Two reactions $R_{1}$ and $R_{2}$ have identical pre-exponential factors. Activation energy of $R_{1}$ exceeds that of $R_{2}$ by 10 kJ mol^-1. If $K_{1}$ and $K_{2}$ are rate constants for reactions R₁ and R₂ respectively at 300 K, then $l$ n ( $K_{2} / K_{1}$ ) is equal to (R = 8.314 J mol¹ K^-1)

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answer is B.

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Detailed Solution

According to the arrhenius equation,
$k = A e^{- E_{a} / R T}$
For reaction $R_{1}; k_{1} = A e^{- E_{a_{1}} / R T}$
In $k_{1}$ = In $A - \frac{E_{a_{1}}}{R T}$
For reaction $R_{2}; k_{2} = A e^{- E_{a_{2}} / R T}$
In $k_{2}$ = In $A - \frac{E_{a_{1}}}{R T}$
Now , in $(\frac{k_{2}}{k_{1}}) = \frac{E_{a_{1}}}{R T} - \frac{E_{a_{2}}}{R T} = \frac{E_{a_{1}} - E_{a_{2}} =}{R T} \frac{Δ E_{a}}{R T} = \frac{10 \times 10^{3}}{8.314 \times 300} = 4$