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Standard XII

Chemistry

Characteristics of Equilibrium Constant

9.2 grams of ...

Question

9.2 grams of $N_{2} O_{4} (g)$ is taken in a closed one litre vessel and heated till the following equilibrium is reached, $N_{2} O_{4} (g) ⇌ {2 N O}_{2} (g)$ . At equilibrium, 50% $N_{2} O_{4} (g)$ is dissociated. What is the equilibrium constant? (in $M$ ) (molecular weight of $N_{2} O_{4} = 92$ )

0.1

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0.2

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Solution

The correct option is D 0.2
Given : $α = 50 % = 0.5$

$m = 9.2 g$ ; $M = 92 g / m o l$

$c = \frac{m}{M} \times \frac{1}{V}$

$= \frac{9.2}{92} \times \frac{1}{1} = 0.1 M$

$[N_{2} O_{4}]_{e q} = \frac{1}{2} (0.1) = 0.05 m o l L^{1}$

$[N O_{2}]_{e q} = 2 \times 0.05 = 0.1 m o l L^{1}$

Equilibrium constant $K_{e} = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$

$K_{e} = \frac{(0.1)^{2}}{(0.05)} = 0.2 m o l L^{- 1}$

Answer D) 0.2

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9.2 grams of N2O4(g) is taken in a closed one litre vessel and heated till the following equilibrium is reached, N2O4(g)⇌2NO2(g). At equilibrium, 50% N2O4(g) is dissociated. What is the equilibrium constant? (in M) (molecular weight of N2O4=92)

The correct option is D 0.2Given : α=50%=0.5m=9.2g; M=92g/molc=mM×1V=9.292×11=0.1M[N2O4]eq=12(0.1)=0.05molL1[NO2]eq=2×0.05=0.1molL1Equilibrium constant Ke=[NO2]2[N2O4]Ke=(0.1)2(0.05)=0.2 molL−1Answer D) 0.2