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The rate law for a reaction between `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 becomesA. (a) `n-m`B. (b) `2^(n-m)`C. (c ) `(1)/(2^(m+n))`D. (d) `m+n` |
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Answer» Correct Answer - b `r_(0)=K[A]^(n)[B]^(m)` `r_(1)=K[2A]^(n)[B//2]^(m)` `r_(1)=K 2^(n-m) [A]^(n)[B]^(m)` or `r_(1)=rxx2^(n-m)` |
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