JEE Chemistry 2026 Question with Solution

Given: Two first-order reactions, $A \to B$ and $C \to D$. For the first reaction, the rate constant at $500 \, \text{K}$ is double that at $300 \, \text{K}$. At $500 \, \text{K}$, $50\%$ reaction is completed in $2 \, \text{h}$, so this is the half-life. The activation energy of the second reaction is half that of the first reaction, and at $500 \, \text{K}$ the rate constant of the second reaction is double that of the first reaction.

Find: The value of the rate constant of the second reaction at $300 \, \text{K}$ in the form $\_\_\_\_\_ \times 10^{-3} \, \text{hour}^{-1}$.

For a first-order reaction,

$$t_{1/2} = \frac{0.693}{k}$$

Given $t_{1/2} = 2 \, \text{h}$ for the first reaction at $500 \, \text{K}$,

$$k_{500}^{(1)} = \frac{0.693}{2} = 0.3465 \, \text{h}^{-1}$$

Now using $k_{500}^{(1)} = 2k_{300}^{(1)}$,

$$k_{300}^{(1)} = \frac{0.3465}{2} = 0.17325 \, \text{h}^{-1}$$

Apply the Arrhenius relation for the first reaction:

$$\ln\left(\frac{k_{500}^{(1)}}{k_{300}^{(1)}}\right) = \frac{E_a^{(1)}}{R}\left(\frac{1}{300} - \frac{1}{500}\right)$$

Since $\frac{k_{500}^{(1)}}{k_{300}^{(1)}} = 2$,

$$\ln 2 = \frac{E_a^{(1)}}{R}\left(\frac{1}{300} - \frac{1}{500}\right)$$ $$0.693 = \frac{E_a^{(1)}}{R} \cdot \frac{200}{150000}$$ $$\frac{E_a^{(1)}}{R} = 5197$$

The second reaction has half this activation energy, so

$$\frac{E_a^{(2)}}{R} = \frac{5197}{2} = 2598.5$$

Also, at $500 \, \text{K}$,

$$k_{500}^{(2)} = 2k_{500}^{(1)} = 2 \times 0.3465 = 0.693 \, \text{h}^{-1}$$

Now use Arrhenius relation for the second reaction:

$$\ln\left(\frac{k_{500}^{(2)}}{k_{300}^{(2)}}\right) = \frac{E_a^{(2)}}{R}\left(\frac{1}{300} - \frac{1}{500}\right)$$ $$\ln\left(\frac{0.693}{k_{300}^{(2)}}\right) = 2598.5 \times \frac{200}{150000} = 3.465$$ $$\frac{0.693}{k_{300}^{(2)}} = e^{3.465} \approx 32$$ $$k_{300}^{(2)} = \frac{0.693}{32} \approx 0.0217 \, \text{h}^{-1}$$

Thus,

$$k_{300}^{(2)} \approx 21.7 \times 10^{-3} \, \text{h}^{-1}$$

Therefore, the nearest integer is 22.

Given: The temperature pair is $300 \, \text{K}$ and $500 \, \text{K}$, and the first reaction obeys $k_{500}^{(1)} = 2k_{300}^{(1)}$.

Find: The corresponding $k_{300}^{(2)}$ when $E_a^{(2)} = \frac{1}{2}E_a^{(1)}$ and $k_{500}^{(2)} = 2k_{500}^{(1)}$.

First evaluate the temperature factor:

$$\frac{1}{300} - \frac{1}{500} = \frac{500-300}{300 \times 500} = \frac{200}{150000} = \frac{1}{750}$$

For reaction $1$,

$$\ln 2 = \frac{E_a^{(1)}}{R} \cdot \frac{1}{750}$$ $$\frac{E_a^{(1)}}{R} = 750 \ln 2 \approx 750 \times 0.693 = 5197$$

Hence for reaction $2$,

$$\frac{E_a^{(2)}}{R} = \frac{1}{2} \times 5197 = 2598.5$$

So for reaction $2$,

$$\ln\left(\frac{k_{500}^{(2)}}{k_{300}^{(2)}}\right) = 2598.5 \times \frac{1}{750} = 3.465$$ $$\frac{k_{500}^{(2)}}{k_{300}^{(2)}} = e^{3.465} \approx 32$$

Now from first-order half-life at $500 \, \text{K}$ for reaction $1$,

$$k_{500}^{(1)} = 0.3465 \, \text{h}^{-1}$$

Therefore,

$$k_{500}^{(2)} = 2 \times 0.3465 = 0.693 \, \text{h}^{-1}$$

Thus,

$$k_{300}^{(2)} = \frac{0.693}{32} \approx 0.0217 \, \text{h}^{-1} = 21.7 \times 10^{-3} \, \text{h}^{-1}$$

So the required nearest integer is 22.