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Question: A vessel of one liter capacity containing 1 mole of \(S{{O}_{2}}\) is heated till a state of equilib...

A vessel of one liter capacity containing 1 mole of SO2S{{O}_{2}} is heated till a state of equilibrium attained.
2SO3(g)⇌2SO2(g)+O2(g)2S{{O}_{3}}(g)\rightleftharpoons 2S{{O}_{2}}(g)+{{O}_{2}}(g)
At equilibrium, 0.6 moles of SO2S{{O}_{2}} had formed. The value of equilibrium constant is:
(A) 0.18
(B) 0.36
(C) 0.45
(D) 0.68

Explanation

Solution

The ratio of products to reactants at equilibrium is called equilibrium constant (K). The two different types of equilibrium reactions are homogeneous and heterogeneous equilibrium. This equilibrium constant assumes that it is familiar with the concept of dynamic equilibrium.

Complete answer:
Equilibrium constants are defined differently for two types of equilibrium separately.
(i) Homogeneous equilibrium: the equilibrium in which all of the reactants and reactants are present in a single-phase or single mixture. A chemical reaction between solutes in liquid solutions belongs to one type of homogeneous equilibrium and the chemical species involved in this equilibrium can be molecules, ions, or a mixture of both.
(ii) Heterogeneous equilibrium: In a system where reactants and products are present in two or more phases. These phases may be solid, liquid, or gas phases or solutions. When solids and pure liquids do not appear in this equilibrium constant expressions.
Consider an equilibrium reaction,
aA+bB⇌cC+dDaA+bB\rightleftharpoons cC+dD
For this reaction, the equilibrium constant K is the ratio of the products of a reaction to the reactants at equilibrium.
Equilibrium constant K = [C]c[D]d[A]a[B]b\dfrac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}
Given reaction,
2SO3(g)⇌2SO2(g)+O2(g)2S{{O}_{3}}(g)\rightleftharpoons 2S{{O}_{2}}(g)+{{O}_{2}}(g)
1.0 mole 0.0 0.0 (initial concentration)
0.4 mole 0.6 0.3 (at equilibrium)
The equilibrium constant for this reaction,
K = [SO2][O2]2[SO3]2=(0.6)2(0.3)(0.4)2\dfrac{[S{{O}_{2}}]{{[{{O}_{2}}]}^{2}}}{{{[S{{O}_{3}}]}^{2}}}=\dfrac{{{(0.6)}^{2}}(0.3)}{{{(0.4)}^{2}}} = 0.675 = (approximately) 0.68
Hence, the equilibrium constant for given reaction = 0.68

So, the correct answer is option D.

Note:
When we consider pure solids or liquids in an equilibrium reaction, calculating the equilibrium constant is not taken into account, because in this case, their concentrations do not change during the reaction. While determining the equilibrium constant for a solution upon the addition of ions that are already present in the solution, the common ion effect must be taken into consideration.