat) = 12 kj To sketch: the reaction profile with and without a catalyst including (1) activation energy (E a ) (2) enthalpy change ()H) for the reaction No Catalyst With Catalyst Energy Energy E a = + 20 kj E a = + 20 kj E a (cat) = + 12 kj )H = kj products )H = kj products reactants reactants Reaction Profile Reaction Profile In the presence of a catalyst, Î the reaction rate increases Ï the rate constant (k) increases Ð t ½ decreases because k increases Ñ E a usually decreases ( k depends on both A and E a ) Ò The enthalpy difference between reactants and products remains unchanged ( this is not a kinetic property)

5 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 5 PART A QUESTION 6 Given: Mechanism studied: O 3 (g) º O 2 (g) + O(g) (rapid equilibrium) O 3 (g) + NO(g) NO 2 (g) + O 2 (g) (slow) NO 2 (g) + O(g) NO(g) + O 2 (g) (fast) Overall: 2 O 3 (g) 3 O 2 (g) To find: the intermediate(s) and the catalyst (1) intermediates: O(g) and NO 2 (g) ( do not appear in the overall reaction) (2) catalyst: NO(g) ( regenerate at the end of the reaction) PART A QUESTION 7 Given: To find: Mechanism consisting of two elementary steps: (1) A + 2 B C + D (slow) (2) D + A C + E (fast) Which of the following statements is incorrect? (a) The species D is an intermediate. TRUE ( The species D does not appear in the overall reaction: 2 A + 2 B 2 C + E.) (b) The first step is the rate-determining step. TRUE ( The slow elementary step is always the rate-determining step.) (c) The second step is a bimolecular elementary step. TRUE ( This elementary step involves two molecules: D and A.) (d) The rate law for the first step is: rate = k [A [B 2. TRUE ( The rate law follows the stoichiometry in the rate-determining reaction.) (e) If the concentration of B is tripled, the rate of reaction increases 6-fold. INCORRECT (The rate of reaction should increase 9-fold.)

6 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 6 PART A QUESTION 8 Given: Reaction studied: 2 H 2 (g) + 2 NO(g) N 2 (g) + 2 H 2 O(g) rate = k exp [H 2 [NO 2 To show: Mechanism (a) or Mechanism (b) is consistent with the experimental rate law The slow elementary step is always the rate-determining step. ˆ rate law for the overall reaction = rate law of the slow step Mechanism (a): (i) H 2 + NO H 2 O + N (slow, ) (ii) N + NO N 2 + O (fast, k 2 ) (iii) O + H 2 H 2 O (fast, k 3 ) From (i): H 2 + NO H 2 O + N (slow, ) (i.e. rate-determining step) rate = [H 2 [NO ˆ Mechanism (a) is not consistent with the experimental data. Mechanism (b): (i) H NO N 2 O + H 2 O (slow, ) (ii) N 2 O + H 2 N 2 + H 2 O (fast, k 2 ) From (i): H NO N 2 O + H 2 O (slow, ) (i.e. rate-detemining step) rate = [H 2 [NO 2 ˆ Mechanism (b) is consistent with the experimental data (where = k exp ).

7 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 7 PART A QUESTION 9 Given: Reaction: 2 A + B D + E (i) A + B C (fast, ) (ii) C A + B (fast, ) (iii) C + A D + E (slow, k 2 ) (1) The slow elementary step is the rate-determining step. From (iii): C + A D + E (slow, k 2 ) rate = k 2 [C [A Î (a) (2) C is the intermediate, [C must be eliminated from the rate law equation and substituted by [reactants and/or [products. use steady-state approach to deduce the rate law for the reaction rate of intermediate C formation = rate of intermediate C disappearance A + B C (fast, ) C A + B (fast, ) C + A D + E (slow, k 2 ) rate = [A [B rate = [C rate = k 2 [C [A [A [B = [C + k 2 [C [A [A [B = [C ( + k 2 [A) ˆ [C [A [B % k 2 [A Ï Substitute Ï into Î: rate k 2 [C [A k 2 ( [A [B % k 2 [A ) [A k 2 [A 2 [B % k 2 [A (b) conditions the rate law would be simplified to rate = k exp [A 2 [B when» k 2 [A, rate ( k 2 ) [A 2 [B if k exp ( k 2 ), rate k exp [A 2 [B

