© Prep 101 Chem 203 Term Test 2 Exam Solutions Prep 101 – Chemistry 203 Term Test 2 Kinetics Problem 1. Solution: v1 k [ I ]1m [ S2O82 ]1n v2 k [ I ]2m [ S2O82 ]2n 1.25 105 k (0.080)m (0.040)n 2 2m m 1 6.25 106 k (0.040)m (0.040)n v1 1.25 105 k (0.080)m (0.040)n 2 2n n 1 v3 6.25 106 k (0.080)m (0.020)n rate k [ I ]1[ S2O82 ]1 k rate [ I ][ S2O82 ] 1.25 105 k1 3.91 103 M 1s 1 (0.080) (0.040) Problem 2. Solution: Rate = k[H2 O]x [CH3 Cl]y Rate1 k (0.0100) x 3.6 104 Rate2 k (0.0200) x 1.44 104 0.5x = .25, therefore x = 2 and the reaction is second-order in H2O Rate1 k (0.0100) x 3.6 104 Rate2 k (0.0200) x 1.44 104 5.54(2/3)y=3.69 (2/3)y = 2/3, therefore y = 1 and the reaction is first-order in CH3Cl Rate = k[H2 O]2 [CH3 Cl]1 Problem 3. Solutions at www.prep101.com/solutions 1 of 12 © Prep 101 Chem 203 Term Test 2 Exam Solutions [ A]0 ln[ A]t1/2 ln[ A]0 kt1/2 , ln kt1/2 [ A]t 1/2 [ A]0 for t1/2 [A]t1/2 2 [ A]0 [ A]0 ln ln ln 2 kt1/2 [ A]t [ A]0 2 1/2 ln 2 t1/2 k Problem 4. Solution: If 20% decomposes, then 80% of the sample remains. ln[ A] ln[ A]0 kt rearrange 0.8[ A]o ln k(50s) [ A]o k 4.463103 sec1 ln 2 t1/ 2 155 sec k Problem 5. Solution: (a) v1 k [CH 3 NNCH 3 ]1n v2 k [CH 3 NNCH 3 ]2n 2.8 106 (5.13 102 ) n 0.25 0.25n n 1 5 1 n 1.1 10 (2.05 10 ) Therefore the reaction is the first order. 2.8 106 M / s Rate = k [CH3 NNCH3(g) ], therefore k = = 5.46 x 10-5 s-1 5.13 102 M (b) ln[A] = ln[A]o – kt ln[0.15.13102] = ln[5.13102] – (5.4610-5)t -5.27 = -2.97 - (5.4610-5)t (5.4610-5)t = 2.30 t = 4.21104 s = 11.7 hours (c) [A]10% = 0.15.13102 = 5.1210-3M Rate = k[A] = (5.4610-5 s-1)( 5.1210-3 M) = 2.8010-7 M/s Solutions at www.prep101.com/solutions 2 of 12 © Prep 101 Chem 203 Term Test 2 Exam Solutions Problem 6. Solution: v2 k [CH 3COCH 3 ]2m [ Br2 ]2n [ H ]l2 v1 k [CH 3COCH 3 ]1m [ Br2 ]1n [ H ]1l 8.00 105 k (0.02)m (0.01)n (0.01)l 2 2m m 1 5 m n l 4.00 10 k (0.01) (0.01) (0.01) v3 8.00 105 k (0.02)m (0.02)n (0.01)l 1 2n 5 m n l v2 8.00 10 k (0.02) (0.01) (0.01) ln 0.5 ln 2n n ln 2 n 0 v4 8.00 105 k (0.01)m (0.02)n (0.02)l 2 2l l 1 v1 4.00 105 k (0.01)m (0.01)n (0.01)l rate k [CH 3COCH 3 ]1[ Br ]0 [ H ]1 k [CH 3COCH 3 ][ H ] k rate [CH 3COCH 3 ]2 [ H ] 4.00 105 k1 4.00 103 M 1s 1 (0.01) (0.01) Problem 7. Solution: Because this reaction is second order 1 1 kt if 1/[A] was plotted vs. t then the y[ A] [ A]o intercept would be 1/[A]0 and the slope would be k 1/[A] slope = +k 1/[A]0 t Solutions at www.prep101.com/solutions 3 of 12 © Prep 101 Chem 203 Term Test 2 Exam Solutions Problem 8. Solution: I is the catalyst as it is used up in the first reaction and is produced in the second. It does not show up in the overall reaction. Problem 9. Problem 10. Solution: Since Hrxn > 0, then Hproducts > Hreactants Thus, Ea rev = 66 – 41 = 25 kJ/mol (draw an energy diagram to see this better) Solutions at www.prep101.com/solutions 4 of 12 © Prep 101 Chem 203 Term Test 2 Exam Solutions Problem 11. Solution: Let T1 = 96 C = 369 K, and T2 = 25 C = 298 K. 𝑘 Using the Arrhenius Equation 𝑙𝑛 (𝑘1 ) = 2 𝐸𝑎 𝑅 1 1 2 1 (𝑇 − 𝑇 ) 3.55 x 1010 s 1 Ea 1 1 ln 10 1 1 1 8.314 JK mol 369 K 298K 1.2 x 10 s and solve for Ea, Ea = 12.8 kJ/mol Problem 12. k E 1 1 Solution: from Arrhenius equation ln 2 a R T2 T1 k1 since vk, then k v E 1 1 ln 2 ln 2 a R T2 T1 k1 v1 E 1 1 ln(3) a Ea 77433.3 J / mol 77.4 kJ / mol 8.314 325 313 Problem 13. 