Question

First order gas phase reaction \(A \to B + C\). \(p_i = \text{initial pressure of gas A}\), \(P_t = \text{total pressure of the reaction mixture at time t}\). Expression of rate constant (k) is

• A. \(\frac{1}{t} \ln \frac{p_i}{2p_i - P_t}\)
• B. \(\frac{1}{t} \ln \frac{2p_i}{p_i - P_t}\)
• C. \(\frac{1}{t} \ln \frac{p_i}{3p_i - 2P_t}\)
• D. \(\frac{1}{t} \ln \frac{3p_i}{4p_i - P_t}\)

Answer 1

The Correct Option is A

To find the expression for the rate constant (\(k\)) for a first-order gas-phase reaction \(A \to B + C\), we can follow these steps:

This is a first-order reaction, and the expression for the rate constant for a first-order reaction is based on the change in pressure over time.

1. Initial Conditions: Initially, only gas A is present with pressure \(p_i\). Therefore, the total initial pressure is \(p_i\).
2. At Time t: Let the change in pressure due to the conversion of A to B and C be \(x\). The total pressure at any time \(t\) is given by: \(P_t = p_i - x + x + x = p_i + x\), as the stoichiometry of the reaction shows that A converts into two moles of products. Therefore, the increase in pressure is equal to \(x\).
3. Using the total pressure: At time \(t\), the expression for total pressure can be rearranged to find \(x\): \(x = P_t - p_i\).
4. Finding the Expression for the Rate Constant: The concentration of A at time \(t\) is related to its pressure and is given by \(p_i - x = 2p_i - P_t\).
5. Form of the First-Order Rate Constant: The rate constant \(k\) for a first-order reaction is expressed as: \(k = \frac{1}{t} \ln \frac{\text{Initial Pressure}}{\text{Pressure at time } t}\).
6. Substituting Values: Hence, \(k = \frac{1}{t} \ln \left( \frac{p_i}{2p_i - P_t} \right)\).

Thus, the correct expression for the rate constant \(k\) is \(\frac{1}{t} \ln \frac{p_i}{2p_i - P_t}\).