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In a reaction , the time required to complete half of the reaction was found to increase 16 times when the initial concentration of the reactant was reduced to 1//4^(th). What is the order of the reaction ? |
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Answer» Solution :`3` `(t_(2))/(t_(1))=((a_(1))/(a_(2)))^(n-1)` `implies16=((1)/((1)/(4)))^(n-1)` `n=3` THIRD order REACTION. |
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