CHEMICAL KINETICS PRACTICE EXAMPLES CHAPTER 14

CHAPTER 14 CHEMICAL KINETICS PRACTICE EXAMPLES 1A (E) The rate of consumption for a reactant is expressed as the negative of the change in molarity divided by the time interval. The rate of reaction is expressed as the rate of consumption of a reactant or production of a product divided by its stoichiometric coefficient.   A    0.3187 M  0.3629 M  1min rate of consumption of A = =  = 8.93  10  5 M s  1  t 8.25 min 60 s rate of reaction = rate of consumption of A2 = 8.93  10  5 M s  1  4.46  10  5 M s  1 2 1B (E) We use the rate of reaction of A to determine the rate of formation of B, noting from the balanced equation that 3 moles of B form (+3 moles B) when 2 moles of A react (–2 moles A). (Recall that “M” means “moles per liter.”) 0.5522 M A  0.5684 M A  3moles B rate of B formation=   1.62  10  4 M s  1 60s  2 moles A 2.50 min  1min 2A (M) (a) The 2400-s tangent line intersects the 1200-s vertical line at 0.75 M and reaches 0 M at 3500 s. The slope of that tangent line is thus 0 M  0.75 M slope = =  3.3  10  4 M s  1 =  instantaneous rate of reaction 3500 s  1200 s The instantaneous rate of reaction = 3.3  10  4 M s  1 . (b) At 2400 s,  H 2 O 2  = 0.39 M. At 2450 s,  H 2 O 2  = 0.39 M + rate   t At 2450 s,  H 2 O 2  = 0.39 M +    3.3  10  4 mol H 2 O 2 L  1 s  1  50s   = 0.39 M  0.017 M = 0.37 M 2B (M) With only the data of Table 14.2 we can use only the reaction rate during the first 400 s,   H 2 O 2  /  t = 15.0  10  4 M s  1 , and the initial concentration,  H 2 O 2  0 = 2.32 M. We calculate the change in  H 2 O 2  and add it to  H 2 O 2  0 to determine  H 2 O 2  100 .   H 2 O 2  = rate of reaction of H 2 O 2   t =  15.0  10  4 M s  1  100 s =  0.15 M  H 2 O 2  100 =  H 2 O 2  0 +   H 2 O 2  = 2.32 M +   0.15 M  = 2.17 M This value differs from the value of 2.15 M determined in text Example 14-2b because the text used the initial rate of reaction  17.1  10  4 M s  1  , which is a bit faster than the average rate over the first 400 seconds. 611Chapter 14: Chemical Kinetics 3A (M) We write the equation for each rate, divide them into each other, and solve for n . R 1 = k     N 2 O 5    1 = 5.45  10  5 M s  1 = k  3.15 M  n n R 2 = k     N 2 O 5    2 = 1.35  10  5 M s  1 = k  0.78 M  n n n k     N 2 O 5    1 k  3.15 M  R 1 5.45  10  5 M s  1 n  3.15  = = 4.04 = = =   =  4.04  n n  5  1   R 2 1.35  10 M s  0.78  k    N 2 O 5   2 k  0.78 M  n n We kept an extra significant figure (4) to emphasize that the value of n = 1. Thus, the reaction is first-order in N 2 O 5 . 3B 1 2  (E) For the reaction, we know that rate = k  HgCl 2    C 2 O 4   . Here we will compare Expt. 4 to Expt. 1 to find the rate. 2 2 1 2  2   0.025 M   0.045 M  rate 4 k   HgCl 2     C 2 O 4   rate 4 = = = 0.0214 = 2 2 1 2    rate 1 k  HgCl   C O  1.8  10  5 M min  1 0.105 M   0.150 M    2    2 4  The desired rate is rate 4 = 0.0214  1.8  10  5 M min  1 = 3.9  10  7 M min  1 . 4A (E) We place the initial concentrations and the initial rates into the rate law and solve for k . 