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InterviewSolution
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A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half? |
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Answer» SOLUTION :RAT `=k[R]^(2)` (i) If R is INCREASED to 2R, then Rate `=k[2R]^(2)=4k[R]^(2)` Thus, the rate increases four times. (II) If R is reduced to `(R )/(2)`, then Rate `=k[(R )/(2)]^(2)= (1)/(4) k[R]^(2)` Thus, the rate is reduced to one-fourth. |
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