8 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 8 PART A QUESTION 10 Given: Reaction studied: COCl 2 (g) CO(g) + Cl 2 (g) rate = k exp [COCl 2 [Cl 2 ½ (i) Cl 2 (g) º 2 Cl(g) (rapid equilibrium, K) (ii) COCl 2 (g) + Cl(g) COCl(g) + Cl 2 (g) (slow, k 2 ) (iii) COCl(g) CO(g) + Cl(g) (fast, k 3 ) To show: the mechanism is consistent with the rate law (rate = k exp [COCl 2 [Cl 2 ½ ) (1) Elementary step (ii) is the slow reaction and is the rate-determining step. rate = k 2 [COCl 2 [Cl Î (2) Cl(g) is the intermediate and must be substituted by [reactants and [products. (a) use Equilibrium Constant (K) to eliminate concentration of an intermediate From (i): Cl 2 (g) º 2 Cl(g) (rapid equilibrium, K) K [Cl 2 [Cl 2 ˆ [Cl = (K [Cl 2 ) ½ Ï Substitute Ï into Î: rate = k 2 [COCl 2 [Cl = k 2 [COCl 2 (K [Cl 2 ) ½ = k 2 K ½ [COCl 2 [Cl 2 ½ = k exp [COCl 2 [Cl 2 ½ where k exp = k 2 K ½ (b) k exp expressed in terms of K and ks k exp = k 2 K ½

9 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 9 PART A QUESTION 11 Given: Reaction studied: 2 H 2 (g) + 2 NO(g) N 2 (g) + 2 H 2 O(g) rate = k exp [H 2 [NO 2 (i) 2 NO º N 2 O 2 (fast equilibrium, K) (ii) N 2 O 2 + H 2 N 2 O + H 2 O (slow, k 2 ) (iii) N 2 O + H 2 N 2 + H 2 (fast, k 3 ) To show: the proposed mechanism is consistent with the experimental rate law rate = k exp [H 2 [NO 2 (1) Elementary step (ii) is the slow reaction and is the rate-determining step. rate = k 2 [N 2 O 2 [H 2 Î (2) N 2 O 2 is the intermediate and must be substituted by [reactants and [products. Use Equilibrium Constant (K) to eliminate concentration of intermediate From (i): 2 NO º N 2 O 2 (fast equilibrium, K) K [N 2 O 2 [NO 2 ˆ [N 2 O 2 = K [NO 2 Ï Substitute Ï into Î: rate = k 2 [N 2 O 2 [H 2 = k 2 (K [NO 2 ) [H 2 = k 2 K [H 2 [NO 2 = k exp [H 2 [NO 2 where k exp = k 2 K ˆ YES, the proposed mechanism is consistent with the experimental rate law.

10 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 10 PART A QUESTION 12 Given: Overall reaction: 2 O 3 (g) 3 O 2 (g) Mechanism: (i) O 3 O + O 2 (fast, ) (ii) O + O 2 O 3 (fast, ) (iii) O + O 3 2 O 2 (slow, k 2 ) (1) Elementary step (iii) is the slow reaction and is the rate-determining step. From (iii): O + O 3 2 O 2 (slow, k 2 ) rate = k 2 [O Î (2) O is the intermediate and must be substituted by [reactants and [products. (a) using Steady-State approach to eliminate an intermediate rate of appearance of intermediate O = rate of disappearance of intermediate O O 3 O + O 2 (fast, ) O + O 2 O 3 (fast, ) O + O 3 2 O 2 (slow, k 2 ) rate = rate = [O [O 2 rate = k 2 [O = [O [O 2 + k 2 [O = [O ( [O 2 + k 2 ) ˆ [O [O 2 % k 2 Ï Substitute Ï into Î: rate = k 2 [O k 2 ( [O 2 % k 2 ) k 2 2 [O 2 % k 2

