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What is the energy of activation of a reaction is its rate doubles when the temperature is raised from 290 K to 300 K ?A. 12 kcalB. 15 kcalC. 10 kcalD. 20 kcal |
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Answer» Correct Answer - A ` log.(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `log2=(E_(a))/(2.303xx2)[(300-290)/(290xx300)]` `E_(a)`=12062 cal =12.0 cal |
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