Step 1 (Formula / Governing Law & Given Data): [1 Mark]

For the first-order gas-phase decomposition reaction:

$$\text{C}_2\text{H}_5\text{Cl(g)} \longrightarrow \text{C}_2\text{H}_4\text{(g)} + \text{HCl(g)}$$

Let the initial pressure of $\text{C}_2\text{H}_5\text{Cl}$ be $P_i = 0.30 \text{ atm}$. At time $t = 30 \text{ s}$, let the decrease in pressure be $x$.

• At $t=0$: $P_{\text{C}_2\text{H}_5\text{Cl}} = P_i$, $P_{\text{C}_2\text{H}_4} = 0$, $P_{\text{HCl}} = 0$. Total Pressure $P_t = P_i = 0.30 \text{ atm}$.
• At $t=30$: $P_{\text{C}_2\text{H}_5\text{Cl}} = P_i - x$, $P_{\text{C}_2\text{H}_4} = x$, $P_{\text{HCl}} = x$.

Total pressure $P_t = (P_i - x) + x + x = P_i + x$. Therefore, $x = P_t - P_i$.

The pressure of the reactant at time $t$ is $P = P_i - x = P_i - (P_t - P_i) = 2P_i - P_t$.

Step 2 (Step-by-Step Mathematical Working): [1.5 Marks]

Given $P_i = 0.30 \text{ atm}$ and $P_t = 0.50 \text{ atm}$ at $t = 30 \text{ s}$:

$$P = 2(0.30) - 0.50 = 0.60 - 0.50 = 0.10 \text{ atm}$$

The rate constant $k$ for a first-order reaction is given by:

$$k = \frac{2.303}{t} \log \left( \frac{P_i}{P} \right)$$ $$k = \frac{2.303}{30} \log \left( \frac{0.30}{0.10} \right)$$ $$k = \frac{2.303}{30} \log(3)$$

Substituting $\log(3) = 0.48$:

$$k = \frac{2.303 \times 0.48}{30}$$ $$k = \frac{1.10544}{30}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

$$k = 0.0368 \text{ s}^{-1}$$

💡 EL10 Concept Insight: Think of this like a bank account where you start with $0.30$ coins. As the reactant turns into two products, the total "pressure" in the room increases because one molecule splits into two; by measuring how much the total pressure grew, we can work backward to see exactly how much reactant is left.