Question

Apply Le Chatelier's principle for the following reaction:
$N_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2N{O}_{\left(g\right)}$ and discuss the effect of pressure and concentration on it.

Answer 1

$N_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2N{O}_{\left(g\right)}$
Le Chatelier's principle: According to this principle, "if a system at equilibrium is subjected to a disturbance or stress, then the equilibrium shifts in the direction that tends to nullify the effect of the disturbance or stress"
Consider the following equilibrium reaction.
$N_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2N{O}_{\left(g\right)}$
(i) Effect of change of pressure: If a system in equilibrium consists of reactants and products in gaseous state, then the concentration of all components can be altered by changing the total pressure of the system.
$N_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2N{O}_{\left(g\right)}$
In the above equilibrium, the number of moles of gaseous products and reactants are equal and hence there is no pressure effect on the equilibrium. By increasing or decreasing the pressure, there will be no change in the equilibrium reaction.
(ii) Effect of change of concentration: At the equilibrium conditions, the reactions mixture contains both the reactant and product molecules that is $N_2,O_2$ and $NO$ molecules. The concentration of the reactant and the product molecules are constant and remain the same at equilibrium state. If the concentration of NO is increased, the equilibrium is disturbed. The excess amount of NO reacts, in the reverse direction to produce back the reactants and this results in the increase in the concentration of $N_2$ and $O_2$. Similarly if the concentrations of reactants such as $N_2$ and $O_2$ are raised, the equilibrium shifts towards the forward reaction. i.e., by increasing the concentration of $N_2$ and $O_2$, more amount of NO will be produced.
(iii) Effect of change temperature:
$N_2{O}_{4\left(g\right)}\rightleftharpoons 2N{O}_{2\left(g\right)}$ $\Delta H+59kJ/Mole$
In this equilibrium the reaction of the product formation $\left(N{O}_2\right)$ is endothermic and reverse reaction of reactant formation $\left(N_2{O}_4\right)$ is exothermic. If the above reaction mixture is heated its temperative will be raised. The equilibrium will shift towards the formation of $N{O}_2$ and subsequently dissociation of $N_2{O}_4$ increases. When the temperature is raised in a chemical equilibrium, endothermic reaction will be favoured. If the temperature is decreased exothermic reaction will be favoured.