For a reaction, consider the plot of lnk versus 1 / T given in the...

For a reaction, consider the plot of $lnk$ 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:

Options

A ${10}^{- 4}{s}^{- 1}$
B ${10}^{- 6}{s}^{- 1}$
C $2 x {10}^{- 4}{s}^{- 1}$
D $4 x {10}^{- 4}{s}^{- 1}$

Correct answer

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

Step-by-step solution

lnk = lnA - \dfrac{Ea}{RT} = lnA - 4606\left(\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}}\right)\left[∴ \dfrac{{E}_{A}}{R} = 4606\right]

\ln \left(\dfrac{k}{{10}^{- 5}}\right) = 4606 \times \left(\dfrac{500 - 400}{500 \times 400}\right)

\ln \left(\dfrac{k}{{10}^{- 5}}\right) = 4606 \times \dfrac{1}{2000} = 2.303

\ln \left(\dfrac{k}{{10}^{- 5}}\right) = \ln 10

k = {10}^{- 4}{s}^{- 1}

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