11 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 11 (b) using Equilibrium Constant (K) method to eliminate an intermediate (Note that K = / ) Recall the rate-determining step: rate = k 2 [O Î From (i): O 3 O + O 2 (fast, ) [O [O K 2 ˆ [O [O 2 Ï Substitute Ï into Î: rate = k 2 [O k 2 ( [O 2 ) k 2 2 [O 2 (c) conditions of the rate laws derived by the methods (a) the same as (b) From Steady-State Approach: rate k 2 2 [O 2 % k 2 From Equilibrium Constant Method: rate k 2 2 [O 2 If [O 2 >> k 2 then the Steady-State rate law will become rate k 2 2 [O 2 which is the same as the rate law derived by Equilibrium Constant method (d) if the experimentally found rate law is: rate = k exp 2 / [O 2, conditions when the proposed mechanism is consistent with the experimental data when k exp k 2 k 2 K

12 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 12 PART B QUESTION 1 Given: Î MgCl 2 (s) + ½ O 2 (g) º MgO(s) + Cl 2 (g) K p = 2.95 atm ½ (at 1000 K) Ï MgCl 2 (s) + H 2 O(g) º MgO(s) + 2 HCl(g) K p = 8.40 atm (at 1000 K) T = 1000 K R = L atm K!1 mol!1 To find: the equilibrium constants K p and K c at 1000 K for the reaction: 2 Cl 2 (g) + 2 H 2 O(g) º 4 HCl(g) + O 2 (g) (!2) Î 2 MgO(s) + 2 Cl 2 (g) º 2 MgCl 2 (s) + O 2 (g) K 1 ( atm ½ )2 2 Ï 2 MgCl 2 (s) + 2 H 2 O(g) º 2 MgO(s) + 4 HCl(g) K 2 = (8.40 atm) 2 Overall: 2 Cl 2 (g) + 2 H 2 O(g) º 4 HCl(g) + O 2 (g) K p K 1 K 2 K p K 1 K 2 ( atm ½ )2 (8.40 atm) atm 8.11 atm )n = 3 n(gaseous products)! 3 n(gaseous reactants) = {(4 + 1) moles}! {(2 + 2) moles} = 5 moles! 4 moles = 1 K p = K c (R T) )n atm = K c {( L atm K!1 mol!1 ) (1000 K)} 1 ˆ K c = !2 mol L!1 = !2 M

13 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 13 PART B QUESTION 2 Given: C(s) + 2 H 2 (g) º CH 4 (g) )H rxn =!75 kj (ˆ exothermic reaction) )n = 1 mole CH 4 (g)! 2 moles H 2 (g) =!1 (a) decrease in temperature For exothermic reaction ()H =!ve), a decrease in temperature shifts the equilibrium forward. For endothermic reaction ()H = +ve), an increase in temperature shifts the equilibrium forward. Since a decrease in temperature favours the exothermic reaction, the equilibrium will shift from LEFT RIGHT. (b) decrease in volume A decrease in the volume of a gaseous system (or an increase in external pressure) favours a decrease in the moles of gas present (i.e. )n =!ve) and the equilibrium shifts to the product side. ( where )n = 3 n (gaseous products)! 3 n (gaseous reactants) ) An increase in the volume of a gaseous system (or a decrease in external pressure) favours an increase in the moles of gas present (i.e. )n = +ve) and the equilibrium shifts to the product side. Since there are two moles of H 2 (g) gaseous reactant converting into one mole of CH 4 (g) gaseous product (i.e. )n =!1), a decrease in volume favours a net decrease in the number of moles of gas. Thus, the equilibrium will shift from LEFT RIGHT. (c) decrease in pressure of H 2 (g) K p p(ch 4 ) p(h 2 ) 2 If Q < K eq, If Q > K eq, If Q = K eq, then the reaction proceeds toward products (i.e. to the right). then the reaction proceeds to form reactants (i.e. to the left). then the system reaches equilibrium. If then p(h 2 ), a reactant, is removed (or decreased) from the system, Q p becomes larger than K p ( denominator becomes smaller), and the equilibrium will shift from RIGHT LEFT. (d) increase in pressure of CH 4 (g) If p(ch 4 ), a product, is added to the system, then Q p becomes larger than K p ( numerator becomes larger), and the equilibrium will shift from RIGHT LEFT.

