Chapter 17.1, Problem 9SSC

Solution

Interpretation:

The homogeneous and heterogenous equilibrium system is to be compared.

Concept introduction:

- The two types of equilibrium system homogenous and heterogenous.

When all the reactants are in same phase, it is a homogenous system. When all the reactants are different phase, then it is a heterogenous system.

For homogenous system

$A(g)+B(g)\rightleftharpoons C(g)$

For heterogenous system.

$P_{(S)}+Q_{(g)}\rightleftharpoons P{Q}_{(S)}$

In homogenous system substances are in same phase while in heterogenous the substance is in different phase.

Equilibrium system is of two types-

1. Homogenous equilibrium: - The word ‘HOMO’ means same so, those equilibrium in which all the substances involved are in same phase are termed as homogenous equilibria.

For e.g. −

$N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$

In this all the substances involved are in gaseous phase so, it is a homogenous equilibrium.

Equilibrium constant for this reaction is

$K_C=\frac{{[N{H}_3(g)]}^2}{[N_2(g)]{[H_2(g)]}^3}$

Or

$K_P=\frac{(P^{2}N{H}_3)}{(P{N}_2)(P^3{H}_2)}$

in gases equilibrium constant is expressed in terms of partial pressure.

2. Heterogenous equilibrium: - The word ‘HETERO’ means different. So, the equilibria in which all the substances involved are in different phases are called heterogenous equilibria.

For example.

$\begin{array}{l}a)H_{2}O(l)\rightleftharpoons H_{2}O(g)\\ b)CaC{O}_3\rightleftharpoons CaO(s)+C{O}_2(g)\end{array}$

Here some substance is in different phase while other are different. So, this system in heterogenous equilibria.

In this system

While writing equilibrium expression, certain conventions are followed.

For example.

$\begin{array}{l}a)BaS{O}_3(s)\rightleftharpoons BaO(s)+C{O}_2(g)\\ K_C=\frac{[BaO(s)][C{O}_2(g)]}{[BaC{O}_3(s)]}\end{array}$

Solids are omitted so, $K_C$ of this reaction

$\begin{array}{l}{K}_C=[C{O}_2(g)]\\ K_p=PC{O}_2(g)\end{array}$

$\begin{array}{l}b)N{H}_3(aq)+H_{2}O(l)\rightleftharpoons N{H}_{4}OH(aq)\\ K_C=\frac{[N{H}_{4}OH(aq)]}{[N{H}_3(aq)][H_{2}O(l)]}\\ \because H_{2}O(l)=1\end{array}$

So,

$\begin{array}{l}{K}_C=\frac{[N{H}_{4}OH(aq)]}{[N{H}_3(aq)]}\\ K_p=\frac{[PN{H}_{4}OH(aq)]}{[PN{H}_3(aq)]}\end{array}$

In homogenous equilibria the substance involved are in same phase while in heterogenous equilibria the substance involved are in different phases.