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What would be the activation energy of a reaction when the temperature is increased from 27^(@)C to 37^(@)C ? |
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Answer» SOLUTION :`ln. (k_(2))/(k_(1))=(E_(a))/(R )[(T_(2)-T_(1))/(T_(1)T_(2))]` `ln 2 = (E_(a))/(R )[(10)/(300xx310)]` `E_(a)=9300` R ln 2 `= 53.4 "KJ mol"^(-1)` |
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