At 200^{circ} C,nitric oxide reacts with oxyg | vedclass

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(C) The equilibrium constant expression for the reaction $2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g)$ is given by $K_C = \frac{[NO_2]^2}{[NO]^2 [O_

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The concentrations of either $NO$ or $O_2$ (or possibly both) will be much smaller than the concentration of $NO_2$.

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(C) The equilibrium constant expression for the reaction $2 NO(g) + O_2(g) ightleftharpoons 2 NO_2(g)$ is given by $K_C = \frac{[NO_2]^2}{[NO]^2 [O_2]}$.Given $K_C = 3 	imes 10^6$,which is a very large value $(K_C \gg 1)$.$A$ large value of $K_C$ indicates that the equilibrium position lies far to the right,meaning the concentration of the product $(NO_2)$ is much higher than the concentrations of the reactants ($NO$ and $O_2$) at equilibrium.Therefore,the concentrations of either $NO$ or $O_2$ (or possibly both) will be much smaller than the concentration of $NO_2$.

At $200^{\circ} C$,nitric oxide reacts with oxygen to form nitrogen dioxide as follows: $2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g)$,$K_C = 3 \times 10^6$. In a mixture of the three species at equilibrium,we can accurately predict that:

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