14 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 14 (e) (f) addition of C(s) The position of a heterogeneous equilibrium does not depend on the amounts of pure solids (e.g. C(s)) or liquids present. Because C(s) does not appear in the K p expression, there is no effect on the position of equilibrium. addition of catalyst A catalyst lowers the activation energy of a reaction by increasing the rate of reaction. Since the catalyst does not appear in the K p expression, the equilibrium remains unchanged. What are the optimal conditions to get a high yield of CH 4 (g)? Factors affecting the equilibrium for the forward reaction: Î temperature Ï pressure An increase in temperature favours the endothermic reaction. A decrease in temperature favours the exothermic reaction. An increase in external pressure (i.e. a decrease in volume) favours a net decrease in the number of moles of gas (i.e. )n =!ve). A decrease in external pressure (i.e. an increase in volume) favours a net increase in the number of moles of gas (i.e. )n = +ve). (1) exothermic reaction ()H =!ve)! favours low temperature (2) 2 moles gaseous reactants 1 mole gaseous product ˆ )n = (1 mole! 2 moles) =!1! favours high pressure PART B QUESTION 3 Given: SO 3 (g) º SO 2 (g) + ½ O 2 (g) endothermic reaction (i.e. )H = +ve) To find: conditions favoured by maximum formation of SO 2 (g) and O 2 (g) (1) 1 mole gaseous reactant 1.5 moles gaseous products )n = (1.5 moles! 1 mole) = +0.5! The forward reaction is favoured by lowering the pressure. (i.e If )n = +ve, then the reaction is favoured by an increase in volume, thus a decrease in external pressure.) (2) endothermic reaction ()H = +ve)! The forward endothermic reaction is favoured by higher temperature.

15 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 15 PART B QUESTION 4 Given: 2 SO 2 (g) + O 2 (g) º 2 SO 3 (g) In an 1.0 L vessel, Initial: SO 2 (g) = 0.80 mol ˆ [SO 2 = (0.80 mol 1.0 L) = 0.80 mol L!1 O 2 (g) = 0.60 mol ˆ [O 2 = (0.60 mol 1.0 L) = 0.60 mol L!1 At Equilibrium: SO 3 (g) = 0.60 mol ˆ [SO 3 = (0.60 mol 1.0 L) = 0.60 mol L!1 To find: K c for the reaction In an 1.0 L vessel, 2 SO 2 (g) + O 2 (g) º 2 SO 3 (g) I: 0.80 mol L! mol L!1! C: s0.60 mol L!1 s½(0.60 mol L!1 ) r0.60 mol L!1 E: 0.20 mol L! mol L! mol L!1 (1) [SO 3 produced at equilibrium = 0.60 mol L!1 (given) (2) [SO 2 reacted (changed) = [SO 3 produced (2 moles SO 2 2 moles SO 3 ) = (0.60 mol L!1 ) (2/2) = 0.60 mol L!1 ˆ [SO 2 leftover at equilibrium = (0.80 mol L!1! 0.60 mol L!1 ) = 0.20 mol L!1 (3) [O 2 reacted (changed) = [SO 3 produced (1 mole O 2 2 moles SO 3 ) = (0.60 mol L!1 ) (½) = 0.30 mol L!1 ˆ [O 2 remained at equilibrium = (0.60 mol L!1! 0.30 mol L!1 ) = 0.30 mol L!1 K c [SO 3 2 [SO 2 2 [O 2 K c (0.60 mol L!1 ) 2 (0.20 mol L!1 ) 2 (0.30 mol L!1 ) 30 ( mol L )!1 30 M!1

16 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 16 PART B QUESTION 5 Given: SO 2 (g) + NO 2 (g) º SO 3 (g) + NO(g) (K c = 3.0 at 400 o C) In a 500 ml (0.500 L) flask, Initial: SO 2 (g) = 1.0 mol ˆ [SO 2 = (1.0 mol L) = 2.0 M NO 2 (g) = 1.0 mol ˆ [NO 2 = (1.0 mol L) = 2.0 M To find: n(so 3 ) and n(no) present at equilibrium In a 500 ml flask, let x = [SO 2 consumed = [NO 2 consumed = [SO 3 produced = [NO produced SO 2 (g) + NO 2 (g) º SO 3 (g) + NO(g) I: 2.0 M 2.0 M!! C: s x s x r x r x E: 2.0! x 2.0! x x x K c [SO 3 [NO [SO 2 [NO (x) (x) (2.0! x) (2.0! x) 3.0 (x) 2 (2.0! x) (x) (2.0! x) x = M At equilibrium in 500 ml flask, (1) x = [SO 3 produced = [NO produced = M ˆ n(so 3 ) = n(no) = (1.268 mol L!1 ) (500 10!3 L) = mol = 0.63 mol (2) [SO 2 remained = [NO 2 remained = (2.0 M! x) = (2.0 M! M) = M ˆ n(so 2 ) = n(no 2 ) = (0.732 mol L!1 ) (500 10!3 L) = mol

