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InterviewSolution
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The rate law for a reaction between the substances A and B is given by Rate = `k[A]^(n)[B]^(m)` On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as:A. (m+n)B. (n-m)C. `2^(n-m)`D. `1/(2^(m+n))` |
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Answer» Correct Answer - C c) `Rate_(1) = k[A]^(n)[B]^(m)` `Rate_(2) = k[2A]^(n)[1//2B]^(m)` `(Rate_(2))/(Rate_(1))= (k[2A]^(n)[1//2B]^(m))/(k[A]^n)[B]^(m)` `=[2]^(n([1//2])^(m))= 2^(n).2^(-m)` `=2(n-m)` |
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