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InterviewSolution
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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. `(1)/(2^(m+n))`B. `2^((n-m))`C. `(m+n)`D. `(n-m)` |
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Answer» Correct Answer - B `Rate_(1)=kk[A]^(n)[B]^(m)` On doubling the concentration of `A` and having the concentration of `B` , we get `Rate_(2)=k[2A]^(n)[B//2]^(m)` Dividing `Rate_(2)` by `Rate_(1)` , we get `(Rate_(2))/(Rate_(1))=(k[2A]^(n)[B//2]^(m))/(k[A]^(n)[B]^(m))=(2)^(n)((1)/(2))^(m)` `=2^(n-m)` |
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