Medium
At equilibrium,the concentrations of $N_{2} = 3.0 \times 10^{-3} \, M$,$O_{2} = 4.2 \times 10^{-3} \, M$ and $NO = 2.8 \times 10^{-3} \, M$ in a sealed vessel at $800 \, K$. What will be $K_{c}$ for the reaction
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$
(0.622) The equilibrium constant $K_{c}$ for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ is given by the expression:
$K_{c} = \frac{[NO]^{2}}{[N_{2}][O_{2}]}$
Substituting the given equilibrium concentrations:
$K_{c} = \frac{(2.8 \times 10^{-3})^{2}}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}$
$K_{c} = \frac{7.84 \times 10^{-6}}{12.6 \times 10^{-6}}$
$K_{c} = 0.622$
$K_{c} = \frac{[NO]^{2}}{[N_{2}][O_{2}]}$
Substituting the given equilibrium concentrations:
$K_{c} = \frac{(2.8 \times 10^{-3})^{2}}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}$
$K_{c} = \frac{7.84 \times 10^{-6}}{12.6 \times 10^{-6}}$
$K_{c} = 0.622$
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