Question

For a first order reaction $A \to P,$ the temperature (T) dependent rate constant (k) was found to followthe equation log k=- (2000) $\frac{1}{T}+6.0.$ The pre-exponential factor A and the activation energy $E_{a,}$ respectively, are

• A. $1.0 \times10^{6} S^{-1} \, and \, 9.2 KJ mol^{-1}$
• B. $6.0^{S-1} and 16.6 KJ mol^{-1}$
• C. $1.0\times10^{6}S^{-1} and 16.6 KJ mol^{-1}$
• D. $1.0\times10^{6}S^{-1} and 38.3 KJ mol^{-1}$

Answer 1

Answer: (D)

$1.0\times10^{6}S^{-1} and 38.3 KJ mol^{-1}$

Answer 2

log k = log A $-\frac{E_{a}}{2.303 R T} \quad\ldots\left(1\right)$
Also given log k = 6.0 - (2000)-$\frac{1}{T}\quad\ldots\left(2\right)$
On comparing equations, (1) and (2)
$log A=6.0 \Rightarrow \quad A=10^{6}S^{-1}$
$and \, \frac{E_{a}}{2.303R}=2000 ;$
$\Rightarrow\quad E_{a}=2000\times2.303\times8.314$
$=\quad38.29 KJ mol^{-1}$