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Standard XII

Chemistry

Characteristics of Equilibrium Constant

The value of ...

Question

The value of $K_{p}$ is $1 \times 10^{- 3} {a t m}^{- 1}$ at $25^{o} C$ for the reaction $2 N O (g) + {C l}_{2} (g) ⇌ 2 N O C l$ . A flask contains $N O$ at $0.02 a t m$ and at $25^{o} C$ . Calculate the $m o l e$ of ${C l}_{2}$ that must be added if $1$ % of the $N O$ is to be converted to $N O C l$ at equilibrium. The volume of the flask is such that $0.2 m o l e$ of gas produces $1 a t m$ pressure at $25^{o} C$ . (Ignore probable association of $N O$ in $N_{2} O_{2}$ )

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Solution

$2 N O (g) + {C l}_{2} (g) ⇌ 2 N O C l$
$t = 0$ $0.02$ $0$ $0$
$t_{e q .}$ $\frac{0.02 \times 99}{100}$ $p$ $\frac{0.02 \times 1}{100}$
$K_{p} = \frac{{[p_{N O C l}]}^{2}}{{[p_{N O}]}^{2} [{C l}_{{C l}_{2}}]} = \frac{{(0.02 \times 0.01)}^{2}}{{(0.02 \times 0.01)}^{2} \times p} = 10^{- 3}$
$p = 0.0102$ atm
$P V = n R T$
$0.102 \times V = n R T . . . (i)$
$1 \times V = 0.2 \times R \times T . . . (i i)$
From eqs (i) and (ii), $n = 0.0204$ (no. of moles of $C l$ at equilibrium)
Pressure of ${C l}_{2}$ involved in reaction
$= \frac{1}{2} \times$ pressure of $N O C l$
$= \frac{1}{2} \times \frac{0.02}{100} = 0.0001 a t m$
$P V = n R T$
$0.0001 \times V - n R T . . . . . . . . . (i i i)$
From eqs. (ii) and (iii), $n = 2 \times 10^{- 5}$ (moles of ${C l}_{2}$ involved in reaction)
Initial moles of ${C l}_{2}$ taken $= 0.0204 + 2 \times 10^{- 5} = 0.0242$

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