First order gas phase reaction A arrow B + C p_i = initial pressure...

The given reaction is $A(g) \rightarrow B(g) + C(g)$

At $t = 0$, the initial pressure of A is $p_i$ and the pressures of B and C are $0$.

At time $t$, let the decrease in pressure of A be $x$.
Pressure of A $= p_i - x$
Pressure of B $= x$
Pressure of C $= x$

Total pressure at time $t$ is $p_t = (p_i - x) + x + x = p_i + x$

From this, we get $x = p_t - p_i$

The partial pressure of A at time $t$ is $P_A = p_i - x = p_i - (p_t - p_i) = 2p_i - p_t$

For a first order reaction, the rate constant $k$ is given by

$k = \dfrac{1}{t}\ln\dfrac{P_{A,0}}{P_A}$

Substituting the values, we get

$k = \dfrac{1}{t}\ln\dfrac{p_i}{2p_i - p_t}$

Answer: $\dfrac{1}{t}\ln\dfrac{p_i}{2p_i - p_t}$