Question

Some process are given below. What happens to the process if it subjected to a change given in the brackets?
(a) Ice $\rightleftharpoons$ water (pressure is increased)
(b) Dissolution of NaOH in water (Temperature is increased)
(c) $N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right),\Delta H=-180.7kJ{}^{-1}$
(Pressure is increased and Temperature is decreased)

Answer 1

a) $Ice\rightleftharpoons water$

If we increase pressure, equilibrium will be shifted in the direction of less volume, hence with rise of pressure. Therefore more ice will melt into water.

b) $NaOH\rightleftharpoons {Na}^{+}+{OH}^{(-)}$

The reaction has a positive change of entropy $(\Delta S>0)$. Hence from the equation $\Delta G=\Delta H-T\Delta S$. So Gibss free energy will decrease with increase of the temperature. As result, the solubility of $NaOH$ increases with increase in temperature.

c) $N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right),\Delta H=-180.7KJ/mol$

Since, ${\Delta n}_g=0$, increase in pressure does not effect the shift in equilibrium. Again it is exothermic reaction decrease in temperature will shift the equilibrium towards right.