Step 1 - Set up a pressure table.
$$C_2H_5Cl(g) \longrightarrow C_2H_4(g) + HCl(g)$$
At $t = 0$ : $\quad p_0 = 0.30$ atm, $\quad 0$, $\quad 0$
At $t = t$ : $\quad (p_0 - x)$, $\quad x$, $\quad x$
where $x$ = pressure of $C_2H_5Cl$ that has decomposed.

Step 2 - Relate $x$ to the total pressure.
$$p_{total} = (p_0 - x) + x + x = p_0 + x$$
At $t = 30$ s, $p_{total} = 0.50$ atm and $p_0 = 0.30$ atm:
$$0.50 = 0.30 + x$$
$$x = 0.20\ \text{atm}$$

Step 3 - Find the pressure of the reactant left at $t = 30$ s.
$$p_A = p_0 - x = 0.30 - 0.20 = 0.10\ \text{atm}$$

Step 4 - Apply the first order integrated rate equation.
For a gaseous first order reaction, partial pressures may be used in place of
concentrations:
$$k = \frac{2.303}{t}\log\frac{p_0}{p_A}$$

Step 5 - Substitute the values.
$$k = \frac{2.303}{30}\log\frac{0.30}{0.10} = \frac{2.303}{30}\log 3$$
$$k = \frac{2.303}{30} \times 0.48 = \frac{1.10544}{30}$$
$$k = 0.03685\ \text{s}^{-1}$$

ANSWER: $k \approx 3.68 \times 10^{-2}\ \text{s}^{-1}$
(Note the unit: for a first order reaction $k$ always has units of $\text{time}^{-1}$.)