Chemistry Daily Challenge 24-July-2015

The $k$ -rate of a reaction is given by: $\begin{aligned} k & = Ae^{-\frac{E_a}{RT}} \end{aligned}$

It is given that:

$\begin{aligned} \log{k} & = -\frac{2000}{T} + 6 \\ \Rightarrow k & = 10^{ -\frac{2000}{T} + 6 } \\ & = e^{ \ln{(10)}(6-\frac{2000}{T})} \\ & = e^{ 6\ln{(10)}}e^{-\frac{2000\ln{(10)}}{T}} \end{aligned}$

$\begin{aligned} \Rightarrow A & = e^{ 6\ln{(10)}} = \boxed{1.0 \times 10^{6}\space s^{-1}} \\ \Rightarrow \dfrac{E_a}{R} & = 2000 \ln{(10)} \\ \Rightarrow E_a & = \dfrac{2000\ln{(10)}R}{1000} = 2\ln{(10)}R \\ & = 2\ln{(10)}\times 8.31446 = \boxed{38.3 \space kJ\space mol^{-1}} \end{aligned}$

The Arrhenius relation between activation energy, temperature and rate constant is given by:

$K = Ae^{-\frac{E_a}{RT}}$

Taking $\log$ on both sides, (not $\ln$ )

$\log K = \log A - \dfrac{E_a}{RT} \log_{10}e$

$\log K = -\dfrac{E_a}{RT} \dfrac{1}{2.303} + \log A$

Comparing terms in the equation given in the question,

$E_a = 2000 \times 2.303 \times 8.314 J mol^{-1} \approx 38.3 KJ mol^{-1}$

$A = 10^6 s^{-1}$

---

Note:

$A$ is pre-exponential factor.

$E_a$ is activation energy.