1. Consider the unbalanced chemical reaction. PH 3 à P 4 + H 2

Transcription

1 1. Consider the unbalanced chemical reaction PH 3 à P 4 + H 2 If mol of phosphorous trihydride is consumed in a 2.0 L container each second of the reaction, what are the rates of the production of the two products? 2.The reaction 2NO (g) + Cl 2 (g) à 2NOCl (g) was studied at -10 C. The following results were obtained with the following initial concentrations [NO] [Cl 2 ] Initial rate (M/s) a. What is the rate law? b. What is the rate constant? (include units) 3. The following data were obtained for the reaction 2ClO 2 (aq) + 2OH - (aq) à ClO 3 - (aq) + ClO 2 - (aq) + H 2 O (l) [ClO 2 ] [OH - ] Initial Rate (M/s) a. Determine the rate law and the value of the rate constant. b. What would be the initial rate for an experiment with [ClO 2 ] = M and [OH - ] = M?

2 4. The rate of the reaction NO 2 (g) + CO (g) à NO (g) + CO 2 (g) depends only on the concentration of nitrogen dioxide below 225 C. At a temperature below 225 C, the following data were collected: Time(s) [NO 2 ] x x x x x a. Determine the rate law and the integrated rate law for the reaction. b. Determine the value of k and its units. c. Calculate the concentration of nitrogen dioxide at time 2.7 x10 4 seconds after the start of the reaction. 5. Nitric oxide (NO) reacts with oxygen to form nitrogen dioxide: 2 NO(g) + O 2 (g) 2 NO 2 (g) a) The following two step mechanism has been proposed for this reaction: NO + O 2 NO 3 (fast) NO 3 + NO 2 NO 2 (slow) Which step is the rate determining step? b) Write a rate equation for the rate determining step of this reaction, assuming it occurs as a single step that depends only on the collision between reactants. c) What happens to the rate if the concentration of NO 3 is halved and the concentration of NO is tripled? d) What is the order of the rate determining step with respect to NO? What is the overall order of this step? 6. Draw an energy diagram for a reaction where ΔH = -40 kj, the activation energy of the uncatalyzed reaction is +120 kj, and the activation energy for a catalyzed reaction is + 80 kj. Indicate the position of the activated complex for both catalyzed and uncatalyzed reactions. 7. List three factors that affect reaction rate and briefly explain the basis for their effects. 8. The breakdown of nitrous oxide gas (N 2 O) to nitrogen gas and oxygen gas is believed to occur in two steps. In the first step, nitrous oxide breaks down to form nitrogen gas and a free oxygen atom. a) Write a balanced equation for the overall reaction (do not use fractional coefficients). b) Write balanced equations for each of the two steps. c) Which substance could be considered a "reaction intermediate?"

3 9. The rate of the reaction, HgCl 2 (aq) + ½ C 2 O 2-4 (aq) Cl - (aq) + CO 2 (g) + ½ Hg 2 Cl 2 (s), is followed by measuring the number of moles of Hg 2 Cl 2 that precipitate per liter per second. The following data are obtained: [HgCl 2 ] [C 2 O 2-4 ] Initial Rate (mol/l s) x x x x 10-7 a) What is the order of the reaction with respect to HgCl 2, with respect to C 2 O 2-4, and overall? b) Write the rate equation for the reaction. c) Calculate k for the reaction. d) When the concentrations of both mercury(ii) chloride and oxalate ion are 0.30 M, what is the rate of the reaction? 10. For the reaction given below, what is the instantaneous rate for each of the reactants and products? 3 A + 2 B 4 C 11. Given the following experimental data, find the rate law and the rate constant for the reaction: NO (g) + NO 2 (g) + O 2 (g) N 2 O 5 (g) Run [NO] o, M [NO 2 ] o, M [O 2 ] o, M Initial Rate, Ms M 0.10 M 0.10 M 2.1 x M 0.10 M 0.10 M 4.2 x M 0.30 M 0.20 M 1.26 x M 0.10 M 0.20 M 2.1 x The half-life of a radioisotope is found to be 4.55 minutes. If the decay follows first order kinetics, what percentage of isotope will remain after 2.00 hours?

