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:

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$10^{- 6} s^{- 1}$

$4 \times 10^{- 4} s^{- 2}$

$10^{- 4} s^{- 1}$

$2 \times 10^{- 4} s^{- 1}$

answer is C.

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Detailed Solution

From Arrhenius equation
In K $= l n A - \frac{E a}{R T}$
Slope = $- \frac{E a}{R T} = - 4606 K$
$\log [\frac{k_{2}}{k_{1}}] = \frac{E_{a}}{2.303 R} [\frac{1}{T} - \frac{1}{T_{2}}]$

$\log [\frac{k_{2}}{10^{- 5}}] = \frac{1}{2.303} \times 4606 [\frac{1}{400} - \frac{1}{500}]$
$= \frac{4606 \times 100}{2.303 \times 400 \times 500} = 1$
$\Rightarrow \log [\frac{k_{2}}{10^{- 5}}] = 1; \frac{k_{2}}{10^{- 5}} = A n t i \log (1)$

$\frac{k_{2}}{10^{- 5}} = 10$ $k_{2} = 10^{- 5} \times 10 = 10^{- 4} s^{- 1}$