JEE Chemistry 2024 Question with Solution

Given:

• First-order reaction:

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

• Total pressure after $23 \, \text{s}$ is $200 \, \text{Torr}$
• Total pressure at complete decomposition is $300 \, \text{Torr}$

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

Let the initial pressure of $A$ be $P_0$. At completion, since $1$ mole of $A$ gives total $3$ moles of gaseous products, we use:

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

So,

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

For a first-order reaction using total pressures:

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

Substituting the values:

$$K = \frac{1}{23} \ln \frac{300 - 100}{300 - 200}$$ $$K = \frac{2.3}{23} \log \frac{200}{100}$$ $$K = \frac{2.3}{23} \log 2$$

Using $\log 2 = 0.301$,

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

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: Rate constant $k$.

If the initial pressure of $A$ is $P_0$ and pressure change corresponding to decomposition is $x$, then at time $t$:

$$P_A = P_0 - x$$

Products formed contribute total pressure $3x$, so total pressure becomes:

$$P_t = (P_0 - x) + 3x = P_0 + 2x$$

Given,

$$P_t = 200$$

At completion, all of $A$ decomposes, so $x = P_0$ and hence:

$$P_{\infty} = P_0 + 2P_0 = 3P_0 = 300$$

Thus,

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

Now apply the first-order expression in pressure form:

$$k = \frac{2.303}{t} \log \left( \frac{P_{\infty} - P_0}{P_{\infty} - P_t} \right)$$ $$k = \frac{2.303}{23} \log \left( \frac{300 - 100}{300 - 200} \right)$$ $$k = \frac{2.303}{23} \log 2$$ $$k \approx \frac{2.303}{23} \times 0.301 \approx 0.0303 \, \text{s}^{-1}$$

So,

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

Therefore, the nearest integer is $3$.