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For the given reaction: `H_(2) + I_(2) rarr 2HI` Given: `{:(T(K),1//T(K^(-1)),log k,),(769,1.3 xx 10^(-3),2.9,),(67,1.5 xx 10^(-3),1.1,):}` The activation energy will beA. `41.4 kcal mol^(-1)`B. `40 kcal mol^(-1)`C. `-41.4 kcal mol^(-1)`D. `-40 kcal mol^(-1)` |
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Answer» Correct Answer - B `k = Ae^(-E_(A)//RT)` `rArr log k = log A - (E_(a))/(2.303 RT)` `{:(log k_(1) = log A-(E_(a))/(2.303RT) ((1)/(T_(1)))),(log k_(2)=logA-(E_(a))/(2.303RT) ((1)/(T_(2)))):}} "Subtract to get" E_(a) = 40 lcal mol^(-1)` |
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