A reaction was run and found to have a half life that got shorter as the initial

Question

A reaction was run and found to have a half life that got shorter as the initial concentration of the reactant ([A]) was increased. When the reaction was run at two different temperatures, the following data sets were generated.

Use the data given to determine the value of k (the rate constant) at each temperature, and then determine the Energy of Activation for this reaction.

k at 530 K = ??? 1/Msec

k at 940 K = ??? 1/Msec

Ea = ??? kJ/mol

At 530 K At 940 K [A] Time (sec.) [A] Time (sec.) 0.285 0.00 0.285 0.00 0.0596 255.0 0.000551 14.0 0.0295 585.0 0.000221 35.0 0.0170 1065.0 0.000164 47.0 0.00637 2950 5.72e-05 135.0

Explanation / Answer

First, we need to determine the order of reaction. Let's plot the data for a first order and second order reaction, The one that got the r2 value closest to 1 will be the one that determines the order of reaction:

1° order: Equation to use: lnA = lnAo - kt; plot lnA vs t

At 530 K:
y = -2.33 - 0.00105x; r2 = 0.7455

It's not a 1° order, let's do it for a second order reaction; 1/A = 1/Ao + kt

At 530 K:
y = 3.4894 + 0.05203x; r2 = 0.99999 --> it's a second order reaction.

At 940 K:
y = 0.92597 + 129.503x; r2 = 0.9999

The value of k, is given by the slope so:
k1 = 0.05203; k2 = 129.503

Now, to calculate the Ea of the reaction, we use the following expression.
k = A exp(-Ea/RT)

So we have two values of T and two values of k so:
k1 = A exp(-Ea/RT1); k2 = A exp(-Ea/RT2)

dividing both, we have:
k1/k2 = exp(-Ea/RT1 + Ea/RT2)
ln(k1/k2) = Ea/R(1/T2 - 1/T1)
R*ln(k1/k2) / (1/T2-1/T1) = Ea

Ea = 8.3144 * ln(0.05203/129.503) / (1/940 - 1/530)
Ea = -65.0156 / -8.2296x10-4
Ea = 790001.88 J/mol or 79 kJ/mol

Hope this helps

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