Chem 6 sample exam 1 (100 points total) @ This is a closed book exam to which the Honor Principle applies. @ The last page contains several equations which may be useful; you can detach it for easy reference. @ Please write clearly and SHOW YOUR WORK. If you need to write on the back of the exam paper, please indicate this clearly. @ Some questions are more challenging than others. Allot your time accordingly, and try to answer EVERY question. NAME
2 1. (8 pts) OCl + I ---> OI + Cl Some initial concentration and rate data for this reaction, in basic aqueous solution, is shown. [OCl ] (M) [I ] (M) [OH ] (M) Rate (Ms 1 ) 0.0040 0.0020 1.00 4.8x10 4 0.0020 0.0040 1.00 4.8x10 4 0.0020 0.0020 1.00 2.4x10 4 0.0020 0.0020 0.50 4.8x10 4 0.0020 0.0020 0.25 9.6x10 4 (a) (3 pts) Find the order of reaction with respect to OCl, I, and OH. Briefly show how you got these results. (b) (2 pts) Give the overall reaction order. (c) (3 pts) Write the rate law and find the value of k (Don't forget the units.)
3 2. (12 pts) At 690 K and constant volume, the total pressure in the reaction C 2 H 4 O (g) -----> CH 4 (g) + CO (g) is measured as a function of time. [At the "infinite" time, you can assume all the C 2 H 4 O has reacted.] time (min) 10 20 40 60 100 200 "infinity" P(torr) 139.14 151.67 172.65 189.15 212.34 238.66 249.88 Find the rate law for the reaction and the value of the rate constant.
4 3. (16 pts) For the reaction H 2 + Br 2 ---> 2HBr the following chain mechanism has been proposed: Br 2 k 1 Br + H 2 H + Br 2 H + HBr 2Br k 2 k 3 k 4 k 5 Br + Br Br 2 HBr + H HBr + Br H 2 + Br (a) (3 pts) Label the steps as initiation (I), propagation (P) or termination (T) in the spaces provided above (you don't need to label the 4th step.) (b) (6 pts) Use the steady-state approximation to find the expressions below: d dt [H] = d dt [Br] = (c) (4 pts) Find [H] and [Br] in terms of [H 2 ], [Br 2 ], and [HBr]. [HINT: to simplify the algebra, first add together the equations you got in part (b).]
5 (d) (3 pts) Give a formula for the rate of production of HBr and briefly explain how you could express it in terms of [H 2 ], [Br 2 ], and [HBr].
6 4. (12 pts) Cl 2 (aq) + H 2 S(aq) ---> S(s) + 2H + (aq) + 2Cl (aq) This reaction is found experimentally to be 2nd order with the rate law: rate = k[cl 2 ][H 2 S] Several mechanisms have been proposed for this reaction. (a) Cl 2 + H 2 S H + + Cl + Cl + + HS (slow) Cl + + HS H + + Cl + S (fast) (b) H 2 S HS + Cl 2 (c) H 2 S H + + Cl 2 Cl + + HS HS + H + (fast equilibrium) 2Cl + S + H + (slow) HS + H + (fast equilibrium) H + + Cl + Cl + (fast equilibrium) H + + Cl + S (slow) Find the rate laws expected for each of these three mechanisms, and state which, if any, of them, are consistent with the experimental rate law.
7 5. (10 points) (a) (4 pts) At 25 C, decomposition of hydrogen peroxide to water and oxygen is catalyzed by Pt metal, which increases the reaction rate by a factor of 4.1x10 4 over the uncatalyzed reaction. By how much (in kj/mol) does the Pt catalyst decrease the activation energy for this reaction? Tungsten catalyzes the decomposition of ammonia: 2NH 3 (g) ---> N 2 (g) + 3H 2 (g) This is a zero-order process, with the rate law: rate = k. (b) (3 pts) Express the concentration of ammonia (c) at a given time (t) during the reaction in terms of its initial concentration (c o ) and the rate constant k. (c) (3 pts) Find an expression for the half-life (t 1/2 ) of this reaction.
8 6. (12 pts) For the reaction A---> B + C In an experiment where the initial concentration of A is [A] o = 0.840 M, the concentration of B is measured as a function of time: time(s) 0 10 20 30 40 50 60 70 80 [B] (M) 0 0.202 0.330 0.412 0.478 0.526 0.562 0.588 0.612 (a) (8 pts) Determine the order of the reaction. (b) (4 pts) Give the rate law, and find the value of the rate constant.
9 7. (10 pts) The activation energy of the elementary reaction NO 2 (g) + CO(g) ---> NO(g) + CO 2 (g) is 132 kj/mol. At 500 K, when [NO 2 ] = [CO] = 0.250 M, the rate of this reaction is 1.26x10 7 Ms 1. Find the rate constant of the reaction at 600 K.
10 8. (10 pts) In interstellar space, there are ~ 10 gaseous atoms (mostly hydrogen) per cubic centimeter, and T~100 K. (a) (3 pts) Assuming that these atoms obey the ideal gas law, find the pressure of the gas (in atm). (b) (7 pts) Compare a gas-phase reaction in interstellar space to the same one on Earth under standard conditions (1 atm, 0 C). Circle your choice: The reaction in space proceeds (a) more quickly than the one on Earth (b) less quickly than the one on Earth (c) at about the same speed as the one on Earth Explain your answer in terms of the collision theory.
11 9. (10 pts, 2 each) Multiple choice. Circle your answers. (a) Reactions which are thermodynamically spontaneous always occur rapidly. TRUE or FALSE. (b) Pick the FALSE statement(s) about half-lives: 1. For a first-order reaction, successive half-lives are identical. 2. For a second-order reaction, successive half-lives become longer. 3. For a first-order reaction, the half-life depends on the initial concentrations. (c) For the reaction A+B--->C, with rate law: rate = k [A][B], the reaction occurs under pseudo-first-order conditions 1. If [A] = 1/[B], 2. If the initial concentration of B is approximately the same as the final concentration of B. 3. if [A] = [B] (d) Pick the TRUE statement(s) about activation energies. 1. Some elementary reactions have negative activation energies. 2. For experiments done around room temperature, if increasing the temperature of a reaction by 10 C doubles the reaction rate, the activation energy is about 50 kj/mol. 3. The activation energy is given by the difference between the energies of the products and the reactants. (e) Pick the FALSE statement(s) about catalysis. 1. A catalyst increases reaction rate by decreasing the activation energy. 2. A catalyst increases reaction rate by decreasing the energy difference between the products and the reactants. 3. In the catalytic reaction of an enzyme with a substrate, under some conditions, the rate does not depend on the concentration of the substrate. 4. To work effectively, catalysts need to be in the same phase as the reaction they affect.
12 Equations, Constants, etc. rate = k[a] [A] = [A] 0 e kt ln[a] = ln[a] 0 kt t 1/2 = 0.693/k rate = k[a] 2 1 1 = + kt [A] [A] 0 t 1/2 1 = k[a] 0 k = e Ea/RT lnk = lna E a RT rate = k 2[E] 0 [S] k -1 + k where K 2 M = [S] + K M k 1 R = 8.314 Jmole 1 K 1 = 0.0821 Latmmole 1 K 1 T( C) + 273 = T(K) pv = nrt N 0 = 6.02x10 23 mole 1 1000 cm 3 = 1L