Question

Integrated rate law equation for a first order gas phase reaction is given by (where \(P_i\)​ is initial pressure and \(P_t\)​ is total pressure at time t)

• A. \( k = \frac{2.303}{t} \times \log \frac{P_i}{(2P_i - P_t)} \)
• B. \( k = \frac{2.303}{t} \times \log \frac{2P_i}{(2P_i - P_t)} \)
• C. \( k = \frac{2.303}{t} \times \log \frac{(2P_i - P_t)}{P_i} \)
• D. \( k = \frac{2.303}{t} \times \log \frac{P_i}{(2P_i - P_t)} \)

Answer 1

The Correct Option is A

The integrated rate law for a first-order reaction can be expressed in terms of pressure, particularly for a gas-phase reaction. Let's derive and understand the equation. 

For a first-order reaction, the general form of the rate law in terms of concentration is:

\(k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}\)

In a gas-phase reaction, the concentration terms can be expressed in terms of pressure:

• \([A]_0\) can be substituted as the initial pressure, \(P_i\).
• The concentration \([A]\) at time \(t\) can be related to the total pressure \(P_t\).

For a first-order reaction where the stoichiometry is \(A \rightarrow B\), the relationship between pressures is given by:

\(P_A = (2P_i - P_t)\)

This indicates that at any time \(t\), the partial pressure of \(A\) is \((2P_i - P_t)\).

Substituting these into the integrated rate equation, we get:

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

Thus, the correct answer is:

Option 1: \( k = \frac{2.303}{t} \times \log \frac{P_i}{(2P_i - P_t)} \)

Let's rule out other options:

• Option 2 and 4 provide the same equation as Option 1, which is indeed correct.
• Option 3 suggests the inversion of the log term, which would not be true for this scenario.

Answer 2

Consider the reaction:

\[ A \rightarrow B + C \]

Initial pressures:

\[ P_i \quad 0 \quad 0 \]

After reaction:

\[ P_i - x \quad x \quad x \]

Total pressure at time \(t\):

\[ P_t = P_i + x \]

Therefore:

\[ P_i - x = P_i - P_t + P_i \] \[ = 2P_i - P_t \]

Hence,

\[ k = \frac{2.303}{t}\times \log \frac{P_i}{2P_i - P_t} \]