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InterviewSolution
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A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half? |
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Answer» SOLUTION :Rate `=k[A]^(2)=KA^(2)`. [Let [A] be REPRESENTED by a] If `[A]=2a, " rate"=k((a)/(2))^(2)=(1)/(4)ka^(2)`. The rate becomes one-fourth. |
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