4. The following graphs were prepared from experimental data for a reactant, A. What is the correct order of A?

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Peregrine Burns
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1 Diebolt Summer 010 CHM 15 HOUR EXAM I KEY (15 pt) Part One: Multiple Choice. 1. Conider the reaction IO 3 + 5I + 6H + 3I + 3H O. If the rate of diappearance of I i M/, what i the rate of appearance of H O? mol I - /L 3 mol H O/5 mol I - A M/ B M/ C M/ D M/ E M/. The rate law for a given reaction i Rate = k[a] 3 [B]. If the concentration of A i doubled and the concentration of B i tripled, the reaction rate would increae by a factor of. Rate = k[a] 3 [3B] = 4 k[a] 3 [B] A. 6 B. 18 C. 4 D. 7 E A firt-order reaction, A Product, ha a rate contant of year -1. How long will it take for 87.5% of a ample of A to be conumed? t 1/ = 0.693/ yr -1 = 100 yr; 1.5% remain after 3 half live; 3 x 100 yr = 300 yr A year B year C. 00. year D year E year 4. The following graph were prepared from experimental data for a reactant, A. What i the correct order of A? 1 [A] [A] O order plot i linear Time Time Time A. zero order B. firt order C. econd order D. inufficient information provided 5. In the following mechanim: tep 1: CH 3 CHO + I CH 3 I + CO + HI tep : CH 3 I + HI CH 4 + I A. I i a catalyt, HI i a catalyt and CH 3 I i an intermediate B. I i an intermediate, HI i an intermediate and CH 3 I i a catalyt C. I i an intermediate, HI i a catalyt and CH 3 I i a catalyt D. I i a catalyt, HI i an intermediate and CH 3 I i an intermediate 6. The reaction NO (g) + F (g) FNO (g) i propoed to occur by the following two tep mechanim. Step 1: NO (g) + F (g) FNO (g) + F(g) low Step : NO (g) + F(g) FNO (g) fat According to thi mechanim, the rate law for the overall reaction i A. rate = k[no ] [F ] B. rate = k[no ][F] C. rate = k[no ][F ] D. rate = k[fno ][F] The law i baed on the reactant from the low tep! CHM15 Summer 10 Exam 1 page 1 of 5

2 7. The potential energy diagram i for a two-tep reaction that i: A. exothermic, firt tep i low and econd tep i fat. B. exothermic, firt tep i fat and econd tep i low. C. endothermic, firt tep i low and econd tep i fat. D. endothermic, firt tep i fat and econd tep i low. Product lower = exo; 1 t tep i higher o it lower. E reaction pathway 8. For the following diagram, which tatement i NOT correct? H P.E F Y Z X G Reaction Path A. X i the heat of reaction. B. G repreent the energy level for the product. C. Z repreent the activation energy for the forward reaction. Z i E a revere rxn D. The reaction i exothermic. E. H repreent the tranition tate. 9. A fat reaction i likely to have: Fat = large rxn rate = large k o t 1/ and E a are mall. A. large k, hort t 1/, large E a B. large k, hort t 1/, mall E a C. large k, long t 1/, mall E a D. mall k, hort t 1/, mall E a E. mall k, long t 1/, large E a 10. Which of the following i a true tatement about chemical equilibrium? A. At equilibrium the total concentration of product equal the total concentration of reactant, that i, [product] = [reactant]. B. Equilibrium i the reult of the ceation of all chemical change. C. There i only one et of equilibrium concentration that equal the K c value. D. Equilibrium depend on the initial concentration of the reactant. E. At equilibrium, the rate of the forward reaction i equal to the rate of the revere reaction 11. For which of the following reaction i K c equal to K p? ame # ga molecule both ide A. N O 4 (g) NO (g) B. CO(g) + O (g) CO (g) C. CS (g) + 3Cl (g) S Cl (g) + CCl 4 (g) D. H (g) + Cl (g) HCl(g) E. all of the above 1. Which of the following i a true tatement about the equilibrium contant, K c, for a given reaction? A. The K c value alway remain the ame under all experimental condition including concentration, preure and temperature. B. The K c value increae when the concentration of one of the product i increaed. C. The K c value depend on the initial concentration of the reactant. D. The K c value change with change in the temperature. E. None of thee tatement are conidered to be true. CHM15 Summer 10 Exam 1 page of 5

