For a reaction with an activation energy of 49.8 kJ mol−1, the ratio – InterviewSolution

1.	For a reaction with an activation energy of 49.8 kJ mol−1, the ratio of the rate constant at 600 K and 300 K, K600K300, is approximately: (R=8.3 J mol−1 K−1) (correct answer +2, wrong answer -0.5)
Answer» For a reaction with an activation energy of $49.8 {kJ mol}^{- 1}$ , the ratio of the rate constant at $600 K$ and $300 K$ , $\frac{K_{600}}{K_{300}}$ , is approximately: $(R = 8.3 {J mol}^{- 1} K^{- 1})$ (correct answer +2, wrong answer -0.5)

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