JEE Chemistry 2026 Question with Solution

Given: Reaction I is first order, so its rate is $R_I = k[A]$. For reaction II, $-\frac{d[Br^-]}{dt} = 2 \times 10^{-4} \, \text{mol L}^{-1} \, \text{min}^{-1}$ at $10.10 \, \text{am}$. Also, the rates of reactions I and II are equal at that time, and $[A] = 10^{-2} \, \text{mol L}^{-1}$.

Find: The first order rate constant $k$ of reaction I.

For reaction II,

$$R_{II} = -\frac{1}{5} \frac{d[Br^-]}{dt}$$

Substituting the given value,

$$R_{II} = \frac{1}{5} \left(2 \times 10^{-4}\right) = 4 \times 10^{-5}$$

Since the rates are same at $10.10 \, \text{am}$,

$$R_I = R_{II} = 4 \times 10^{-5}$$

For the first order reaction I,

$$R_I = k[A]$$

So,

$$k \left(10^{-2}\right) = 4 \times 10^{-5}$$

Therefore,

$$k = 4 \times 10^{-3} \, \text{min}^{-1}$$

Hence, the correct option is B.

Given: In reaction II, the stoichiometric coefficient of $Br^-$ is $5$. The disappearance rate of $Br^-$ is given. The rates of reactions I and II are equal at $10.10 \, \text{am}$.

Find: The value of $k$ for reaction I.

The rate of a reaction is obtained by dividing the rate of disappearance of a reactant by its stoichiometric coefficient. Therefore, for reaction II,

$$R_{II} = -\frac{1}{5} \frac{d[Br^-]}{dt}$$

Using the given data,

$$R_{II} = -\frac{1}{5} \left(-2 \times 10^{-4}\right) = 4 \times 10^{-5} \, \text{mol L}^{-1} \, \text{min}^{-1}$$

Now the question states that

$$R_I = R_{II}$$

Hence,

$$R_I = 4 \times 10^{-5} \, \text{mol L}^{-1} \, \text{min}^{-1}$$

For a first order reaction,

$$R_I = k[A]$$

With $[A] = 10^{-2} \, \text{mol L}^{-1}$,

$$k \times 10^{-2} = 4 \times 10^{-5}$$

Dividing both sides by $10^{-2}$,

$$k = \frac{4 \times 10^{-5}}{10^{-2}} = 4 \times 10^{-3} \, \text{min}^{-1}$$

Therefore, the first order rate constant is $4 \times 10^{-3} \, \text{min}^{-1}$.