1. Identify the reaction type

For the given first-order gas phase reaction, the pressure form of the integrated rate law is typically derived for a reaction like:

$A(g) \rightarrow 2B(g)$

Let initial pressure of $A$ be $P_i$.

At time $t$, suppose pressure decrease due to decomposition is $x$.

Then:

• Pressure of $A$ at time $t$ = $P_i - x$
• Pressure of $B$ formed = $2x$

So total pressure at time $t$ is:

$P_t = (P_i - x) + 2x = P_i + x$

Thus,

$x = P_t - P_i$

Hence pressure of reactant $A$ left at time $t$ is:

$P_A = P_i - x = P_i - (P_t - P_i) = 2P_i - P_t$

2. Use the first-order integrated rate law

For a first-order reaction:

$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$

For gases at constant temperature, concentration is proportional to partial pressure, so:

$k = \frac{2.303}{t} \log \frac{P_i}{P_A}$

Substitute $P_A = 2P_i - P_t$:

$k = \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t}$

3. Match with the options

• A: $k=\frac{2.303}{t} \times \log \frac{P_i}{(2P_i-P_t)}$ ✅
• B: inverse of correct ratio ❌
• C: missing logarithm ❌
• D: numerator should be $P_i$, not $2P_i$ ❌

4. Final answer

The correct integrated rate law is:

$\boxed{k=\frac{2.303}{t}\log\frac{P_i}{2P_i-P_t}}$

So the correct option is A.