Question

Which of the following represents integrated rate law equation for gas phase first order reaction, $\text{A}_{(\text{g})} \rightarrow \text{B}_{(\text{g})} + \text{C}_{(\text{g})}$ if $P_i =$ initial pressure of A and $P =$ total pressure of reaction mixture at time $t$?

• A. $k = \frac{2.303}{t} \times \log_{10} \frac{P_i}{2P_i - P}$
• B. $k = \frac{2.303}{t} \times \log_{10} \frac{P_i}{2P_i - P}$
• C. $k = \frac{1}{t} \ln \frac{2P_i - P}{P_i}$
• D. $k = \frac{2.303}{t} \times \log_{10} \frac{P_i - P}{P_i}$

Hint:

For any gas phase first-order decomposition of the type $\text{A} \rightarrow n\text{B} + \text{C}$, the denominator term representing the remaining reactant pressure will always take the general pattern of $[P_i(n - 1) - P]$. For a simple 1-to-2 molecule splitting, it simplifies perfectly to $2P_i - P$!

Answer 1

The Correct Option is B

Step 1: Start from the first-order law.
For $\text{A} \rightarrow$ products, $k = \frac{2.303}{t}\log_{10}\frac{P_i}{P_A}$, where $P_A$ is the pressure of A still present at time $t$.
Step 2: Set up a pressure table.
Initially A is $P_i$ and B, C are zero. After loss $x$, A is $P_i - x$, B is $x$, and C is $x$.
Step 3: Write the total pressure.
$P = (P_i - x) + x + x = P_i + x$.
Step 4: Solve for the progress variable.
Rearranging, $x = P - P_i$.
Step 5: Find the remaining pressure of A.
$P_A = P_i - x = P_i - (P - P_i) = 2P_i - P$.
Step 6: Substitute into the rate law.
$k = \frac{2.303}{t}\log_{10}\frac{P_i}{2P_i - P}$.
Step 7: Match the option.
This matches the form in option (2).
\[ \boxed{k = \dfrac{2.303}{t}\log_{10}\dfrac{P_i}{2P_i - P}} \]