Ncert Exercises & Solutions

Consider the first-order gas-phase decomposition reaction given below.

A(g) → B(g) + C(g)

The initial pressure of the system before the decomposition of A was $P_{i}$ . After the lapse of time t, the total pressure of the system increased by X units and became $P_{t}$ . The rate constant k for the reaction is:

1. $k = \frac{2 .303}{t} \log \frac{P_{i}}{P_{i} - x}$

2. $k = \frac{2 .303}{t} \log \frac{P_{i}}{2 P_{i} - P_{t}}$

3. $k = \frac{2 .303}{t} \log \frac{P_{i}}{2 P_{i} + P_{t}}$

4. $k = \frac{2 .303}{t} \log \frac{P_{i}}{P_{i} + x}$

HINT: Use First order gas-phase reaction

Explanation:

Initially At time t

$A (g)$ $\to B (g)$ $+$ $C (g)$
$P_{i}$ $0$ $0$
$P_{i} - x$ $x$ $x$

For first-order reaction

$P_{t}$ $= P_{i}$ $- x$ $+$ $x$ $= x$ $=$ $P_{i}$ $+$ $x$
$x$ $=$ $P_{t}$ $-$ $P_{i}$
$k$ $=$ $\frac{2.303}{t} \log$ $\frac{P_{i}}{P_{i} - x}$
$= \frac{2.303}{t} \log \frac{P_{i}}{P_{i} -(P_{t} - P_{i})}$
$= \frac{2.303}{t} \log$ $\frac{P_{i}}{2 P_{i}}$ $- P_{t}$

Hence, option second is the correct answer.

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