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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 becomes

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m + n

2^(n-m)

2^1/(m+n)

m - n

answer is D.

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Detailed Solution

$Rate_{old} = k [A]^{n} [B]^{m}$

$\begin{array}{l} {Rate}_{new} = k (2 [A])^{n} {(\frac{[B]}{2})}^{m} = k \cdot 2^{n - m} [A]^{n} [B]^{m} \\ \Rightarrow \frac{{Rate}_{new}}{{Rate}_{old}} = 2^{n - m} \end{array}$

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