2 2 rate = k  A   B  = 4.78  10  2 M s  1 = k  1.12 M   0.87 M  4.78  10  2 M s  1 k = 4B 5A  1.12 M  = 4.4  10  2 M  2 s  1 0.87 M 2 1 2  (E) We know that rate = k    HgCl 2      C 2 O 4   and k = 7.6  10  3 M  2 min  1 . Thus, insertion of the starting concentrations and the k value into the rate law yields: 1 2 Rate = 7.6  10  3 M  2 min  1  0.050 M   0.025 M  = 2.4  10  7 M min  1 (E) Here we substitute directly into the integrated rate law equation. ln A t =  kt + ln A 0 =  3.02  10  3 s  1  325 s + ln 2.80 =  0.982 +1.030 = 0.048 b g A t = e 5B 2 0.048 = 1.0 M (M) This time we substitute the provided values into text Equation 14.13.  HO   0.443 1.49 M   2 2   t k = ln  =  kt =  k  600 s = ln =  0.443 = 7.38  10  4 s  1   600 s HO 2.32 M   2 2   0 Now we choose    H 2 O 2    0 = 1.49 M,    H 2 O 2    t = 0.62, ln    H 2 O 2    t     2   0 H 2 O =  kt =  k  1200 s = ln 0.62 M =  0.88 1.49 M t = 1800 s  600 s = 1200 s k =  0.88 = 7.3  10  4 s  1  1200 s These two values agree within the limits of the experimental error and thus, the reaction is first- order in [H 2 O 2 ]. 612Chapter 14: Chemical Kinetics 6A (M) We can use the integrated rate equation to find the ratio of the final and initial concentrations. This ratio equals the fraction of the initial concentration that remains at time t .  A  t =  kt =  2.95  10  3 s  1  150 s =  0.443 A   0  A  t = e  0.443 = 0.642; 64.2% of  A  0 remains. A   0 ln 6B (M) After two-thirds of the sample has decomposed, one-third of the sample remains. Thus    H 2 O 2    t =    H 2 O 2    0  3 , and we have   H 2 O 2    t    2   0 ln   H 2 O t = 7A   =  kt = ln  H 2 O 2    0  3    H 2 O  2   0 = ln  1/ 3  =  1.099 =  7.30  10  4 s  1 t  1.099 1 min = 1.51  10 3 s  = 25.1 min  4  1  7.30  10 s 60 s (M) At the end of one half-life the pressure of DTBP will have been halved, to 400 mmHg. At the end of another half-life, at 160 min, the pressure of DTBP will have halved again, to 200 mmHg. Thus, the pressure of DTBP at 125 min will be intermediate between the pressure at 80.0 min (400 mmHg) and that at 160 min (200 mmHg). To obtain an exact answer, first we determine the value of the rate constant from the half-life. k = ln 0.693 0.693 = = 0.00866 min  1 80.0 min t 1/2  P DTBP  t  P DTBP  0 =  kt =  0.00866 min  1  125 min =  1.08  P DTBP  t = e  1.08 = 0.340  P DTBP  0  P DTBP  t = 0.340   P DTBP  0 = 0.340  800 mmHg = 272 mmHg 7B (M) (a) We use partial pressures in place of concentrations in the integrated first-order rate equation. Notice first that more than 30 half-lives have elapsed, and thus the ethylene oxide 30 pressure has declined to at most 0.5 = 9  10  10 of its initial value. b g ln P 30 P 3600 s =  kt =  2.05  10  4 s  1  30.0 h  =  22.1 30 = e  22.1 = 2.4  10  10 P 0 P 0 1 h P 30 = 2.4  10  10  P 0 = 2.4  10  10  782 mmHg = 1.9  10  7 mmHg 613Chapter 14: Chemical Kinetics P ethylene oxide initially 782 mmHg  1.9  10 7 mmHg (~ 0). Essentially all of the ethylene oxide is converted to CH 4 and CO. Since pressure is proportional to moles, the final pressure will be twice the initial pressure (1 mole gas  2 moles gas; 782 mmHg  1564 mmHg). The final pressure will be 1.56  10 3 mmHg. (b) (D) We first begin by looking for a constant rate, indicative of a zero-order reaction. If the rate is constant, the concentration will decrease by the same quantity during the same time period. If we choose a 25-s time period, we note that the concentration decreases  0.88 M  0.74 M =  0.14 M during the first 25 s,  0.74 M  0.62 M =  0.12 M during the second 25 s,  0.62 M  0.52 M =  0.10 