2. The rate kinetics for the following reaction at 298 K are shown below. Answer

Question

2. The rate kinetics for the following reaction at 298 K are shown below. Answer the following questions regarding the reaction: 3A (g) + 2B (g) C (g) Alo (M 1.03 3.91 5.06 3.92 8.32 [Blo (M 0.456 0.247 0.458 1.51 0.661 Initial Rate (M/s 0.0153 0.0171 0.792 0.640 0.260 Experiment 3 4 a. What is the overall rate law for the reaction? Include the rate constant. (3) b. What is the overall order of the reaction? (3) c. If the frequency factor is 1.56x1015 M2 s1, what is the rate constant be 100 °C? (3) d. How much faster/slower would we expect the reaction to be at 250 °C than at 25 °C? Assume the concentrations of A and B are the same at both 25 °C and 250 °C. (3)

Explanation / Answer

Let the rate be rate= k CAa CBb

Consider experiments 2 and 4. While CA is almost the same in both, dividing the rates in both, we get

r4/r2 = (CB4/CB2)b

Substituting values, we get

0.640/0.0171 = (1.51/0.247)b'

Taking logarithm on both sides, we get b=2

Now, substitute this value of b for rates in any of the experiments. Consider experiments 1 and 2

r2/r1 = (CA2/CA1)a (CB2/CB1)b

Substituting values, we get

0.0171/0.0153 = (3.91/1.03)a (0.247/0.456)2 --- we get a=1

Now, for eqn 1, to find rate constant,

0.0153 M/s = k x 1.03 M x (0.456 M)2

k = 0.07144 M-2 s-1

Thus, rate= (0.07144 M-2 s-1) CACB2 --- a   -- valid for the given temperature of 298 K

Overall order of reaction = a+ b = 3 ---- b

We have k= ko exp(-c/T), where c is a constant

Substituting ko = 1.56 x 1015 M-2s-1 and T=298 K and k = 0.07144 M-2 s-1

0.07144 = 1.56 x 1015 x exp(-c/298)

c= 11211.463 K

Now, for 100 C= 373 K, getting rate constant by substituing constants,

k = 137.816 M-2 s-1    ---- c

Now, as done in part c, calculating rate constant for T= 250 C = 523 K, we get

k = 4961452.815 M-2 s-1

Since reaction rate is directly proportional to rate constant, the reaction will be 6.94 x 107 times faster at 250 C

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