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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

Rate law for a reaction between A and B is given by
\(\mathrm{R}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}\)
If concentration of A is doubled and concentration of \(B\) is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction \(\left(\frac{r_2}{r_1}\right)\) is

  1. A \(2^{(n-m)}\)
  2. B \((n-m)\)
  3. C \((m+n)\)
  4. D \(\frac{1}{2^{m+n}}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2^{(n-m)}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{r}_1=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}\) Now A is doubled \& B is halved in concentration \(\Rightarrow \mathrm{r}_2=\mathrm{k} 2^{\mathrm{n}}[\mathrm{~A}]^{\mathrm{n}} \cdot \frac{[\mathrm{~B}]^{\mathrm{m}}}{2^{\mathrm{m}}}\) Now…
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