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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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$4 \times 10^{- 4} S^{- 1}$

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

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

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

answer is C.

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Detailed Solution

From Arrhenius equation

lnK=lnA-Eq/RT

Slope = -Eq/R = -4606K

$\log (\frac{k_{2}}{k_{1}}) = \frac{E q}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$\log (\frac{k_{2}}{10^{- 5}}) = \frac{4606}{2.303 R} [\frac{1}{400} - \frac{1}{500}]$

$\log (\frac{k_{2}}{10^{- 5}}) = 1 (\frac{k_{2}}{10^{- 5}}) = 10$

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

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