Question

Given,1 mole of A, 1.5 mole of B and 2 mole of C are taken in a vessel of volume one litre. At equilibrium concentration of C is 0.5 mole/L. Equilibrium constant for the reaction,
${{A}_{(g)}}+{{B}_{(g)}}\rightleftarrows {{C}_{(g)}}$ is:
A. 0.66
B. 0.066
C. 66
D. 6.6

Answer 1

Hint: Equilibrium is the state of reaction in which the rates of forward direction is equal to backward direction and concentrations of the reactant and products remain constant. Equilibrium is a dynamic process.

Complete answer:
The equilibrium constant of a chemical reaction which is represented by K tells us about the relationship between the products and reactants when a chemical reaction reaches equilibrium. It can be represented by another term ${{K}_{c}}$ which can be defined as the ratio of the concentration of reactants to the concentration of the products each raised to their respective stoichiometric coefficients.
${{A}_{(g)}}+{{B}_{(g)}}\rightleftarrows {{C}_{(g)}}$
In this equation equilibrium constant is given by
${{K}_{eq}}=\dfrac{[C]}{[A][B]}$
 ${{A}_{(g)}}+{{B}_{(g)}}\rightleftarrows {{C}_{(g)}}$
At t = 0, 1 1.5 2
At equilibrium, 1-x 1.5-x 2+x = 0.5
Hence x = 0.5-2 = -1.5
2 + x = 2 – 1.5 = 0.5; Concentration of C = 0.5
Concentration of molecule A is 1-x i.e. 1-(-1.5) = 2.5
Concentration of molecule B is 1.5-x i.e. [1.5-(-1.5)] = 3
By putting the value in equilibrium constant reaction we get
${{K}_{eq}}=\dfrac{[0.5]}{[2.5][3]}=0.0666$

Hence we can say that option B is the correct answer.

Note:
Equilibrium constant is reaction specific and it is fixed at a constant temperature. A catalyst changes the rate of forward and backward reactions equally but will not affect the value of equilibrium constant. Also changes in concentration, pressure and temperature may affect the equilibrium favoring but not the equilibrium constant.