Question

The rate constant of a first-order reaction is 3.75
x 10-4
s
-1 at 300°C. If the activation energy is 101
kJ/mol, calculate the temperature at which its rate
constant is 7.50 x 10-4
s
-1
.

Answer 1

$$k = Ae^{ { -E_a \over RT}}$$

$$ln({k_2 \over k_1}) = {E_a(T_2-T_1) \over RT_2T_1}$$

$$ln({7.50*10^{-4} \over 3.75*10^{-4}}) = {101 000(x-573) \over 8.314*573x}$$

$$0.69 = {101000x - 57873000 \over 4764x}$$

$$3287.16x = 101000x - 57873000$$

$$57873000 = 97712.84x$$

$$x = 592.27 K = 592K$$

$$T = 592K = 319^0C$$