Question

The initial pressure of the system for the reaction \[ A(g) \rightarrow B(g)+C(g) \] was \(P_i\). Total pressure at time \(t\) is \(P_t\). The rate constant \(k\) is:

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

Hint:

Always calculate the partial pressure of the reactant first before substituting into the first-order rate equation.

Answer 1

The Correct Option is A

Concept: For gaseous first-order reactions, pressure terms can be directly used in place of concentration because pressure is proportional to concentration.

Step 1: Assume decomposition of \(A\). Initially, \[ \text{Pressure of }A = P_i \] Suppose pressure decrease due to decomposition is \(x\). Then at time \(t\), \[ A = P_i-x \] Since one mole of \(A\) forms one mole each of \(B\) and \(C\), \[ B=x,\qquad C=x \]

Step 2: Calculate total pressure at time \(t\). Total pressure becomes: \[ P_t=(P_i-x)+x+x \] \[ P_t=P_i+x \] Thus, \[ x=P_t-P_i \]

Step 3: Find remaining pressure of reactant \(A\). \[ P_A=P_i-x \] Substituting \(x=P_t-P_i\), \[ P_A=P_i-(P_t-P_i) \] \[ P_A=2P_i-P_t \]

Step 4: Apply first-order rate equation. For first-order reactions: \[ k=\frac{2.303}{t}\log\frac{P_i}{P_A} \] Substituting \(P_A=2P_i-P_t\), \[ k=\frac{2.303}{t}\log\frac{P_i}{2P_i-P_t} \] Hence, option (1) is correct.