Q62 / 225 · JEE Main 2003 · Chemistry

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

  • A
    $(m+n)$
  • B
    $(n-m)$
  • Correct answer
    C
    $2^{(\mathrm{n}-\mathrm{m})}$
  • D
    $\frac{1}{2^{(\mathrm{m}+\mathrm{n})}}$

Solution

Rate $_1=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}} ; \quad$ Rate $_2=\mathrm{k}[2 \mathrm{~A}]^{\mathrm{n}}[1 / 2 \mathrm{~B}]^{\mathrm{m}}$ $\therefore \frac{\text { Rate }_2}{\text { Rate }_1}=\frac{\mathrm{k}[2 \mathrm{~A}]^n[1 / 2 \mathrm{~B}]^m}{\mathrm{k}[\mathrm{A}]^n[\mathrm{~B}]^m}=[2]^n[1 / 2]^m=2^{\mathrm{n}} \cdot 2^{-m}=2^{\mathrm{n}-\mathrm{m}}$

Hence, the correct answer is option: 'C'.

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