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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. `2^(m - n)`C. `1//2^(n - m)`D. `2^(n - m)` |
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Answer» Correct Answer - D `r_(1) = k[A]^(n) [B]^(m)` `r_(2) = k [2A]^(n) [(B)/(2)] = 2^(n-m) k [A]^(n) [B]^(m)` `(r_(1))/(r_(2)) = 2^(n - m)` |
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