Determine Partial Pressure of Product in Gas Phase Kinetics

To find the partial pressure of $\text{C(g)}$ at $t = 30\text{ minutes}$, we consider the stoichiometry of the given gas phase reaction carried out in a closed vessel:

$2\text{A(g)} \longrightarrow 4\text{B(g)} + \text{C(g)}$

Let $P_0$ be the initial pressure of $\text{A(g)}$ at $t = 0$.

At any time $t$, let the extent of reaction be represented by $x$, where $2x$ is the decrease in the partial pressure of $\text{A(g)}$. The partial pressures of all species at different time intervals are given as follows:

$$\begin{array}{rccc} & 2\text{A(g)} & \longrightarrow & 4\text{B(g)} & + & \text{C(g)} \\ \text{At } t = 0: & P_0 & & 0 & & 0 \\ \text{At } t = 30\text{ min}: & P_0 - 2x & & 4x & & x \\ \text{At } t = \infty: & 0 & & 2P_0 & & \frac{P_0}{2} \end{array}$$

1. Calculation of Initial Pressure ($P_0$):

At $t = \infty$, the reaction goes to completion ($\text{A(g)}$ is completely consumed, i.e., $2x = P_0 \implies x = \frac{P_0}{2}$).

The total pressure at infinite time ($P_\infty$) is: $P_\infty = P_{\text{A}} + P_{\text{B}} + P_{\text{C}} = 0 + 2P_0 + \frac{P_0}{2} = \frac{5}{2} P_0$

Given that $P_\infty = 600\text{ mm Hg}$: $\frac{5}{2} P_0 = 600 \implies P_0 = \frac{600 \times 2}{5} = 240\text{ mm Hg}$

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2. Calculation of Partial Pressure of $\text{C(g)}$ at $t = 30\text{ min}$:

The total pressure at $t = 30\text{ min}$ ($P_{30}$) is: $P_{30} = (P_0 - 2x) + 4x + x = P_0 + 3x$

Given that $P_{30} = 300\text{ mm Hg}$ and $P_0 = 240\text{ mm Hg}$: $240 + 3x = 300$ $3x = 60 \implies x = 20\text{ mm Hg}$

The partial pressure of $\text{C(g)}$ at $30\text{ minutes}$ is: $P_{\text{C}} = x = 20\text{ mm Hg}$