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Derivation of Kp and Kc

The decomposi...

Question

The decomposition of $N_{2} O_{4}$ to ${N O}_{2}$ is carried at $280 K$ in chloroform. When equilirbrium has been established, $0.2$ mole of $N_{2} O_{4}$ and $2 \times 10^{- 3}$ mole of ${N O}_{2}$ are present in $2$ litre solution. The equilibrium constant for the reaction,
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ is:

A

1 \times 10^{- 2}

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B

2 \times 10^{- 3}

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C

1 \times 10^{- 5}

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D

2 \times 10^{- 5}

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Solution

The correct option is B $1 \times 10^{- 5}$
The decomposition of $N_{2} O_{4}$ to ${N O}_{2}$ is carried at $280 K$ in chloroform. When equilibrium has been established, $0.2$ mole of $N_{2} O_{4}$ and $2 \times 10^{- 3}$ mole of ${N O}_{2}$ are present in $2$ litre solution.
$[N_{2} O_{4}] = \frac{0.2 m o l}{2 L} = 0.1 M$
$[N O_{2}] = \frac{2 \times 10^{- 3} m o l}{2 L} = 1 \times 10^{- 3} M$
$K_{c} = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$
$K_{c} = \frac{(1 \times 10^{- 3})^{2}}{0.1}$

$K_{c} = 1 \times 10^{- 5}$
The equilibrium constant for the reaction,

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ is $1 \times 10^{- 5}$

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