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For the decomposition of `HI` the following logarithmic plot is shown `: [R=1.98 cal// mol - K]` The activation energy of the reaction is aboutA. `45600 cal`B. `13500 cal`C. `24600 cal`D. `32300 cal` |
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Answer» Correct Answer - 1 `logk=-(E_(a))/(2.303R)(1)/(T)+"constant"=-(E_(a))/(2.303R)xx10^(-3)xx(10^(3))/(T)+"constant"` Thus slope of graph will be `-(E_(a)xx10^(-3))/(2.303R)= -(4)/(0.4)` `implies" "E_(a)=2.303xx1.98xx10^(4)=45600cal` |
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