M during the third 25 s, and  0.52 M  0.44 M =  0.08 M during the fourth 25-s period. This is hardly a constant rate and we thus conclude that the reaction is not zero-order. We next look for a constant half-life, indicative of a first-order reaction. The initial concentration of 0.88 M decreases to one half of that value, 0.44 M, during the first 100 s, indicating a 100-s half-life. The concentration halves again to 0.22 M in the second 100 s, another 100-s half-life. Finally, we note that the concentration halves also from 0.62 M at 50 s to 0.31 M at 150 s, yet another 100-s half-life. The rate is established as first-order. The rate constant is 0.693 0.693 k = = = 6.93  10  3 s  1 . t 1/2 100 s That the reaction is first-order is made apparent by the fact that the ln[B] vs time plot is a straight line with slope = -k (k = 6.85  10 3 s 1 ). Plot of 1/[B] versus Time Plot of ln([B]) versus Time Plot of [B] versus Time 7 0 -0.2 0.75 -0.4 0.65 -0.6 0 200 6 5 -0.8 0.55 0.45 -1 -1.2 0.35 400 0.85 8A 4 3 -1.4 2 -1.6 0.25 -1.8 0.15 0 100 200 Time(s) 300 1 -2 0 Time(s) 614 100 200 Tim e (s) 300Chapter 14: Chemical Kinetics 8B (D) We plot the data in three ways to determine the order. (1) A plot of [A] vs. time is linear if the reaction is zero-order. (2) A plot of ln [A] vs. time will be linear if the reaction is first-order. (3) A plot of 1/[A] vs. time will be linear if the reaction is second-order. It is obvious from the plots below that the reaction is zero-order. The negative of the slope of the line equals k =   0.083 M  0.250 M   18.00 min = 9.28  10  3 M/min (k = 9.30  10 3 M/min using a graphical approach). Plot of [A] versus Time 0.25 0 10 Plot of ln([A]) versus Time -1.5 0.23 0.21 20 12 Plot of 1/[A] versus Time 11 10 -1.7 0.17 0.15 0.19 -1.9 9 8 7 -2.1 0.13 6 0.11 -2.3 5 0.09 y = -0.00930x + 0.2494 4 0.07 0 9A 10 Time (min) -2.5 20 0 Time (min) 10 Time (min) 20 (M) First we compute the value of the rate constant at 75.0  C with the Arrhenius equation. We know that the activation energy is E a = 1.06  10 5 J/mol, and that k = 3.46  10  5 s  1 at 298 K. The temperature of 75.0 o C = 348.2 K. ln F G H I J K F G H I J K k 2 k 2 E a 1 1 1.06  10 5 J / mol 1 1  = ln = =  = 6.14  5  1  1  1 k 1 3.46  10 s R T 1 T 2 8.3145 J mol K 298.2 K 348.2 K k 2 = 3.46  10  5 s  1  e +6.14 = 3.46  10  5 s  1  4.6  10 2 = 0.016 s  1 t 1/2 = 9B 0.693 0.693 = = 43 s at 75  C k 0.016 s  1 (M) We use the integrated rate equation to determine the rate constant, realizing that one-third remains when two-thirds have decomposed.   N 2 O 5    t    5   0 ln   k = N 2 O   = ln  N 2 O 5    0  3    N 2 O  5   0 1 = ln =  kt =  k  1.50 h  =  1.099 3 1.099 1 h = 2.03  10  4 s  1  1.50 h 3600 s 615Chapter 14: Chemical Kinetics Now use the Arrhenius equation to determine the temperature at which the rate constant is 2.04  10 4 s 1 . F G H F G H I J K k 2 2.04  10  4 s  1 E a 1 1 1.06  10 5 J / mol 1 1  ln = ln = 1.77 = =   5  1  1  1 k 1 3.46  10 s R T 1 T 2 8.3145 J mol K 298 K T 2 1 1 1.77  8.3145 K  1 = = 3.22  10  3 K  1  5 1.06  10 T 2 298 K I J K T 2 = 311 K 10A (M) The two steps of the mechanism must add, in a Hess's law fashion, to produce the overall reaction. Overall reaction: CO + NO 2   CO 2 + NO or CO + NO 2   CO 2 + NO Second step:   NO 3 + CO   NO 2 + CO 2  or +  NO 2 +