# The rate constant for the reaction 2 C (g) -> D (g) is 0.231 s^-1 at 2°C and 11.6 s^-1 at 77°C. What is the the value of A in the Arrhenius equation?

May 17, 2018

Consider,

$2 C \left(g\right) r i g h t \le f t h a r p \infty n s D \left(g\right)$, where

${k}_{1} = 0.231 {\text{M"^-1"s}}^{-} 1$ at ${T}_{1} = 275.15 \text{K}$, and

${k}_{2} = 11.6 {\text{M"^-1"s}}^{-} 1$ at ${T}_{2} = 350.15 \text{K}$

Recall the Arrhenius equation,

k = Ae^((-E_"a")/(RT))" " (1)

$\implies \ln k = \ln A - {E}_{\text{a}} / \left(R T\right)$

If we make two equations with each set of parameters and subtract them, we arrive at,

$\ln \left({k}_{2} / {k}_{1}\right) = - {E}_{\text{a"/R(1/T_2 - 1/T_1) " }} \left(2\right)$

Let's derive the activation energy of the reaction from your data,

$\implies {E}_{\text{a" = (R * ln(k_1/k_2))/(1/T_2 - 1/T_1) approx 41.8"kJ}}$

With all of this data, we can calculate the frequency factor using $\left(1\right)$ and either set of conditions. Hence,

=> A = k/(e^((-E_"a")/(RT))) approx 2.00*10^7"M"^-1"s"^-1