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: Understanding the Question:
The question requires us to determine the integrated rate equation for a first-order gas-phase reaction where a single gaseous reactant decomposes into two gaseous products. The expression must be formulated in terms of initial pressure ($P_i$) and total pressure ($P$) at an elapsed time $t$.

Step 2: Key Formula or Approach:
The standard first-order integrated rate law equation is:
$$k = \frac{2.303}{t} \log_{10} \left( \frac{[\text{A}]_0}{[\text{A}]_t} \right) = \frac{2.303}{t} \log_{10} \left( \frac{P_i}{P_A} \right)$$ Where $P_A$ represents the partial pressure of reactant A remaining at time $t$.

Step 3: Detailed Explanation:
Let's set up an atmospheric pressure ledger for the reaction:
$$ \begin{array}{lccccc} & \text{A}_{(\text{g})} & \longrightarrow & \text{B}_{(\text{g})} & + & \text{C}_{(\text{g})} \\ \text{Initial pressure (at } t = 0\text{):} & P_i & & 0 & & 0 \\ \text{Pressure at time } t\text{:} & P_i - x & & x & & x \end{array} $$ The total pressure $P$ inside the vessel at time $t$ is the sum of all individual partial pressures:
$$P = P_A + P_B + P_C$$ $$P = (P_i - x) + x + x$$ $$P = P_i + x$$ Rearranging this to solve explicitly for the progress variable $x$:
$$x = P - P_i$$ Now, substitute this expression for $x$ back to evaluate the remaining partial pressure of reactant A ($P_A$) at time $t$:
$$P_A = P_i - x$$ $$P_A = P_i - (P - P_i)$$ $$P_A = 2P_i - P$$ Substitute $P_A$ into the core first-order rate formula:
$$k = \frac{2.303}{t} \log_{10} \left( \frac{P_i}{2P_i - P} \right)$$ This corresponds perfectly to option (B).

Step 4: Final Answer:
The correct integrated rate law equation is $k = \frac{2.303}{t} \times \log_{10} \frac{P_i}{2P_i - P}$, corresponding to option (B).