The following water gas shift reaction is an important industrial process for the production of hydrogen gas.
$\mathrm{CO}\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$
At a given temperature ${\mathrm{K}}_{\mathrm{p}}=2.7$. If $0.13 \mathrm{mol}$ of $\mathrm{CO},0.56 \mathrm{mol}$ of water, $0.78 \mathrm{mol}$ of ${\mathrm{CO}}_2$ and $0.28 \mathrm{mol}$ of ${\mathrm{H}}_2$ are introduced into a $2 \mathrm{L}$ flask, and find out in which direction must the reaction proceed to reach equilibrium.

1. Physical and chemical equilibrium:

(i) In equilibrium state all the properties of the system like temperature, pressure and composition become constant and no change seems to occur in it.

(ii) In a state of static equilibrium, ‘no change’ at all occurs in the system. This type of equilibrium is attained when different kinds of forces acting on the system balance each other and no ‘net’ force acts on it.

(iii) In a state of dynamic equilibrium, two opposite changes occur in the system simultaneously and at the same rate, which results in no ‘net change’ occurring in the system.

(iv) When equilibrium is reached in a physical process, it is called physical equilibrium.

(v) When equilibrium is reached in a chemical process, it is called chemical equilibrium.

2. Kp and Kc relation:

(i) For a general gas phase reaction $\mathrm{mA}\left(\mathrm{g}\right)+\mathrm{nB}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{xC}\left(\mathrm{g}\right)+\mathrm{yD}\left(\mathrm{g}\right)$ the expressions for ${\mathrm{K}}_{\mathrm{c}}$and ${\mathrm{K}}_{\mathrm{p}}$ are ${\mathrm{K}}_{\mathrm{c}}=\frac{\left[\mathrm{C}{]}^{\mathrm{x}}\right[\mathrm{D}{]}^{\mathrm{y}}}{\left[\mathrm{A}{]}^{\mathrm{n}}\right[\mathrm{B}{]}^{\mathrm{n}}}$ and the ${\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{p}}_{\mathrm{C}}^{\mathrm{x}}\times {\mathrm{p}}_{\mathrm{D}}^{\mathrm{y}}}{{\mathrm{p}}_{\mathrm{A}}^{\mathrm{m}}\times {\mathrm{p}}_{\mathrm{B}}^{\mathrm{n}}}\mathrm{ }$

where $\left[\mathrm{A}\right], \left[\mathrm{B}\right], \left[\mathrm{C}\right]$ and $\left[\mathrm{D}\right]$ are the equilibrium concentrations of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ respectively and ${\mathrm{P}}_{\mathrm{A}}, {\mathrm{P}}_{\mathrm{B}}, {\mathrm{P}}_{\mathrm{C}}$and ${\mathrm{P}}_{\mathrm{D}}$ are the equilibrium partial pressures of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ respectively and $\mathrm{m}, \mathrm{n}, \mathrm{x}$ and $\mathrm{y}$ are their respective stoichiometric coefficients in the balanced chemical equation representing the reaction.

(ii) Kp=Kc(RT)Δngwhere, ${\mathrm{\Delta n}}_{\mathrm{g}} =$ change in number of moles of gases during the reaction.

3. Heterogeneous and homogeneous equilibria:

If an equilibrium is established in a reaction in which all the reactants and products are present in the same phase, it is called homogeneous equilibrium and if present in more than one phase, it is called heterogeneous equilibrium.

4. Relationship between equilibrium constant, reaction quotient and gibbs energy:

If reaction quotient value is less than the corresponding equilibrium constant, i.e.,

Qc<Kc or Qp<Kp, it indicates that the reaction must be proceeding in the forward direction.

${\mathrm{Q}}_{\mathrm{c}}$ and ${\mathrm{K}}_{\mathrm{p}}$ of a reaction are related to Gibbs energy of the reaction, ${\mathrm{\Delta }}_{\mathrm{r}}\mathrm{G}$ and standard Gibbs energy of the reaction, ${\mathrm{\Delta rG}}^{⊝}$ as

ΔrG!=-RTlnKp

ΔrG=ΔrG!+RTlnQp

and ΔrG=RTlnQpKp

5. Le Chatelier’s principle:

Le Chatelier’s principle states, “If a system in equilibrium is subjected to a change of concentration, temperature or pressure, equilibrium shifts in the direction that tends to undo the effect of the change imposed.”

6. The relation between standard free energy change $(\mathrm{\Delta G}°)$ and equilibrium constant:

The relation between standard free energy change $(\mathrm{\Delta G}°)$ and equilibrium constant is

$\mathrm{\Delta G}° = –\mathrm{RTln} \mathrm{K}$

$–\mathrm{RTln} \mathrm{K} = \mathrm{\Delta H}° – \mathrm{T\Delta S}°$