JEE Chemistry 2025 Question with Solution

Given: The plot of $t_{1/2}$ versus $[A]_0$ is a straight line through the origin with slope $76.92$. Initial concentration is $[A]_0 = 2.5 \, \text{mol L}^{-1}$ and time is $10 \, \text{min}$.

Find: The concentration of $A$ after $10 \, \text{min}$ in the form _____ \times 10^{-3} \, \text{mol L}^{-1}.

For a zero-order reaction,

$$t_{1/2} = \frac{[A]_0}{2K}$$

So the slope of the graph is

$$\text{slope} = \frac{1}{2K} = 76.92$$

Hence,

$$K = \frac{1}{2 \times 76.92} = \frac{1}{153.84}$$

Now use the zero-order integrated rate law,

$$[A] = [A]_0 - Kt$$

Substituting the values,

$$[A] = 2.5 - \frac{1}{153.84} \times 10$$ $$[A] = 2.435 \, \text{mol L}^{-1}$$

Therefore,

$$2.435 \, \text{mol L}^{-1} = 2435 \times 10^{-3} \, \text{mol L}^{-1}$$

So, the correct option is A.

Given: A graph of $t_{1/2}$ versus $[A]_0$ is linear and passes through the origin. The slope is $76.92$. Also, $[A]_0 = 2.5 \, \text{mol L}^{-1}$.

Find: Concentration of $A$ after $10 \, \text{min}$.

For a zero-order reaction, the half-life depends directly on initial concentration:

$$t_{1/2} \propto [A]_0$$

with formula

$$t_{1/2} = \frac{[A]_0}{2k}$$

Comparing with the straight-line form,

$$t_{1/2} = \left(\frac{1}{2k}\right)[A]_0$$

therefore,

$$\frac{1}{2k} = 76.92$$

So,

$$k = \frac{1}{153.84} \, \text{mol L}^{-1} \text{min}^{-1}$$

Now apply the integrated rate law for zero-order kinetics:

$$[A]_t = [A]_0 - kt$$

At $t = 10 \, \text{min}$,

$$[A]_{10} = 2.5 - \frac{10}{153.84}$$ $$[A]_{10} \approx 2.5 - 0.06500$$ $$[A]_{10} \approx 2.435 \, \text{mol L}^{-1}$$

Expressing this in the required form,

$$2.435 \, \text{mol L}^{-1} = 2435 \times 10^{-3} \, \text{mol L}^{-1}$$

Therefore, the correct option is A, that is $2435$.