Difficult
For reaction $A \to B$,rate constant $K_1 = A_1 e^{-E_{a_1}/RT}$ and for the reaction $X \to Y$,rate constant $K_2 = A_2 e^{-E_{a_2}/RT}$. If $A_1 = 10^8, A_2 = 10^{10}$ and $E_{a_1} = 600 \ cal \ mol^{-1}$,$E_{a_2} = 1800 \ cal \ mol^{-1}$,then the temperature at which $K_1 = K_2$ is (given: $R = 2 \ cal \ K^{-1} \ mol^{-1}$):
- 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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