\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \)
\(\mathrm{K}_{\mathrm{c}}=\frac{4 x^{2}}{(a-x)(b-x)}\)
This expression doesn't involve, V. So, increase or decrease of volume will not affect the equilibrium and hence the yield of the product.

This condition is not static and is dynamic, because both the forward and reverse reactions are still occurring with the same rate. No macroscopic change is observed.

If we know the value of k_c and Q, the reaction quotient, we can predict the direction of a reaction
If Q = Kc; the reaction is in equilibrium state.
If Q > Kc: the reaction will proceed in the reverse direction i.e., formation of reactants.
If Q < Kc: the reaction will proceed in the forward direction i.e., formation of products.

If Q > K_c; the reaction will proceed in the reverse direction (ie) formation of reactants to proceed to reach equilibrium.

i) \(K_{p}=K_{c}(R T)^{\Delta n_{g}}\)
K_p = Equilibrium constant in term of partial Pressures.
K_c = Equilibrium constant in term of concentration.
R = Gas constant; T = Temperature
\(\Delta \mathrm{n}_{\mathrm{g}}\) = Difference between the sum of number of moles of products and the sum of number of moles of reactants in gas phases.
ii) Synthesis of HI:
\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \)
\(\Delta n_{g}=0 \therefore K_{p}=K_{c}(R T) \Delta n_{g}\)
\(K_{p}=K_{c}(R T)^{\circ} \)
\(K_{p}=K_{c} \text {. }\)