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The correct expression for Arrhenius equation showing the effect of temperature on rate constant is $[T_{2} > T_{1}]$

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$\log_{10} (\frac{K_{2}}{K_{1}}) = \frac{R}{2.303 E_{a}} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

$\log_{10} (\frac{K_{2}}{K_{1}}) = \frac{E_{a}}{R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

$\log_{10} (\frac{K_{2}}{K_{1}}) = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

$\log_{10} (\frac{K_{2}}{K_{1}}) = \frac{E_{a}}{2.303 R} [\frac{T_{1} T_{2}}{T_{2} - T_{1}}]$

answer is D.

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Detailed Solution

For a reaction with the effect of temperature on rate constant at two different temperature $T_{1} and T_{2}$ Arrhenius equation is as below.

$\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

The modified Arrhenius equation is

$\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

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