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 :
Kp for the following reaction is 3.0 at 1000 K.
CO2(g) + C(s) $$\rightleftharpoons$$ 2CO(g)
What will be the value of Kc 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, Kc 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