Question

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 $\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ is

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

Hint:

The rate law depends on the concentrations of the reactants raised to their respective orders.

Answer 1

The Correct Option is A

1. Initial rate law: \[ \mathrm{r}_{1} = \mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}} \]
2. New concentrations: - Concentration of A is doubled: $2[\mathrm{A}]$ - Concentration of B is halved: $\frac{[\mathrm{B}]}{2}$
3. New rate law: \[ \mathrm{r}_{2} = \mathrm{k}(2[\mathrm{A}])^{\mathrm{n}} \left( \frac{[\mathrm{B}]}{2} \right)^{\mathrm{m}} \] \[ \mathrm{r}_{2} = \mathrm{k} \cdot 2^{\mathrm{n}} [\mathrm{A}]^{\mathrm{n}} \cdot \frac{[\mathrm{B}]^{\mathrm{m}}}{2^{\mathrm{m}}} \]
4. Ratio of new rate to initial rate: \[ \frac{\mathrm{r}_{2}}{\mathrm{r}_{1}} = \frac{\mathrm{k} \cdot 2^{\mathrm{n}} [\mathrm{A}]^{\mathrm{n}} \cdot \frac{[\mathrm{B}]^{\mathrm{m}}}{2^{\mathrm{m}}}}{\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}} = 2^{\mathrm{n}} \cdot \frac{1}{2^{\mathrm{m}}} = 2^{(\mathrm{n}-\mathrm{m})} \] Therefore, the correct answer is (1) $2^{(\mathrm{n}-\mathrm{m})}$.