Question

The rate constant for a first order reaction is given by the following equation :
$ln\ k=33.24−\frac {2.0×10^4\ K}{T}$
The activation energy for the reaction is given by ______ $kJ mol^{–1}$. (In nearest integer)
(Given : $R = 8.3 JK^{–1} mol^{–1}$)

Answer 1

Given that,
$ln\ k=33.24−\frac {2.0×10^4\ K}{T}$
So, $\frac {E_a}{R} = 2 \times 10^4$
The activation energy for the reaction,
$E_a = 2 × 10^4 × 8.3$
$E_a = 166 × 10^3\ J/mol$
$E_a= 166\ kJ/mol$

So, the answer is $166\ kJ/mol$.