In: Chemistry
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.
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 |
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
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