Question

The following data was collected in a laboratory experiment conducted at 305.2 K.

A.) Determine the complete rate law for this simple chemical process A → B

Time	Concentration
0	4.804
0.5	3.26799191
1	2.279801588
1.5	1.573137404
2	1.047454686
2.5	0.713650161
3	0.52422989
3.5	0.366551798
4	0.218181335
4.5	0.183699478
5	0.132508896
5.5	0.097326158
6	0.033145439
6.5	0.036526291
7	0.005104128
7.5	0.017253798
8	0.01185835
8.5	0.008150117
9	0.025601488
9.5	0.016150157
10	0.002645956
10.5	0.018181463
11	0.018750139
11.5	0.020859016
12	0.020590393
12.5	0.01959423
13	0.019721118
13.5	0.000191672
14	0.000131734
14.5	0.01990946
15	0.019937773

B.) For the set of kinetic data in the problem above, if the rate doubles when the temperature increases to 415.4 K, what is the activation energy for the process?

Answer 1

a)

This appears to be first order, since t vs- ln(A) is a straight line

Rate = k*[A]

k = 0.4687

b)

Apply Ahrrenius equation

ln(k2/K1) = E/R*(1/T1- 1/T2)

ln(2k1 / k1) = E/8.314*(1/(305.2 ) - 1/415.4))

ln(2) = E * 0.0001045491

E = ln(2) /0.0001045491

E = 6629.8722 J/mol