4 13. The mechanism of a reaction is shown below. a) What is the overall reaction? b) Which compounds are intermediates? c) Predict the rate law based on this mechanism. d) What is the overall order of the reaction? HOOH + I HOI + OH (slow) HOI + I I 2 + OH (fast) 2OH + 2H 3 O + 4 H 2 O (fast) 14. For the reaction A + B C, the rate constant at 215 o C is 5.0 x 10-3 and the rate constant at 452 o C is 1.2 x a) What is the activation energy in kj/mol? b) What is the rate constant at 100 o C. Answers: mol/2.0 L = M 4PH 3 à P 4 + 6H 2 1- Δ[PH 3 ] = + Δ[P 4 ] = +6[H 2 ] 2 Δt Δt Δt Δ[P 4 ] = M/4 = M/s Δt Δ [H 2 ] = 6 ( M/s)/4 = M/s 2. Expt 1, = 2 = [NO] x [0.20] y = 2y y= [NO] x [0.10] y Expt 2, = 4 [0.20] x [Cl] y = 2 x x = [0.10] x [Cl] y rate = k [NO] 2 [Cl] Expt 1 k = [(0.18M/s)/(0.1M) 2 (.1M)] = 180 M -2 s -1 rate = 180 M -2 s -1 [NO] 2 [Cl]

5 3. Expt 1,2 4 = = [OH - ] y (0.100) x = 2 x x = [OH - ] y (0.0500) x Expt 2,3 2 = = [ClO 2 ] x [0.100] y = 2 y y = [ClO 2 ] x [0.0500] y rate = k[clo 2 ] 2 [OH-] Expt 1 k = [(0.0575M/s) / ( M) 2 (0.100 M)] = 230 M -2 s -1 rate = 230 M -2 s -1 [ClO 2 ] 2 [OH-] b. rate = 230 M -2 s -1 (0.175 M) 2 ( M) = M/s 4. Note: One cannot solve this problem by comparing the different experiment and determining the order by the change in rate, since that only works with initial rates and not instantaneous rates, which are given. A graphical method must be used. t [NO 2 ], M ln [NO 2 ] 1/[NO 2 ] x x x x x Note that: For a zero order reaction, as shown in the following figure, the plot of [A] versus time is a straight line with k = - slope of the line. Other graphs are curved for a zero order reaction.

6 For a first order reaction, as shown in the following figure, the plot of the logrithm of [A] versus time is a straight line with k = - slope of the line. Other graphs are curved for a first order reaction. For a second order reaction, as shown in the following figure, the plot of 1/[A] versus time is a straight line with k = slope of the line. Other graphs are curved for a second order reaction.

7 So this reaction is second order, based on the graphs. rate = k[no 2 ] 2 Slope of line is positive k k = = 2.08 x10-4 M -1 s x10 4 s 0s 1/[NO 2 ] t = + (2.08 x10-4 M -1 s -1 )(27,000s) + 1/0.500 = = 7.61 = M 5. a. Rate determining step (RDS) = step 2 (the slow step) b. If only considered a single step, than rate = k 2 [NO 3 ][NO] HOWEVER, nitrogen trioxide is an intermediate and should not be included in the rate law Therefore, rate = k[no][no][o 2 ] is a better answer c. If nitrogen trioxide concentration is halved, there would be no effect overall from this change as it does not appear in the rate law. If the [NO] is tripled, the rate of reaction would triple. d. The RDS alone is 2 nd order, with NO being first order in that step. Note that the overall reaction order is 3 rd Usually, inceasing the concentration of a reactant will increase the rate of a reaction (at least, the initial rate). A greater density of reactant particles allows for more collisions per second, and henceforth more reactions/s. However, this is not always the case as reactants can be 0 th order.

8 Generally, increasing the temperature of a reaction increases the value of the rate constant, thereby speeding up a reaction. Since reactant particles have more average KE, they will collide with each other more often and with more energy, thereby initiating more reactions/s. The greater the surface area of a solid, the more reactant particles that are exposed and can undergo a reaction. The larger the chunk of a solid, the less surface area it has, thereby decreasing how quickly it can react. 8. N 2 O à N 2 + O 2O à O 2 2N 2 O à 2N 2 + O 2 Clearly, oxygen is the intermediate (since it is neither at the beginning nor end of the reaction) 9. Expt 3,2 1x10-6 = 2 = (0.20) x [C 2 O 2-4 ] y x = 1 5.2x10-7 (0.10) x [C 2 O 2-4 ] y Expt 2,1 5.2x10-7 = 4 = [HgCl 2 ] x (0.20) y y = 2 1.3x10-7 [HgCl 2 ] x (0.10) y rate = k[hgcl 2 ] 2 [C 2 O 4 2- ] Expt 1 k = [1.3x10-7 M/s / (0.l0 M)(0.10 M) 2 ] = 1.3x10-4 M -2 s -1 rate = (1.3x10-4 M -2 s -1 ) (0.30 M)(0.30 M) 2 = 3.5 x10-6 M/s Δ[A] - 3 Δt = 1-2 Δ[B] Δt = 1 4 Δ[C] Δt 11. Rate = k[no] [NO 2 ] k = 2.1 M -1 s k = min -1 A t = 1.15 x 10-6 % (not much!!) 13. a) Overall reaction: HOOH + 2 I + 2 H 3 O + I H 2 O b) Intermediates: OH and HOI c) Predicted mechanism: Rate = k [HOOH][I ] d) Overall order: 2 nd order 14. a) 39.4 kj/mol b) 2.50 x 10-4 s -1