How will an increase in pressure and affect each of the following equilibria? An increase in temperature
(i) `2NH_(3)(g)hArrN_(2)(g)+3H_(g)DeltaH=932kJ`
(ii) `N_(2)(g)+O(2)(g)hArr2NO(g)DeltaH=181kJ`
(iii) `2O_(3)(g)hArr3O_(2)(g)DeltaH=-285kJ`
(iv) `CaO(s)+CO_(2)(g)hArr CaCO_(3)(s)DeltaH=-176kJ`

Explanation

According to Le Chattelier's principle increase in pressure shifts the equuilibrium in the direction where the number of gaseous moleculses are less.
Increase in temperature favolurs the endothermic reaction while decrease in temperature favours the exothermic reaction.
(i) `2NH_(3)(g)hArrN_(2)(g)+3H_(2)(g)DeltaH=92kJ`
The number of gaseous reactant molecule (2) is less than the sum of the total number of product molecules (4). So increase in pressure shits the equilibrium in the backward direction.
Forward reaction is endothermic while the backward reaction is exothermic. So increase temperature shifts the equilibrium in the foward direction.
(ii) `N_(2)(g)+O_(2)(g)hArr2NO(g) DeltaH=181kJ`
The sum of the gaseous reactant molecules is equal to the sum of the gaseous product molecules. So pressure has no affect on the equilibrium.
Forward reaction is endothermic while backward reaction is exothermic. So increase in temperature favours the reaction.
(iii) `2O_(3)9g)hArr3O_(2)(g) DeltaH=-285kJ`
The number of gaseous reactant molecules
(2) is less than the number of gaseous product molecules. So increase in pressure favours the formation of `O_(3)`.
The forward reaction is exothermic. While the backward reaction is endothermic.
So increase in temperature favours the backward reaction.
(iv) `CaO(s)+CO_(2)(g)hArrCaCO_(3)(s)DeltaH=-176kJ`
There is only one gaseous reactant molecule in this reaction. No gaseous product molecules. So increase in the pressure shift the equilibrium in forward direction.
The forward reaction is exothermic while backward reaction is endothermic.
So increase in pressure shifts the equilibrium in the backward direction.