Question

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is $4 \times 10^{10} s^{-1}$ . Calculate $k$ at 318 K and $E_a$.

Answer 1

For a first order reaction,
$t =\frac { 2.303}{k} log\ \frac { a}{a-x}$
$At \ 298 \ K$
$t =\frac { 2.303}{k} log\ \frac { 100}{90}$
$t = \frac {0.1054}{K}$
$At\ 308 \ k$
$t' =\frac { 2.303}{k'} log\ \frac { 100}{75}$
$t' = \frac {2.2877}{k'}$
According to the question
$t= t'$
$\frac {0.1054}{K} = \frac {2.2877}{k'}$
$\frac {k'}{k }= 2.7296$
From Arrhenius equation, we obtain
$log \ \frac {k_2}{k_1} = \frac {E_a}{2.303\ R} (\frac {T_2-T_1}{T_1T_2})$
Therefore, $log \ (2.7296) = \frac {Ea}{2.303\times8.314} (\frac {308-298}{298\times 308})$
⇒ $E_a = \frac {0.6021\times 2.303\times 8.314\times 298\times 308 \times log (2.7296)}{308-298}$
⇒ $Ea = 76640.096 \ J mol^{-1}$
⇒ $Ea = 76.6\ kJ mol^{-1}$
$To\ calculate \ k\ at \ 318 \ K,$
$It\ is\ given\ that,$
$A = 4\times 10^{10} s^{-1} \ and \ T = 318\ K$
$Again,\ from \ Arrhenius \ equation,\ we \ obtain$
$log \ k = log\ A- \frac {E_a}{2.303\ RT}$
$log\ k = log\ (4\times10^{10}) - \frac {76.64\times 10^3}{2.303\times 8.314×318}$
$log \ k = (0.6021+10)-12.5876$
$log \ k =-1.9855$
$Therefore,$
$k = antilog \ (-1.9855)$
$k = 1.034\times10^{-2} s^{-1}$