Consider the following first-order gas-phase reaction: $A_{(g)} \to B_{(g)} + C_{(g)}$. At time $t$,the total pressure is $p_t \ atm$. Derive the integrated rate equation for this reaction.

(N/A) The reaction is: $A_{(g)} \to B_{(g)} + C_{(g)}$
| Time | $A_{(g)}$ | $B_{(g)}$ | $C_{(g)}$ |
| :--- | :--- | :--- | :--- |
| $t = 0$ | $p_i \ atm$ | $0 \ atm$ | $0 \ atm$ |
| $t = t$ | $(p_i - x) \ atm$ | $x \ atm$ | $x \ atm$ |
Here,$p_i$ is the initial pressure of $A$ at $t = 0$,and $x$ is the decrease in pressure of $A$ at time $t$.
The total pressure $p_t$ at time $t$ is given by the sum of partial pressures:
$p_t = (p_i - x) + x + x = p_i + x$
From this,we can express $x$ in terms of $p_t$ and $p_i$:
$x = p_t - p_i$
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 integrated rate equation is:
$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$
Substituting the partial pressures for concentrations:
$k = \frac{2.303}{t} \log \frac{p_i}{p_A} = \frac{2.303}{t} \log \frac{p_i}{2p_i - p_t}$