17 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 17 PART B QUESTION 6 NCl 5 (g) º NCl 3 (g) + Cl 2 (g) K c = !3 M (at 298 K) (a) [Cl 2 (g) when 2.00 moles of NCl 5 (g) introduced to 1.00 L container (i.e. [NCl 5 = (2.00 moles 1.00 L) = 2.00 M) Let x = [Cl 2 produced = [NCl 3 produced = [NCl 5 consumed NCl 5 (g) º NCl 3 (g) + Cl 2 (g) I: 2.00 M!! C: s x r x r x E: 2.00! x x x K c [NCl 3 [Cl 2 [NCl !3 M (x) (x) (2.00! x) 1 st approximation: x = !2 M 2 nd approximation: x = !2 M 3 rd approximation: x = !2 M (b) ˆ x = [Cl 2 produced = !2 M = !2 M [Cl 2 (g) when moles of NCl 5 (g) introduced to 1.00 L container (i.e. [NCl 5 = (0.100 moles 1.00 L) = M) Let x = [Cl 2 produced = [NCl 3 produced = [NCl 5 consumed NCl 5 (g) º NCl 3 (g) + Cl 2 (g) I: M!! C: s x r x r x E: 0.100! x x x K c [NCl 3 [Cl 2 [NCl !3 M (x) (x) (0.100! x) 1 st approximation: x = !2 M 2 nd approximation: x = !2 M 3 rd approximation: x = !2 M ˆ x = [Cl 2 produced = !2 M = !2 M

18 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 18 PART B QUESTION 7 Given: PCl 3 (g) + Cl 2 (g) º PCl 5 (g) (K c = 1.60 at 544 K) In a 3.00 L container, initial: PCl 3 (g) = 6.00 moles ˆ [PCl 3 = (6.00 mol 3.00 L) = 2.00 M Cl 2 (g) = 3.00 moles ˆ [Cl 2 = (3.00 mol 3.00 L) = 1.00 M To find: [Cl 2 leftover at equilibrium Let x = [Cl 2 consumed = [PCl 3 consumed = [PCl 5 produced ˆ [Cl 2 leftover at equilibrium = (1.00 M! x) PCl 3 (g) + Cl 2 (g) º PCl 5 (g) I: 2.00 M 1.00 M! C: s x s x r x E: 2.00! x 1.00! x x K c [PCl 5 [PCl 3 [Cl (x) (2.00! x) (1.00! x ) 1.60 x 2! 5.80 x = 0 (quadratic equation) x! b ± b 2! 4 a c 2 a a = 1.6 b =!5.80 c = 3.20 x! (!5.80) ± (!5.80) 2! 4 (1.60) (3.20) 2 (1.60) ˆ At equilibrium, x = M OR M (neglected) [Cl 2 leftover = (1.00 M! x) = (1.00 M! M) = M = 0.32 M

19 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 19 PART B QUESTION 8 Given: CO(g) + H 2 O(g) º CO 2 (g) + H 2 (g) At equilibrium: [CO 2 = [H 2 = M [CO = [H 2 O = 4.00 M (a) K c K c [CO 2 [H 2 [CO [H 2 O K c (0.900 M) (0.900 M) (4.00 M) (4.00 M) (b) Q c when an additional 1.00 M of CO(g) and H 2 O(g) are added to the container [CO = [H 2 O = (4.00 M M) = 5.00 M Q c [CO 2 [H 2 [CO [H 2 O Q c (0.900 M) (0.900 M) (5.00 M) (5.00 M) (c) In which direction will reaction go? Since Q c < K c, (i.e < ) ˆ The reaction will proceed from LEFT RIGHT. If Q < K eq, If Q > K eq, If Q = K eq, then the reaction proceeds toward products (i.e. to the right). then the reaction proceeds to form reactants (i.e. to the left). then the system reaches equilibrium.

