Question

For a first order reaction : \[ \text{A(g)} \rightarrow \text{B(g)} + \text{C(g)} \] If initial pressure is \( P_0 \) and total pressure at time \( t \) is \( P_t \), then the expression of rate constant \( K \) is:

• A. \( K = \frac{1}{t}\ln\left(\frac{P_0}{P_0 - P_t}\right) \)
• B. \( K = \frac{1}{t}\ln\left(\frac{P_0}{2P_0 - P_t}\right) \)
• C. \( K = \frac{1}{t}\ln\left(\frac{2P_0}{P_0 - P_t}\right) \)
• D. \( K = \frac{1}{t}\frac{2P_0}{2P_0 - P_t} \)

Answer 1

The Correct Option is B

Concept:
For a first-order reaction: \[ K = \frac{1}{t}\ln\left(\frac{[\text{Initial}]}{[\text{Remaining}]}\right) \] When pressure is used instead of concentration, the same formula applies since pressure is proportional to concentration. Step 1: Set up the reaction and pressures.
\[ \text{A} \rightarrow \text{B} + \text{C} \] At \( t = 0 \): \[ P_0 \quad 0 \quad 0 \] At time \( t \): \[ P_0 - x \quad x \quad x \] Step 2: Express total pressure.
\[ P_t = (P_0 - x) + x + x = P_0 + x \] \[ \Rightarrow x = P_t - P_0 \] Step 3: Apply first-order rate law.
\[ K = \frac{1}{t}\ln\left(\frac{P_0}{P_0 - x}\right) \] Substitute \( x = P_t - P_0 \): \[ K = \frac{1}{t}\ln\left(\frac{P_0}{P_0 - (P_t - P_0)}\right) \] \[ K = \frac{1}{t}\ln\left(\frac{P_0}{2P_0 - P_t}\right) \]