JEE Chemistry 2026 Question with Solution

Given: The reaction is $A \to nB$ and the initial $10 \, \text{min}$ data are to be used.

Find: The value of $n$.

From the graph, concentration of $A$ decreases from $0.05 \, \text{M}$ to $0.04 \, \text{M}$ in $10 \, \text{min}$.

$$-\Delta [A] = 0.01 \, \text{M}$$

From the graph, concentration of $B$ increases from $0$ to $0.04 \, \text{M}$ in $10 \, \text{min}$.

$$\Delta [B] = 0.04 \, \text{M}$$

For $A \to nB$,

$$\text{Rate} = -\frac{d[A]}{dt} = \frac{1}{n}\frac{d[B]}{dt}$$

So, using the ratio of changes in the initial interval,

$$n = \frac{\Delta [B]}{-\Delta [A]} = \frac{0.04}{0.01} = 4$$

Therefore, the value of $n$ is $4$ and the correct option is A.

Given: Initial concentration changes are read from the concentration-time plot.

Find: The stoichiometric coefficient of product $B$.

In a reaction of the form $A \to nB$, if the stoichiometric coefficient of reactant $A$ is $1$, then the product formed is $n$ times the reactant consumed.

During the first $10 \, \text{min}$,

• reactant consumed = $0.05 - 0.04 = 0.01 \, \text{M}$
• product formed = $0.04 - 0 = 0.04 \, \text{M}$

Hence,

$$n = \frac{\text{product formed}}{\text{reactant consumed}} = \frac{0.04}{0.01} = 4$$

Therefore, the dissociation reaction is consistent with $A \to 4B$ in the initial phase.