20 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 20 PART B QUESTION 9 Given: 2 N 2 O 4 (g) + 6 H 2 O(g) º 4 NH 3 (g) + 7 O 2 (g) Initial: [N 2 O 4 = 2.20 M [H 2 O = 4.20 M At equilibrium: [O 2 = 1.96 M (i.e. (1.96 M 7 moles O 2 ) = M per mole of O 2 produced) To find: [N 2 O 4, [H 2 O and [NH 3 at equilibrium 2 N 2 O 4 (g) + 6 H 2 O(g) º 4 NH 3 (g) + 7 O 2 (g) I: 2.20 M 4.20 M!! C: s2 (0.280 M) s6 (0.280 M) r 4 (0.280 M) r7 (0.280 M) = M = 1.68 M = 1.12 M = 1.96 M E: 1.64 M 2.52 M 1.12 M 1.96 M (1) [N 2 O 4 reacted = [O 2 formed (2 moles N 2 O 4 7 moles O 2 ) = (1.96 M) (2/7) = M ˆ [N 2 O 4 remained at equilibrium = (2.20 M! M) = 1.64 M (2) [H 2 O reacted = [O 2 formed (6 mole H 2 O 7 moles O 2 ) = (1.96 M) (6 / 7) = 1.68 M ˆ [H 2 O remained at equilibrium = (4.20 M! 1.68 M) = 2.52 M (3) [NH 3 produced = [O 2 formed (4 mole NH 3 7 moles O 2 ) = (1.96 M) (4 / 7) = 1.12 M

21 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 21 PART B QUESTION 10 Given: 2 NH 3 (g) º N 2 (g) + 3 H 2 (g) In a 2.00 L container, Initial: NH 3 (g) = 5.00 moles ˆ [NH 3 = (5.00 moles 2.00 L) = 2.50 M At equilibrium: 30% of NH 3 (g) dissociated (i.e. [NH 3 dissociated (changed) = 2.50 M 30% = M ) (a) the equilibrium concentrations of NH 3 (g), N 2 (g) and H 2 (g) 2 moles of NH 3 (g) dissociates into 1 mole of N 2 (g) and 3 moles of H 2 (g) 2 NH 3 (g) º N 2 (g) + 3 H 2 (g) I: 2.50 M!! C: s M r ½ (0.750 M) r3/2 (0.750 M) E: 1.75 M M M At equilibrium, [NH 3 remained = (initial [NH 3! [NH 3 dissociated) = (2.50 M! M) = 1.75 M [N 2 produced = ([NH 3 dissociated) (1 mole N 2 2 moles NH 3 ) = (0.750 M) (½) = M (b) K c of the reaction at 500 K [H 2 produced = ([NH 3 dissociated) (3 mole H 2 2 moles NH 3 ) = (0.750 M) (3/2) = M K c [N 2 [H 2 3 [NH 3 2 At equilibrium, [NH 3 = 1.75 M [N 2 = M [H 2 = M K c (0.375 M) (1.125 M) 3 (1.75 M) M M 2

22 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 22 PART D QUESTION 1 )T = k f m )T = freezing-point depression ( o C) k f = molal freezing-point depression constant ( o C kg mol!1 ) m = molality (mol kg!1 ) (a) 45.0 g sucrose C 12 H 22 O 11 in 100 g water M(C 12 H 22 O 11 ) = 342 g mol!1 n(c 12 H 22 O 11 ) = mass(c 12 H 22 O 11 ) M(C 12 H 22 O 11 ) = 45.0 g 342 g mol!1 = mol m(c 12 H 22 O 11 ) = n(c 12 H 22 O 11 ) mass(water) = mol !3 kg = mol kg!1 )T = k f m = (1.86 o C kg mol!1 ) ( mol kg!1 ) = o C ˆ Freezing point of 45.0 g sucrose C 12 H 22 O 11 in 100 g water is!2.45 o C. (b) mol FeCl 3 in 1 kg water FeCl 3 dissociates into four ions per formula unit. n(fecl 3 ) = mol n(ions) = n(fecl 3 ) (4 moles ions 1 mole FeCl 3 ) = (0.100 mol) (4) = mol m(ions) = n(ions) mass(water) = (0.400 mol) (1 kg water) = mol kg!1 )T = k f m = (1.86 o C kg mol!1 ) (0.400 mol kg!1 ) = o C ˆ Freezing point of mol FeCl 3 in 1 kg water is!0.744 o C. (c) mol BaCl 2 O dissolved in 52.2 g water BaCl 2 dissociates into three ions per formula unit. n(ions) = n(bacl 2 ) (3 moles ions 1 mole BaCl 2 ) = ( mol) (3) = mol n(h 2 O) = n(bacl 2 O) (2 moles H 2 O 1 mole BaCl 2 O) = ( mol) (2) mass(h 2 O) = ( mol 2) (18.02 g mol!1 ) = g total mass(h 2 O) = 52.2 g g = g = !3 kg m(ions) = n(ions) total mass(water) = ( mol) ( !3 kg) = mol kg!1 )T = k f m = (1.86 o C kg mol!1 ) ( mol kg!1 ) = o C ˆ Freezing point of mol BaCl 2 O dissolved in 52.2 g water is!2.11 o C.

