At $$700 \mathrm{~K}$$, the Equilibrium constant value for the formation of $$\mathrm{HI}$$ from $$\mathrm{H}_2$$ and $$\mathrm{I}_2$$ is 49.0 . 0.7 mole of $$\mathrm{HI}(\mathrm{g})$$ is present at equilibrium. What will be the concentrations of $$\mathrm{H}_2$$ and $$\mathrm{I}_2$$ gases if we initially started with $$\mathrm{HI}(\mathrm{g})$$ and allowed the reaction to reach equilibrium at the same temperature?

Consider the following equilibrium,

$$\begin{aligned} & 2 \mathrm{No}(g) \rightleftharpoons \mathrm{N}_2+\mathrm{O}_2 ; \mathrm{K}_{\mathrm{G}}=2.4 \times 10^{20} \\ & \mathrm{No}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NoBr}(\mathrm{g}) ; \mathrm{K}_{\mathrm{C}_2}=1.4 \end{aligned}$$

Calculate $$K_C$$ for the reaction,

$$\frac{1}{2} \mathrm{~N}_2(g)+\frac{1}{2} \mathrm{O}_2(g)+\frac{1}{2} \mathrm{Br}_2(g) \rightleftharpoons \mathrm{NOBr}(g)$$

The equilibrium constants for the reactions $$a, b$$, and $$c$$ are as given:

a) $$\mathrm{N}_2+3 \mathrm{H}_2=2 \mathrm{NH}_3: \mathbf{K}_1$$

b) $$\mathrm{N}_2+\mathrm{O}_2=2 \mathrm{NO}: \mathrm{K}_2$$

c) $$2 \mathrm{H}_2+\mathrm{O}_2=2 \mathrm{H}_2 \mathrm{O}: \mathbf{K}_3$$

What would be the Equilibrium constant for the reaction:

$$4 \mathrm{NH}_3+5 \mathrm{O}_2=4 \mathrm{NO}+6 \mathrm{H}_2 \mathrm{O} ; \mathbf{K}_{\mathbf{x}}$$

For the reaction, H$$_2$$(g) + I$$_2$$(g) $$\rightleftharpoons$$ 2HI(g) the position of equilibrium can be shifted to the right by