Question

Some process is given below. What happens to the process if it is subjected to a change given in the brackets?
(a) $Ice\leftrightarrows water$ ( pressure is increased)
(b) Dissolution of NaOH in water (Temperature is increased)
(c) ${{N}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2NO(g)\text{ }\Delta H=-180.7k{{J}^{-1}}$ (Pressure is increased and temperature is decreased)

Answer 1

Hint: At equilibrium the concentration of both reactants and products remains constant and on increasing pressure, equilibrium will shift to the direction in which there is decrease in volume and on decreasing temperature , equilibrium will shift to the direction in forward direction ( in case of exothermic reactions) and in backward directions (in case of endothermic reactions). Now answer the given statements.

Complete Solution :
> Equilibrium is a state or condition in which the rate of forward reaction is equal to the rate of backward reaction and there is no change in the concentrations of the reactants and the products at equilibrium.
> When we increase the pressure at equilibrium , then the reaction will shift in that direction in which there is decrease in the number of moles or decrease in volume and vice -versa.
> When the reaction exothermic i.e. heat is released then on decreasing the temperature, the equilibrium will shift towards the right side i.e. the reaction will proceed in the forward direction and vice- versa.

Now considering the statement one by one:
(a) $Ice\rightleftharpoons water$ ( pressure is increased)
As we know, if pressure is increased it will shift in that direction in which there is decrease in the volume. So, the reaction will proceed in a forward direction and more ice will melt into the water.

(b) Dissolution of NaOH in water (Temperature is increased)
Consider the dissolution reaction of NaOH as:
$NaOH\rightleftharpoons N{{a}^{+}}+O{{H}^{-}}$
In this heat is released during dissolution i.e. entropy of the system is positive and when temp is increased , Gibbs energy of system will decrease according to the formula as: $\Delta G=\Delta H-T\Delta S$ and hence the solubility of NaOH will increases as the reaction will proceed in the forward direction.

(c) ${{N}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2NO(g)\text{ }\Delta H=-180.7k{{J}^{-1}}$
(pressure is increased and temperature is decreased)
On increasing pressure , there is no effect on state of equilibrium as number of moles are same but on decreasing temperature because of the exothermic reaction ($\Delta H=negative$) so , the equilibrium shifts towards the right side i.e. the reaction will proceed in forward direction.

Note: Equilibrium can only be attained in a closed system i.e. when the reaction occurs in a closed vessel or container and neither the products nor the reactants are lost from the container and is the necessary condition to attain equilibrium and can be attained in the open systems.