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A graph plotted between `log k` versus `1//T` for calculating activation energy is shown byA. B. C. D. |
Answer» Correct Answer - B According to Arrhenius equation, `k prop T` and `k = Ae^(-Ea//RT)` `log k = log A - (E_(a))/(2.303 xx R) xx(1)/(T)` (`y = c + mx`, equation of straight line) The value of `log k` should increase uniformly with `T` or decrease with `1//T`. Hence, the answer is (b). |
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