Question

For a reaction $A+B\to C$, the rate law is written as $r=k{\left[A\right]}^2\left[B\right]$ . Doubling the concentration of ‘$A$’ without changing concentration of ‘$B$’ increases the rate of reaction by

• A. 2 times
• B. 4 times
• C. 8 times
• D. 16 times

Answer 1

The correct option is B

4 times

The explanation for the Correct option: (B)

1. The rate of reaction is defined as the drop in reactant concentration over time or the increase in product concentration over time.
2. The rate of reaction is determined by the reactant concentration at the start.
3. For the given reaction, $A+B\to C$, the rate of reaction is as follows; $r=k{\left[A\right]}^2\left[B\right]$
4. Thus, the rate of reaction is proportional to ${\left[A\right]}^2$.
5. According to the given conditions;

The concentration of $A$ is doubled i.e. $A=2A$

The concentration of $B$ remains the same.

Thus, rate becomes; ​$$\begin{gathered} r_1=k{\left[2A\right]}^2\left[B\right] \\ {r}_1=4r \end{gathered}$$​

6. Hence, the rate will be increased by 4 times.

The explanation for Incorrect option:

(A) On doubling the concentration rate becomes 4 times so this option is incorrect.

(C) On doubling the concentration rate becomes 4 times so this option is incorrect.

(D) On doubling the concentration rate becomes 4 times so this option is incorrect.

Conclusion: Hence, option (B) is correct.