Question

For a reaction $2A+B\to 2AB$, it is found that doubling the concentration of both the reactants increases the rate to eight times that of initial rate but doubling the concentration of B alone doubles the rate. Then the order of the reaction with respect to A and B is

• A. 0, 3
• B. 0, 2
• C. 2, 1
• D. 2, 2

Answer 1

The correct option is C

2, 1

The explanation for the correct option: (C)

1. Order of the reaction: The sum of the exponents or powers to which molar concentration terms are elevated in the rate law determines the reaction order.
2. The order might be expressed in terms of the specific order of the reactants or the overall order of the reaction.
3. Given reaction; $2A+B\to 2AB$
4. Let the order of reaction w.r.t A be m and order of reaction w.r.t. B be n.
5. Thus, the rate law is written as: $\mathrm{Rate},\mathrm{r}=\mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{m}}.{\left[\mathrm{B}\right]}^{\mathrm{n}}$
6. Condition 1: Now, according to the first given condition;

Concentration of $A=2A$

Concentration of $B=2B$

Thus, the Rate of this Reaction : $$\begin{gathered} {\mathrm{r}}_1=\mathrm{k}{\left[2\mathrm{A}\right]}^{\mathrm{m}}{\left[2\mathrm{B}\right]}^{\mathrm{n}} \\ =\mathrm{k}{\left(2\right)}^{\mathrm{m}+\mathrm{n}}{\left[\mathrm{A}\right]}^{\mathrm{m}}{\left[\mathrm{B}\right]}^{\mathrm{n}} \\ =8\mathrm{r} \end{gathered}$$

7. Condition 2: Now, according to the second given condition;

Concentration of $A=A$

Concentration of $B=2B$

Thus, the Rate of this Reaction : $$\begin{gathered} {\mathrm{r}}_2=\mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{m}}{\left[2\mathrm{B}\right]}^{\mathrm{n}} \\ =\mathrm{k}{\left(2\right)}^{\mathrm{n}}{\left[\mathrm{A}\right]}^{\mathrm{m}}{\left[\mathrm{B}\right]}^{\mathrm{n}} \\ =2\mathrm{r} \end{gathered}$$

8. Divide rate 1 by rate 2:

$$\begin{gathered} \frac{\mathrm{r}1}{\mathrm{r}2}=\frac{8\mathrm{r}}{2\mathrm{r}}=\frac{\mathrm{k}{\left(2\right)}^{\mathrm{m}+\mathrm{n}}{\left[\mathrm{A}\right]}^{\mathrm{m}}{\left[\mathrm{B}\right]}^{\mathrm{n}}}{\mathrm{k}{\left(2\right)}^{\mathrm{n}}{\left[\mathrm{A}\right]}^{\mathrm{m}}{\left[\mathrm{B}\right]}^{\mathrm{n}}} \\ \mathrm{Solving}\mathrm{this}\mathrm{we}\mathrm{get}, \\ 4=\frac{{\left(2\right)}^{\mathrm{m}+\mathrm{n}}}{2^{\mathrm{n}}}={\left(2\right)}^{\mathrm{m}+\mathrm{n}-\mathrm{n}}={\left(2\right)}^{\mathrm{m}}{\left(2\right)}^2={\left(2\right)}^{\mathrm{m}}\therefore \mathrm{m}=2 \end{gathered}$$

Hence, order of reaction with respect to A is 2.

9. As on doubling the concentration of B , the rate also doubles up, which means that the rate is directly proportional to the concentration of B. Hence, the order of reaction with respect to B is 1.

10. Finally, order w.r.t A is 2 and order w.r.t B is 1.

Conclusion: Hence, option (C) is correct.