Difficult
For reaction $A \to B$,the rate constant $k_1 = A_1 e^{-E_{a_1} / (RT)}$ and for the reaction $X \to Y$ the rate constant $k_2 = A_2 e^{-E_{a_2} / (RT)}$. If $A_1 = 10^8$,$A_2 = 10^{10}$ and $E_{a_1} = 600 \ cal/mol$,$E_{a_2} = 1800 \ cal/mol$,then the temperature at which $k_1 = k_2$ is (Given : $R = 2 \ cal/K \cdot mol$)
- A$1200 \ K$
- B$1200 \times 4.606 \ K$
- C$\frac{1200}{4.606} \ K$
- D$\frac{600}{4.606} \ K$
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