2 Solution: Since we have an elementary process, then rate of the reaction is rate k[ A] , This reaction is second order therefore 1 1 kt [ A] [ A]o 1 1 k(35.2 min) 0.1M 0.2 M 5 M-1 = k(35.2 min) k = 1.42 x 10-1 M-1min-1 Problem 14. The steps are elementary, and the first step is the rate determining step. So the overall rate law only depends on the first step: rate = k[O3][NO] Solutions at www.prep101.com/solutions 5 of 12 © Prep 101 Chem 203 Term Test 2 Exam Solutions Problem 15. Solution: Slow step rate = k2[E][B] For the fast equilibrium, rate = k1[A][B] = k-1[E] Rearrange to: [E] k1 kk [A][B] , therefore, rate 1 2 [ A][B]2 This is inconsistent with the rate law. k1 k 1 (ii): slow step rate = k2[E][A] and fast equilibrium rate = k1[A][B] = k-1[E] [E] kk k1 [A][B] , therefore, rate 1 2 [ A]2[B] This is consistent with the rate law k1 k1 (iii) The first step is the rate determining step. The rate law would be rate = k1[A]2 This is inconsistent with the proposed rate law. Problem 16. Solution: N2O2 is an intermediate, therefore cannot be present in the rate law. Rate is found from the slower of the two reactions: Rate = k2[N2O2][O2], where k-1[N2O2] = k1[NO]2 Therefore, rate k1k2 [NO]2[O2 ] k [NO]2[O2 ] k1 Solutions at www.prep101.com/solutions 6 of 12 © Prep 101 Chem 203 Term Test 2 Exam Solutions Equilibrium Problem 17. Solution: 1. CO2 3eq K eq C3H 8 1eq O2 5eq 2. Keq CH 3COO Na eq eq Problem 18. Solution: (a) HBr(g) ½ H2(g) + ½ Br2(g) From H2(g) + Br2(g) 2 HBr(g), K1 = 3.5 x 104, you must reverse the reaction: 2 HBr(g) H2(g) + Br2(g), K-1 = 1/K1 = 2.9 x 10-5, and then half the reaction: HBr(g) ½ H2(g) + ½ Br2(g), K = K 1 = 5.3 x 10-3 (b) H2(g) + ½ Cl2(g) + ½ Br2(g) HCl(g) + HBr(g) This reaction is the sum of half of each of the reactions above. H2(g) + Br2(g) 2 HBr(g), with K1 = 3.5 x 104 ½ H2(g) + ½ Br2(g) HBr(g), with K3 = K1 = 187 H2(g) + Cl2(g) 2 HCl(g), with K2 = 9.5 x 103 ½ H2(g) + ½ Cl2(g) HCl(g), with K 4 = K 2 = 97 H2(g) + ½ Cl2(g) + ½ Br2(g) HCl(g) + HBr(g), K = K3 x K4 = 1.8 x 104 Solutions at www.prep101.com/solutions 7 of 12 © Prep 101 Chem 203 Term Test 2 Exam Solutions Problem 19. Solution: The expression for K P4O10 eq P4 eq O2 5eq 3 P4O10 init 1 M 1 3 , the reaction proceeds to the right. 5 5 1 M 1 M P4 init O2 init P4O10 init 5.0 M In (b) Q 3.7 3 , the reaction proceeds to the left. 5 P4 init O2 init 8.0 M 0.70 M 5 In (a) Q Problem 20. Solution: To determine the final concentrations, the first thing needed are the initial reactant concentrations and an expression for the reaction coefficient Q. [A2(g) ] = and Q 1.00 mol 2.00 mol = 4.00 M, [AB(g) ] = [B2(g) ] = = 8.00 M, 0.250 L 0.250 L 2 AB init A2 init B2 init (8.00)2 2.0 (4.00) (8.00) The direction of the reaction needs to be determined. To do this, we compare Q vs K. Since Q = 2.0 > K = 0.5, the reaction proceeds to the left. To determine what final concentrations will be from initial concentrations, a handy tool – the Initial, Change, Equilibrium (ICE) table can be used. All compounds involved in the reaction are included in an ICE table as follows, with the species that will be consumed on the left side, and the species that will be produced on the right side. Since the reaction proceeds to the left, A2 and B2 will be formed and AB will be consumed. A2(g) B2(g) Concentration (M) 2 AB(g) Initial 8.00 M 4.00 M 8.00 M Change -2x +x +x Equilibrium 8.00 – 2x 4.00 + x 8.00 + x As the reaction proceeds, x moles of A2(g) and B2(g) are formed as 2x moles of AB(g) are consumed. Make sure to consider stoichiometric coefficients appropriately. The equilibrium values are simply the