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Standard XII

Chemistry

Le Chatelier's Principle for Del N Greater Than Zero

9.2g of N2O4g...

Question

$9.2 g$ 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 $50 %$ of $N_{2} O_{4} (g)$ is dissociated. What is the equilibrium constant?
[in ${mole lit}^{- 1}$ ] [M.wt. of $N_{2} O_{4} is 92$ ]

0.1

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Solution

The correct option is B $0.2$
$9.2 g$ of $N_{2} O_{4} (g)$ corresponds to $\frac{9.2}{92} = 0.1 moles$
Out of these $0.05 moles$ ( $50 % of 0.1 moles)$ dissociate to form $2 \times 0.05 = 0.1 moles of N O_{2} (g)$
The equilibrium constant is $K_{c} = \frac{[N O_{2}]^{2}}{N_{2} O_{4}} = \frac{(0.1)^{2}}{0.05} = 0.2$
Hence the correct option is B.

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9.2g 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% of N2O4(g) is dissociated. What is the equilibrium constant? [in mole lit−1] [M.wt. of N2O4 is 92]

The correct option is B 0.29.2 g of N2O4(g) corresponds to 9.292=0.1 moles Out of these 0.05 moles (50% of 0.1 moles) dissociate to form 2×0.05=0.1 moles of NO2(g) The equilibrium constant is Kc=[NO2]2N2O4=(0.1)20.05=0.2 Hence the correct option is B.