For A(g) $\longrightarrow$ B(g) + C(g) at constant volume the first-order law in pressures (Eq. 6.13) is $k = \dfrac{2.303}{t}\log_{10}\dfrac{P_i}{2P_i - P}$, with $P_i$ the initial and $P$ the total pressure at time t. …

Step 1. For the first order gas-phase decomposition at constant volume, the chapter's Eq. (6.13) gives $k = \dfrac{2.303}{t}\log_{10}\dfrac{P_i}{2P_i - P}$, where 2$P_i$ - P is the reactant's partial pressure at time t.

Step 2. From the data, $P_i$ = 0.5 bar and P = 0.6 bar at t = 100 s, so 2$P_i$ - P = 2 $\times$ 0.5 - 0.6 = 0.4 bar. …

Apply the pressure form of the first-order integrated law (Eq. 6.13): the reactant's partial pressure at time t is …

• Plugging the TOTAL pressure P = 0.6 bar in as the reactant pressure -- the reactant's partial pressure is 2Pi - P = 0.4 bar.
• Using P/Pi instead of Pi/(2Pi - P) inside the logarithm. …