Decompasition of A is a first order reaction at T(K) and is given by $$A(g) \rightarrow B(g)+C(g)$$.
In a closed 1 L vessel, 1 bar A(g) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5bar. What is the rate constant (in $$min^{-1})$$ of the reaction ? (log 2 = 0.3)

First order decomposition: $$A(g) \to B(g) + C(g)$$

Initial: $$P_A = 1$$ bar. At time t: $$P_A = 1 - x$$, $$P_B = x$$, $$P_C = x$$.

Total pressure = $$1 - x + x + x = 1 + x = 1.5$$ bar. So $$x = 0.5$$.

$$P_A = 1 - 0.5 = 0.5$$ bar at t = 100 min.

For first order: $$k = \frac{2.303}{t}\log\frac{P_0}{P_A} = \frac{2.303}{100}\log\frac{1}{0.5} = \frac{2.303 \times 0.3}{100} = \frac{0.6909}{100} = 6.9 \times 10^{-3}$$ min⁻¹

The answer is Option 3: $$6.9 \times 10^{-3}$$ min⁻¹.