or $2.303\,\log \,\frac{{{k_2}}}{{{k_1}}}\, = \,\frac{{{E_a}}}{R}\,\left[ {\frac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right]$

$2.303\,\log \,\left[ {\frac{{1.667\, \times \,{{10}^{ - 4}}}}{{1.667 \times \,{{10}^{ - 6}}}}} \right]\, = $ $\, - \frac{{{E_a}}}{R}\left[ {\frac{1}{{1844}} - \frac{1}{{1000}}} \right]$

$2.303\, \times \,2\, = \,\frac{{{E_a}}}{R} \times \,\frac{{844}}{{1844 \times 1000}}$   .... $(1)$

$\therefore \,\frac{{{E_a}}}{R}\, = \,\frac{{4.606\, \times \,1844 \times 1000}}{{844}}$

$2.303\,\log \,\left[ {\frac{{{k_3}}}{{1.667\, \times \,{{10}^{ - 6}}}}} \right]\,$ $ = \,\frac{{{E_a}}}{R} \times \,\frac{{1423 - 1000}}{{1423 \times 1000}}$

$ = \,\frac{{{E_a}}}{R}\, \times \,\frac{{423}}{{1423 \times 1000}}$        .... $(2)$

Dividing equation $(2)$ by equation $(1)$

$\frac{{\log \left[ {\frac{{{k_3}}}{{1.667 \times {{10}^{ - 6}}}}} \right]}}{2}$

$ = \,\frac{{423}}{{1423 \times 1000}}\, \times \,\frac{{1844 \times 1000}}{{844}}$

$\therefore \,\,\log \left[ {\frac{{{k_3}}}{{1.667 \times {{10}^{ - 6}}}}} \right]$

$ = \,2\, \times \,\frac{{423 \times 1844}}{{1423 \times 844}}\, = \,1.299$

On taking antilog , $k_3=19.9$

${k_3}\, = \,19.9\, \times 1.667 \times \,{10^{ - 6}}\, = \,3.318\, \times \,{10^{ - 5}}\,{s^{ - 1}}$