Question

A first order reaction occuring at 300 K has an activation energy of 2.484 $kcal/mol$. When the temperature is doubled then the rate of the reaction is increased by n times. Find the value of n. (Take ln 2 = 0.69 and R = 2 $cal/mol.K$)

Answer 1

$ln\frac{k_2}{k_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$
$=\frac{2484}{2}(\frac{1}{300}-\frac{1}{600})$
$=1242\times \frac{1}{600}=2.07$
$=3\times 0.69=3ln2=ln{2}^3$

$\Rightarrow \frac{k_2}{k_1}=8$
$\Rightarrow k_2=8k_1$
On doubling the temperature, k increases and since rate $\propto$ k, n = 8