$A$ plot of $ \frac{1}{T} $ vs. $ \ln k $ for a reaction gives the slope $ -1 \times 10^{4} \ K $. The energy of activation for the reaction is (Given $ R = 8.314 \ J \ K^{-1} \ mol^{-1} $)
- A$ 1.202 \ kJ \ mol^{-1} $
- B$ 83.14 \ kJ \ mol^{-1} $
- C$ 8314 \ J \ mol^{-1} $
- D$ 12.02 \ J \ mol^{-1} $
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