3 13. H = +57. kj for the forward reaction of Heat + N O 4 (g) NO (g). The amount of N O 4 (g) can be maximized by carrying out the reaction at Want left hift V = P hift to ide with le mole of ga; T hift left if heat i removed A. high T, high P B. high T, low P C. low T, high P D. low T, low P 14. If K c = 0.50 for the following reaction at 8 C, what i the value of K p? PCl 3 (g) + Cl (g) PCl 5 (g) K p = 0.50(0.0806*501 K) -1 = 6.08x10-3 A B C. 1.34x10 - D. 6.08x10-3 E Conider the reaction A(g) 3B(g), where K c = 15. If 5.00 mole of A and 5.00 mole of B are introduced into a.00 L flak, which of the following tatement mut be true? A. The initial reaction mixture i already at equilibrium. B. [A] decreae and [B] increae to attain equilibrium. C. [A] increae and [B] decreae to attain equilibrium. D. [A] = [B] at equilibrium mol [ B].5 [A] = [B] = =.50 M; Q = = 6.5; Q < K o hift, [A] and [B].00L.5 Part Two. Short anwer (15 point) 1. Conider the following equilibrium reaction: (5 mol) 4HCl(g) + O (g) (4 mol) Cl (g) + H O(g) + heat H = ᅳ 113 kj For a-c, predict the effect of the following change on the poition of equilibrium (left, right, no change), when each of the following change i made (15 pt): a. Removing O (g) r left (removing reactant) b. Decreaing the volume right (hift to ide with fewer mole) c. Adding HCl(g) right (adding reactant) d. Decreaing the temperature right (hift toward heat removed) e. What happen to the value of K c (increae, decreae or tay the ame), if the temperature i decreaed? Increae (eq hift, [product], o K ) Part Three. Numerical Problem. (50 point) 1. Conider the following reaction: SO Cl SO + Cl. The reaction i firt order with repect to SO Cl and the rate contant, k, i 4.67x at a certain temperature. How many hour will it take for the concentration of SO Cl to decreae from M to 0.10 M? (8 pt) t 0.10 = - kt = -(4.67x )t (0.103) = -(4.67x )t -.7 = -(4.67x )t. 7 t = x10 = 4.86x10 4 1min 1hour = 13.5 hour 60 60min CHM15 Summer 10 Exam 1 page 3 of 5

4 . Given the following reaction: CO + Cl COCl [CO] (M) [Cl ] (M) Rate (M/) 1.00x x x x x x x x x 10 5 a) Determine the order for CO and Cl and write the rate law for thi reaction. (7 pt) CO order = 1 Cl order = Rate = k[co][cl ] b) Calculate the rate contant, k, and include the proper unit. (4 pt) Rate.64x10 4 M/ k = = 6.60x10 9 M - -1 [CO][Cl ] (1.00x10 M)(.00x10 M) 3. The activation energy for a reaction i 6.0 kj/mol. If the rate contant i 1.0 x at C, what i the rate contant at C? (11 pt) k 1 = 1.0 x T 1 = 773 K E a = 6.0 kj/mol T = 873 K J R = Kmol 1kJ = x 10-3 kj/kmol 1000 J k k 1 E = a 1 1 R T T k 4 1 = kj 6.0 mol 3 kj x10 Kmol K 773 K k 4 1 =.718x10 4 K(-1.48x10-4 K -1 ) = k k = = e = = 6.74x CHM15 Summer 10 Exam 1 page 4 of 5

5 4. A mixture of M H and M F i placed in a container and undergoe the following reaction. What are the equilibrium concentration of H, F and HF? (10 pt) K c = H (g) + F (g) HF(g) K c = 9.09 [HF] [H ][F ] (x) (x) 9.09 = = x0.375 x x Take quare root both ide: = x 3.015( x) = x x = x 1.13 = 5.015x thu, x = = x perfect quare Equilibrium Conc: [H ] = [F ] = = M; [HF] = (0.5) = M 5. A ample of 1.00 mole NOCl i placed in a.00 L container. At a given temperature, the following reaction ha K c = 1.6 x Calculate the concentration of all pecie preent at equilibrium. (10 pt) NOCl(g) NO(g) + Cl (g) Initial [NOCl] = 1.00 mol/.00 L = M NOCl(g) NO(g) + Cl (g) I, M C, M -x +x x E, M x x x K c = [NO] [Cl [NOCl] ] Small K o try approximation method: 1.6x10-5 (x) (x) = (0.500) 4.0x10-6 = 4x 3 H (g) + F (g) HF(g) Initial, M Change, M -x -x +x Equilibrium, M x x x x 3 = 1.0x10-6 x = (1x10-6 ) 1/3 = 1.0x10 - check approximation = (1.0x10 - /.50) 100% = % (< 5% o good approximation) Equilibrium: [NOCl] = (1.0x10 - ) = 0.48 M NOCL [NO] = x = 1.0x10 - = 0.00 M NO [Cl ] = x = M Cl CHM15 Summer 10 Exam 1 page 5 of 5

a. rate = k[no] 2 b. rate = k([no][o 2 ] c. rate = k[no 2 ] 2 [NO] -2 [O 2 ] -1/2 d. rate = k[no] 2 [O 2 ] 2 e. rate = k([no][o 2 ]) 2

a. rate = k[no] 2 b. rate = k([no][o 2 ] c. rate = k[no 2 ] 2 [NO] -2 [O 2 ] -1/2 d. rate = k[no] 2 [O 2 ] 2 e. rate = k([no][o 2 ]) 2 General Chemistry III 1046 E Exam 1 1. Cyclobutane, C 4 H 8, decomposes as shown: C 4 H 8 (g)! 2 C 2 H 4 (g). In the course of a study of this reaction, the rate of consumption of C 4 H 8 at a certain