JEE Chemistry 2024 Question with Solution

Given: The reaction is

$$\text{A(g)} \rightarrow 2\text{B(g)} + \text{C(g)}$$

The total pressure after $23 \, \text{s}$ is $200 \, \text{Torr}$ and the total pressure after complete decomposition is $300 \, \text{Torr}$.

Find: The value of the rate constant in the form _____ × $10^{-2} \, \text{s}^{-1}$.

For this reaction, if the initial pressure of $A$ is $P_0$, then at completion the total pressure becomes $3P_0$. Thus,

$$P_\infty = 3P_0 = 300$$

So,

$$P_0 = 100 \, \text{Torr}$$

Using the first-order pressure relation,

$$K = \frac{1}{t} \ln \frac{P_\infty - P_0}{P_\infty - P_t}$$

Substituting the values,

$$K = \frac{1}{23} \ln \frac{300 - 100}{300 - 200}$$ $$K = \frac{2.3}{23} \log \frac{200}{100}$$ $$K = \frac{2.3}{23} \times 0.301$$ $$K = 0.0301 = 3.01 \times 10^{-2} \, \text{s}^{-1}$$

Therefore, the required nearest integer is 3.

Given:

• Reaction:

$$\text{A(g)} \rightarrow 2\text{B(g)} + \text{C(g)}$$

• $P_t = 200 \, \text{Torr}$ at $t = 23 \, \text{s}$
• $P_\infty = 300 \, \text{Torr}$

Find: The nearest integer in _____ × $10^{-2} \, \text{s}^{-1}$.

If the initial pressure is $P_0$, then complete decomposition of one mole of $A$ produces total three moles of gaseous species, so the final pressure is three times the initial pressure:

$$P_\infty = 3P_0$$

Hence,

$$3P_0 = 300$$ $$P_0 = 100 \, \text{Torr}$$

Now use the integrated first-order expression written in terms of total pressure:

$$K = \frac{1}{t} \ln \left(\frac{P_\infty - P_0}{P_\infty - P_t}\right)$$

Substitute:

$$K = \frac{1}{23} \ln \left(\frac{300 - 100}{300 - 200}\right)$$ $$K = \frac{1}{23} \ln 2$$

Using $\ln 2 = 2.303 \times 0.301$,

$$K = \frac{2.3 \times 0.301}{23}$$ $$K = 0.0301 \, \text{s}^{-1}$$

So,

$$K \approx 3 \times 10^{-2} \, \text{s}^{-1}$$

Therefore, the nearest integer is 3.