For a reaction, consider the plot of  ln K  versus  1/T  given in the figure if the rate constant of this reaction at 400K is 10^-5 S^-1, then the rate constant at 500K is:
        
 

 From Arrhenius equation
 In K $=lnA-\frac{Ea}{RT}$
Slope = $-\frac{Ea}{RT}=-4606K$
$\log \left[\frac{k_2}{k_1}\right]=\frac{E_a}{2.303R}[\frac{1}{T}-\frac{1}{T_2}]$

$\log \left[\frac{k_2}{{10}^{-5}}\right]=\frac{1}{2.303}\times 4606[\frac{1}{400}-\frac{1}{500}]$
$=\frac{4606\times 100}{2.303\times 400\times 500}=1$
$\Rightarrow \log \left[\frac{k_2}{{10}^{-5}}\right]=1;\frac{k_2}{{10}^{-5}}=Anti\log \left(1\right)$

$\frac{k_2}{{10}^{-5}}=10$    $k_2={10}^{-5}\times 10={10}^{-4}{s}^{-1}$