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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 reducing the concentration of B to half, the ratio of the new rate to the earlier rate of the reaction will be as

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$(n - m)$

$1 / 2^{(m + n)}$

$2^{(n - m)}$

$(m + n)$

answer is A.

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Detailed Solution

$r_{1} = k [A]^{n} [B]^{m} and r_{2} = k [2 A]^{n} {(\frac{[B]}{2})}^{m} = 2^{(n - m)} k [A]^{n} [B]^{m}$

Hence, $\frac{r_{2}}{r_{1}} = 2^{(n - m)}$ .

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