Rate law for a reaction between A and B is given by 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 is [2025]

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Solution

Q.
Rate law for a reaction between A and B is given by $R = k {[A]}^{n} {[B]}^{m}$

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 $(\frac{r_{2}}{r_{1}})$ is [2025]

1 $2^{(n - m)}$

2 $(n - m)$

3 $(m + n)$

4 $\frac{1}{2^{m + n}}$

Ans.
(1)

$r_{1} = k {[A]}^{n} {[B]}^{m}$

$r_{2} = k {[2 A]}^{n} {[\frac{B}{2}]}^{m}$

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

$\frac{r_{2}}{r_{1}} = 2^{n - m}$

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