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Le Chatelier's Principle for Del N Equals to 0

For the react...

Question

For the reaction at equilibrium: $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ , the concentration of $N_{2} O_{4}$ and $N O_{2}$ at equilibrium are $4.8 \times 10^{- 2}$ and $1.2 \times 10^{- 2}$ mol/L respectively. The value of $K_{c}$ for the reaction is:

A

3 \times 10^{- 3} M

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B

3 \times 10^{3} M

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C

3.3 \times 10^{2} M

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D

3 \times 10^{- 1} M

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Solution

The correct option is A $3 \times 10^{- 3} M$
The equilibrium reaction is $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ .
The expression for the equilibrium constant is $K_{c} = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]} = \frac{(1.2 \times 10^{- 2})^{2}}{4.8 \times 10^{- 2}} = 3 \times 10^{- 3} M$ .

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