Question

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If the uncatalyzed reaction takes about 2400 years to occur, about how long will the catalyzed reaction take? Assume the frequency factor A is constant, and assume the initial concentrations are the same.

Short answer

The time required for the catalyzed reaction is 3.1* 10^-8 years.

Step-by-step solution

Describing the Ratio of Rate Constants of the Uncatalyzed to a Catalyzed Reaction

Given data -

Activation energy on catalyzed surface $\left(E_1\right)$= 59 kJ/mol

Activation energy on without catalyzed surface $\left(E_2\right)$= 184 kJ/mol

Temperature$\left(T\right)$= 600K

Gas constant, = 8.314 J/mol K

Time is taken by the uncatalyzed reaction $\left(t_2\right)$= 2400 years

The pre-exponential factor $\left(A\right)$is the same for each reaction

Here we need to find the time$\left(t_1\right)$

The Arrhenius equation is given as

$k=e^{-E/RT}$

On catalyst surface with rate constant,

$k_1=e^{-E_1/RT}..........\left(1\right)$

Without catalyst surface with rate constant,

$k_1=e^{-E_2/RT}..........\left(2\right)$

Dividing (1) by (2), we get

$$\begin{gathered} \frac{k_1}{k_2}=e^{\left(E_2-E_1\right)/RT} \\ \frac{k_1}{k_2}=e^{\left(\frac{1}{4988.4}\left(125\right)\left({10}^3\right)\right)} \\ \frac{k_1}{k_2}=e^{\left(25.058\right)} \\ \frac{k_1}{k_2}=7.6\times {10}^{10} \end{gathered}$$

Describing the time needed for the catalyzed reaction

The reaction proceeds very fast than the uncatalyzed reaction. The time required for the catalyzed reaction is

$\frac{2400years}{7.6\times {10}^{10}}=3.15\times {10}^{-8}years$

The time required for the catalyzed reaction to take place is 3.1*10^-8 years.