Question

For a first order reaction $A(g) \rightarrow 2B(g) + C(g)$, the rate constant in terms of initial pressure $p_0$ and pressure at time $t$ ($p_t$), is given by

• A. $\dfrac{1}{t}\ln\dfrac{p_0}{p_t-p_0}$
• B. $\dfrac{1}{t}\ln\dfrac{2p_0}{3p_0-p_t}$
• C. $\dfrac{1}{t}\ln\dfrac{3p_0}{p_t-p_0}$
• D. $\dfrac{1}{t}\ln\dfrac{3p_0}{3p_t-p_0}$

Hint:

In gas-phase first-order reactions, total pressure changes help track reactant loss.

Answer 1

The Correct Option is B

Step 1: Relation between pressure and extent of reaction.
For reaction: \[ A \rightarrow 2B + C \] 1 mole of A produces 3 moles of products. If $x$ is the amount decomposed: \[ p_t = p_0 + 2x \Rightarrow x = \frac{p_t - p_0}{2} \] Step 2: First-order kinetics.
\[ k = \frac{1}{t}\ln\frac{[A]_0}{[A]_t} = \frac{1}{t}\ln\frac{p_0}{p_0 - x} \] \[ = \frac{1}{t}\ln\frac{p_0}{p_0 - \frac{p_t - p_0}{2}} = \frac{1}{t}\ln\frac{p_0}{\frac{3p_0 - p_t}{2}} \] \[ = \frac{1}{t}\ln\left( \frac{2p_0}{3p_0 - p_t} \right) = \frac{1}{t}\ln\left( \frac{3p_0}{p_t - p_0} \right) \] This matches option (C).