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)
- \( k = \frac{2.303}{t} \times \log \frac{P_i}{(2P_i - P_t)} \)
- \( k = \frac{2.303}{t} \times \log \frac{2P_i}{(2P_i - P_t)} \)
- \( k = \frac{2.303}{t} \times \log \frac{(2P_i - P_t)}{P_i} \)
- \( k = \frac{2.303}{t} \times \log \frac{P_i}{(2P_i - P_t)} \)
The Correct Option is A
Approach Solution - 1
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.
Approach Solution -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,
\[ k = \frac{2.303}{t}\times \log \frac{P_i}{2P_i - P_t} \]
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