A reaction can take place by two paths. $k_{1}$ and $k_{2}$ are rate constants for the two paths & $E_{1}$ and $E_{2}$ are their respective activation energies.
At temperature $T_{a} : k_{1} > k_{2}, E_{1} < E_{2}$
Its temperature is raised to $T_{b}$ , the rate constants change to $k_{1}^{'}$ & $k_{2}^{'}$ . Which relation is correct between $k_{1}, k_{2}, k_{1}^{'}$ & $k_{2}^{'}$ (considering activation energy does not change with temperature)?

\frac{k_{1}^{'}}{k_{1}} > \frac{k_{2}^{'}}{k_{2}}

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\frac{k_{1}^{'}}{k_{1}} = \frac{k_{2}^{'}}{k_{2}}

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\frac{k_{1}^{'}}{k_{1}} < \frac{k_{2}^{'}}{k_{2}}

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\frac{k_{1}^{'}}{k_{2}^{'}} > \frac{k_{1}}{k_{2}}

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Solution

The correct option is B $\frac{k_{1}^{'}}{k_{1}} < \frac{k_{2}^{'}}{k_{2}}$
$k_{1} = A e^{\frac{- E}{R T_{a}}} k_{2} = A e^{\frac{- E_{2}}{R T_{a}}}$

$k_{1}^{'} = A e^{\frac{- E_{1}}{R T_{b}}} k_{2}^{'} = A e^{\frac{- E_{2}}{R T_{b}}}$

$\frac{k_{1}}{k_{1}^{'}} = \frac{A e^{\frac{- E_{1}}{R T_{b}}}}{A e^{\frac{- E_{1}}{R T_{b}}}}$

$\frac{k_{1}}{k_{1}^{'}} = e^{\frac{- E_{1}}{R} ⎛ ⎝ \frac{1}{T_{a}} - \frac{1}{T_{b}} ⎞ ⎠}$

$\Rightarrow \frac{j_{1}^{'}}{k_{1}} = e^{\frac{E_{1}}{R} ⎛ ⎝ \frac{1}{T_{a}} - \frac{1}{T_{a}} ⎞ ⎠}$

Similarly $\frac{k_{2}^{'}}{k_{2}} = e^{\frac{E_{2}}{R} ⎛ ⎝ \frac{1}{T_{a}} - \frac{1}{T_{b}} ⎞ ⎠}$

$E_{1} < E_{2}$

$\Rightarrow e^{\frac{E_{1}}{R} ⎛ ⎝ \frac{1}{T_{a}} - \frac{1}{T_{b}} ⎞ ⎠} < e^{\frac{E_{2}}{R} ⎛ ⎝ \frac{1}{T_{a}} - \frac{1}{T_{b}} ⎞ ⎠}$

$\Rightarrow \frac{k_{1}^{'}}{k_{1}} < \frac{k_{2}^{'}}{k_{2}}$

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