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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 having the concentration of B, the ratio of the new rate to the earlier rate of the reactio will be:A. `(1)/(2^(m)+n)`B. (m+n)C. (n-m)D. `2^((n-m)` |
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Answer» Correct Answer - D `"Initial rate"=k[A]^(n)[B]^(n)` when[A]=[2A] and `[B][(1)/(2)B]`, then `"New rate"=k[2A]^(n)[1//28]^m` On dividing Eq.(ii)by Eq.(i), we get `"New rate"/"Initial rate"=(k[2A]^(n)[1//2B]^(m))/(k[A]^(n)[B}^(m))` `"New rate"/"Initial rate"=(2)^(n)(1//2)^(m)=(2)^(n)(2)^(-m)=2(n-m)` Hence, the ratio of new rate fo the reaction is `2^(n-m).` |
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