Question

Reaction A(g) → 2B(g) + C(g) is a first-order reaction. It was started with pure A.
The following table shows the pressure of the system at different times: 

Which of the following options is incorrect?

• A. Initial pressure of A is 80 mm Hg
• B. The reaction never goes to completion
• C. Rate constant of the reaction is 0.693 min\(^{-1}\)
• D. Partial pressure of A after 10 minutes is 40 mm Hg

Hint:

For first-order reactions, use the integrated rate law to calculate the rate constant. The reaction does not go to completion but asymptotically approaches the equilibrium state.

Answer 1

The Correct Option is C

Step 1: Analyze Statement (1) The total pressure at \(t = \infty\) is 240 mm Hg, which is the pressure due to the products of the reaction (2B + C). The initial pressure of A was 80 mm Hg, and since 2 moles of B and 1 mole of C are produced per mole of A, the increase in total pressure from the initial 80 mm Hg to the final 240 mm Hg indicates that 160 mm Hg pressure is due to the products. So, the initial pressure of A is indeed 80 mm Hg. Therefore, statement (1) is correct.
Step 2: Analyze Statement (2)
For a first-order reaction, the reaction does not go to completion. It asymptotically approaches a state where the concentration of the reactant is very low but not zero. As time progresses, the pressure will increase, but it will not go to 240 mm Hg immediately. The reaction approaches this value asymptotically. Therefore, Statement (2) is correct.
Step 3: Analyze Statement (3) - Rate Constant Calculation
For a first-order reaction, the integrated rate law is given by: \[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \] Where: \([A]_0\) is the initial pressure of A (80 mm Hg), \([A]\) is the pressure of A after 10 minutes (the difference between the initial and the total pressure), \(k\) is the rate constant, \(t\) is the time in minutes. At \(t = 10\) minutes, the total pressure is 160 mm Hg, and the pressure of A is: \[ P_A = 80 - (160 - 240) = 40 \text{ mm Hg} \] Now, applying the rate law: \[ \ln \left( \frac{80}{40} \right) = k \times 10 \] \[ \ln(2) = k \times 10 \] \[ k = \frac{\ln(2)}{10} = \frac{0.693}{10} = 0.0693 \, \text{min}^{-1} \] The rate constant is approximately \(0.0693 \, \text{min}^{-1}\), not 1.693 min\(^{-1}\).
Thus, Statement (3) is incorrect.
Step 4: Analyze Statement (4)
The partial pressure of A after 10 minutes is given as 40 mm Hg, which we calculated earlier. Therefore, Statement (4) is correct.

Answer 2

The given reaction is: \[ \text{A(g)} \rightarrow 2\text{B(g)} + \text{C(g)} \] At \( t = 0 \), the initial pressure is \( P_0 \).
At \( t \to \infty \), the final pressure is \( P_{\infty} = 3P_0 = 240 \).
The pressure at \( t = 0 \) is \( P_0 = 80 \, \text{mm of Hg} \).
We can use the equation: \[ K t = \ln \left( \frac{P_{\infty} - P_0}{P_{\infty} - P_t} \right) \] Substitute the values: \[ K \times 10 = \ln \left( \frac{240 - 80}{240 - 160} \right) \] Solving for \( K \): \[ K = \frac{\ln 2}{10} = 0.0693 \, \text{min}^{-1} \] Thus, the rate constant \( K \) is \( \boxed{0.0693 \, \text{min}^{-1}} \).