Question

Consider the following gas phase reaction being carried out in a closed vessel at 25°C: \[ 2A(g) \longrightarrow 4B(g) + C(g) \] The total pressure of the system at different time intervals is given as:

We need to find the pressure of \( C(g) \) at the 30 minutes time interval.}

Answer 1

Correct Answer: 5

Step 1: Understanding the reaction and the total pressure.
The reaction involves the conversion of 2 moles of \( A \) into 4 moles of \( B \) and 1 mole of \( C \). Since the reaction is occurring in a closed vessel, the total pressure is proportional to the number of moles of the gases present, assuming constant temperature and volume. Let the initial moles of \( A \) be \( x \), and the moles of \( B \) and \( C \) at time \( t \) be \( 0 \) and \( 0 \), respectively. At any time \( t \), the change in the number of moles of \( A \) is \( -2y \), where \( y \) is the extent of the reaction. The number of moles of \( B \) formed is \( 4y \), and the number of moles of \( C \) formed is \( y \).
Step 2: Relating the total pressure to the change in moles.
The total pressure at any time \( t \) is the sum of the partial pressures of \( A \), \( B \), and \( C \). Assuming ideal gas behavior, the total pressure is directly proportional to the total number of moles of gas. Therefore: \[ P_{\text{total}} = P_A + P_B + P_C \] At equilibrium (when \( t = \infty \)): - The total pressure is 600 mm Hg, corresponding to the final moles of gas after the reaction is complete.\]\ - Since 2 moles of \( A \) produce 4 moles of \( B \) and 1 mole of \( C \), the total number of moles at equilibrium is \( 4y + y = 5y \).
- At equilibrium, the total pressure is proportional to 5 times the number of moles.

Step 3: Calculate the moles at the given time interval.
At 30 minutes, the total pressure is 300 mm Hg. Using the ratio of the total pressures, we can find the mole fraction of \( C \). Let the total moles of gas at time \( t = 30 \) min be \( 5y \), and the moles of \( C \) formed be \( y \). Using the proportionality of pressure to moles, we get the ratio: \[ \frac{P_{\text{total at 30 min}}}{P_{\text{total at equilibrium}}} = \frac{5y}{5y_{\text{eq}}} \] Substituting the given pressures: \[ \frac{300}{600} = \frac{5y}{5y_{\text{eq}}} \] \[ y_{\text{eq}} = \frac{600}{5} = 120. \] Finally, the partial pressure of \( C \) at 30 minutes is: \[ P_C = \frac{y}{y_{\text{eq}}} \times P_{\text{total at equilibrium}} = \frac{y}{120} \times 600. \] Simplifying: \[ P_C = 5 \, \text{mm Hg}. \] Thus, the pressure of \( C(g) \) at 30 minutes is: \[ \boxed{5} \, \text{mm Hg}. \]