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Equilibrium Constant and Standard Free Energy Change

One mole of ...

Question

One mole of $N_{2} O_{4}$ in a $1 L$ flask decomposes to attain the equilibrium $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
At the equilibrium the mole fraction of $N O_{2}$ is $1 / 2$ . Hence $K_{C}$ will be :

A

1 / 3

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B

1 / 2

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C

2 / 3

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D

1

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Solution

The correct option is C $2 / 3$

$N_{2} O_{4} - - - \to 2 N O_{2}$

$t = 01 m o l 0$

$t = t_{e q} . 1 - x 2 x$

mole fraction of $N O_{2}$ = $2 x / (1 - x + 2 x) = 2 x / (x + 1) = 1 / 2$

solving, we get 4x=x+1

$3 x = 1$

$x = 1 / 3$

$= (4 / 9) / (2 / 3) = 4 X 3 / 9 X 2 = 2 / 3$

Hence,2/3 is the answer.

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