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
- A \(2^{(n-m)}\)
- B \((n-m)\)
- C \((m+n)\)
- D \(\frac{1}{2^{m+n}}\)
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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