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The rate law for a reaction between the substances $\mathrm{A}$ and $\mathrm{B}$ is given by Rate $=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$ 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 will be as
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$2^{(\mathrm{n}-\mathrm{m})}$
$2^{(\mathrm{n}-\mathrm{m})}$
Rate $_1=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}} ; \quad$ Rate $_2=\mathrm{k}[2 \mathrm{~A}]^{\mathrm{n}}[1 / 2 \mathrm{~B}]^{\mathrm{m}}$
$\therefore \frac{\text { Rate }_2}{\text { Rate }_1}=\frac{\mathrm{k}[2 \mathrm{~A}]^n[1 / 2 \mathrm{~B}]^m}{\mathrm{k}[\mathrm{A}]^n[\mathrm{~B}]^m}=[2]^n[1 / 2]^m=2^{\mathrm{n}} \cdot 2^{-m}=2^{\mathrm{n}-\mathrm{m}}$
$\therefore \frac{\text { Rate }_2}{\text { Rate }_1}=\frac{\mathrm{k}[2 \mathrm{~A}]^n[1 / 2 \mathrm{~B}]^m}{\mathrm{k}[\mathrm{A}]^n[\mathrm{~B}]^m}=[2]^n[1 / 2]^m=2^{\mathrm{n}} \cdot 2^{-m}=2^{\mathrm{n}-\mathrm{m}}$
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