sum of the initial + change concentrations. Solutions at www.prep101.com/solutions 8 of 12 © Prep 101 Chem 203 Term Test 2 Exam Solutions Substitute the equilibrium concentrations into K, and solve for x (remember that K is defined for the reaction in the way that it was initially described): K AB 2eq A2 eq B2 eq (8.00 2 x )2 0.5 (4.00 x ) (8.00 x ) so, (8.00 2 x )2 0.5(4.00 x )(8.00 x ) 64 32 x 4 x 2 0.5(32 12 x x 2 ) 64 32 x 4 x 2 16 6 x 0.5x 2 3.5x 2 38 x 48 0 To solve for x, the quadratic formula must be used, x b b2 4ac 38 ( 38)2 4(3.5)(48) 1.46 or 9.39, 2a 2(3.5) Since 2 x 9.39 = 18.78 > 8.00 (which would leave the equilibrium concentration of AB at equilibrium negative), then the x = 1.46 So, final concentrations are: [A2] = 4.00 + 1.46 = 5.46 M [B2] = 8.00 + 1.46 = 9.46 M [AB] = 8.00 – 2(1.46) = 5.08 M To check, substitute these concentrations into the Equilibrium constant expression, AB 2eq (5.08)2 K 0.5 , matches up. A2 eq B2 eq (5.46) (9.46) Problem 21. [ NO ]4 Answer: K c [ H ]4 [ NO3 ]4 Solution: Pure solids and pure liquids and solvents are not included in the expression. Products go over Reactants, and coefficients in the equation are written as superscripts. Problem 22. Solution: K c = [CO2(g) ] Remember: Solids and liquids are not included in the expression. Solutions at www.prep101.com/solutions 9 of 12 © Prep 101 Chem 203 Term Test 2 Exam Solutions Problem 23. Solution: Equation 2 is equal the double and reverse of equation1 therefore 1 K p2 = K −2 ) 2 = 40.6 p1 = ( 0.157 Problem 24. Answer: 1 −1/2 Equilibrium Constant = K1 = √K1 Solution: The second equation if reversed (1/K) and halved (K1/2). Combining these two gives 1/K1/2. Problem 25. Answer: So the reaction equations is: 2 CO(g) + O2(g) ⇌ 2 CO2(g) K c = 3.3 × 1091 The relationship between K c and K p is: K p = K c (RT)∆n gas In this case there are 3 moles of gas in the reactants and 2 moles of gas in the products, so ∆n = -1 So solving for K p : −1 L atm K p = K c (RT)∆n gas = (3.3 × 1091 ) [(0.0821 ) (298 K)] = 1.35 × 1090 mol K Problem 26. Solution: N2(g) + i: c: e: 1.00 +x 1.00+ x Q=1>K so then rxn goes to the left C2 H2(g) ⇌ 1.00 +x 1.00 + x 2 HCN(g) 1.00 -2x 1.00-2x (1.00 2 x )2 (1.00 2 x ) Kc Kc 2 (1.00 x ) (1.00 x ) x 1 Kc 2 Kc 0.488 Solutions at www.prep101.com/solutions 10 of 12 © Prep 101 Chem 203 Term Test 2 Exam Solutions Problem 27. Solution: COCl2(g) initial: change: equil: x2 Kc = 0.04 x x2 + Kcx – 0.04Kc = 0 x 0.04 -x 0.04-x ⇌ CO(g) +Cl2(g) 0 +x x 0 +x x Kc Kc 2 (4)(0.04)( Kc) 5.29 103 2 Problem 28. Solution: Kc I C E [CO2 ][ H 2 ] 5.10 [CO][ H 2O] CO + 0.1M -x 0.1-x H2 O ⇌ 0.1 -x 0.1-x CO2 0.1 +x 0.1+x +H2 0.1 +x 0.1+x (0.1 x )2 0.01 0.2 x x 2 Kc 5.10 (0.1 x )2 0.01 0.2 x x 2 0.01 0.2 x x 2 0.051 1.02 x x 2 0 0.041 1.22 x 1.22 x 0.041 x 0.034 M Therefore at equilibrium [H2] = 0.1 + x = 0.1 + 0.034M = 0.134M Problem 29. Solution: Only C is true An increase in volume will shift the equilibrium to the right thus causing an increase in the total moles of CO at equilibrium. Solutions at www.prep101.com/solutions 11 of 12 © Prep 101 Chem 203 Term Test 2 Exam Solutions Problem 30. Solution: a) shift to the left b) no effect c) shift to the right d) shift to the right Problem 31. Solution: The reaction will shift left forming Ni(CO)4(g) to reach equilibrium Problem 32. Answer: C Solution: Q = [NO]2[Cl2]/[NOCl]2 = (1.2)2(0.56)/(1.3)2 = 0.51 = K, therefore we are already at equilibrium. Solutions at www.prep101.com/solutions 12 of 12