Consider the following reaction in a sealed vessel at equilibrium with concentrations of $$\mathrm{N}_2=3.0 \times 10^{-3} \mathrm{M}, \mathrm{O}_2=4.2 \times 10^{-3} \mathrm{M}$$ and $$\mathrm{NO}=2.8 \times 10^{-3} \mathrm{M}$$.

$$2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$$

If $$0.1 \mathrm{~mol} \mathrm{~L} \mathrm{~L}^{-1}$$ of $$\mathrm{NO}_{(\mathrm{g})}$$ is taken in a closed vessel, what will be degree of dissociation ($$\alpha$$) of $$\mathrm{NO}_{(\mathrm{g})}$$ at equilibrium?

For a weak acid HA, the percentage of dissociation is nearly 1% at equilibrium. If the concentration of acid is 0.1 mol L$$^{-1}$$, then the correct option for its K$$_a$$ at the same temperature is :

K_p for the following reaction is 3.0 at 1000 K.

CO₂(g) + C(s) $$\rightleftharpoons$$ 2CO(g)

What will be the value of K_c for the reaction at the same temperature?

(Given : R = 0.083 L bar K^$$-$$1 mol^$$-$$1)

3O₂(g) $$\rightleftharpoons$$ 2O₃(g)

for the above reaction at 298 K, K_c is found to be 3.0 $$\times$$ 10^$$-$$59. If the concentration of O₂ at equilibrium is 0.040 M then concentration of O₃ in M is