23 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 23 PART D QUESTION 2 Given: To find: 500 g (500 10!3 kg) of water freezing point =!2.14 o C (i.e. )T = 2.14 o C) mass of CaCl 2 required )T = k f m )T = freezing-point depression ( o C) k f = molal freezing-point depression constant ( o C kg mol!1 ) m = molality (mol kg!1 ) Let x = n(cacl 2 ) CaCl 2 dissociates into three ions per formula unit. n(ions) = n(cacl 2 ) (3 moles ions 1 mole CaCl 2 ) = 3 x m(ions) = n(ions) mass(water) = (3 x) (500 10!3 kg) = 6.00 x kg!1 )T = k f m 2.14 o C = (1.86 o C kg mol!1 ) (6.00 x kg!1 ) x = mol = n(cacl 2 ) M(CaCl 2 ) = [ (35.453) g mol!1 = g mol!1 ˆ mass of CaCl 2 required = n(cacl 2 ) M(CaCl 2 ) = ( mol) ( g mol!1 ) = g = 21.3 g

24 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 24 PART D QUESTION 3 Given: osmotic pressure (B) = 7.6 atm T = 37 o C = ( ) K = 310 K R = atm L K!1 mol!1 To find: the concentration of NaCl solution in mol L!1 B = c R T B = osmotic pressure (atm) c = molar concentration (mol L!1 ) R = L atm K!1 mol!1 T = absolute temperature (K) B = c(ions) R T 7.6 atm = c(ions) ( atm L mol!1 K!1 ) (310 K) c(ions) = mol L!1 NaCl dissociates into two ions (Na + and Cl! ) per formula unit. ˆ c(nacl) = c(ions) (1 mole NaCl 2 moles ions) = ( mol L!1 ) (½) = mol L!1 = 0.15 mol L!1

25 CHEM*130 (F 01) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 25 PART D QUESTION 4 Given: To find: ml water sample mol L!1 EDTA ml of EDTA to reach the Eriochrome Black indicator endpoint hardness of water as expressed in: (a) mol L!1 of Ca 2+ (b) ppm of Ca 2+ Assume the Ca 2+ is the only cation that will be titrated against EDTA 4!. The reaction is: Ca 2+ (aq) + EDTA 4! (aq) (Ca!EDTA) 2! (aq) n(edta 4! ) = c(edta 4! ) V(EDTA 4! ) = ( mol L!1 ) ( !3 L) = !4 mol 1 mole of Ca 2+ (aq) reacts with 1 mole of EDTA 4! (aq) n(ca 2+ ) = n(edta 4! ) (1 mole Ca 2+ 1 mole EDTA 4! ) = ( !4 mol) (1) = !4 mol (a) c(ca 2+ ) in mol L!1 = n(ca 2+ ) V(Ca 2+ present in water sample) = ( !4 mol) ( !3 L) = !3 mol L!1 = !3 mol L!1 M(Ca 2+ ) = g mol!1 = mg mol!1 1 ppm = 1 mg L!1 (b) c(ca 2+ ) in ppm = ( !3 mol L!1 ) ( mg mol!1 ) = mg L!1 = ppm