Question

A(g) $ \rightarrow $ B(g) + C(g) is a first order reaction.

The reaction was started with reactant A only. Which of the following expression is correct for rate constant k ?

• A. \( k = \frac{1}{t} \ln \frac{2(P_\infty - P_t)}{P_t} \)
• B. \( k = \frac{1}{t} \ln \frac{P_\infty}{P_t} \)
• C. \( k = \frac{1}{t} \ln \frac{P_\infty}{2(P_\infty - P_t)} \)
• D. \( k = \frac{1}{t} \ln \frac{P_\infty}{(P_\infty - P_t)} \)

Hint:

For gas-phase reactions where pressure is measured, it's often convenient to express the rate law and integrated rate law in terms of partial pressures. Remember to use the stoichiometry of the reaction to relate the change in pressure of the reactant to the total pressure of the system at any time.

Answer 1

The Correct Option is C

Step 1: Set up the stoichiometry and partial pressures.
Let the initial pressure of A be \( P_0 \) at time \( t = 0 \). 
Since only A is present initially, \( P_t \) at \( t=0 \) is \( P_0 \). 
At time \( t \), let the pressure of A reacted be \( p \). 
Then the partial pressures of A, B, and C at time \( t \) are: \( P_A = P_0 - p \) \( P_B = p \) \( P_C = p \) 
The total pressure of the system at time \( t \) is \( P_t = P_A + P_B + P_C = (P_0 - p) + p + p = P_0 + p \). 
From this, we get \( p = P_t - P_0 \). So, \( P_A = P_0 - (P_t - P_0) = 2P_0 - P_t \). At time \( t = \infty \), the reaction goes to completion, so the pressure of A becomes zero. \( P_A(\infty) = 0 \) \( P_B(\infty) = P_0 \) \( P_C(\infty) = P_0 \) 
The total pressure at \( t = \infty \) is \( P_\infty = P_A(\infty) + P_B(\infty) + P_C(\infty) = 0 + P_0 + P_0 = 2P_0 \). 
From this, we have \( P_0 = P_\infty / 2 \). 
Now, substitute \( P_0 \) in the expression for \( P_A \): \( P_A = 2(P_\infty / 2) - P_t = P_\infty - P_t \). 
Step 2: Apply the first-order rate law in terms of partial pressure.
For a first-order reaction \( A \rightarrow products \), the rate law is \( -\frac{dP_A}{dt} = k P_A \). Integrating this equation from \( t = 0 \) to \( t \) and from \( P_A(0) = P_0 \) to \( P_A(t) \): \[ \int_{P_0}^{P_A} \frac{dP_A}{P_A} = -k \int_{0}^{t} dt \] \[ \ln \frac{P_A}{P_0} = -kt \] \[ k = \frac{1}{t} \ln \frac{P_0}{P_A} \] 
Step 3: Substitute the expressions for \( P_0 \) and \( P_A \) in terms of \( P_t \) and \( P_\infty \).
We have \( P_0 = P_\infty / 2 \) and \( P_A = P_\infty - P_t \). Substituting these into the rate constant expression: \[ k = \frac{1}{t} \ln \frac{P_\infty / 2}{P_\infty - P_t} \] \[ k = \frac{1}{t} \ln \frac{P_\infty}{2(P_\infty - P_t)} \] 
Step 4: Match the derived expression with the given options.
The derived expression for the rate constant \( k \) matches option (3).