Question

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

Answer 1

Hint: The Le Chatelier's principle states that 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. So to solve this question we have to consider the changes taking place in the system and the system will react in such a way that it opposes the change.

Complete answer:
We have to consider the effect of change of pressure and concentration in the reaction system. The given reaction is as follows:
${{N}_{2(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{(g)}}$

Effect of change of pressure: If a system in equilibrium consists of reactants and products in gaseous state necessarily, then the concentration of all components in the system can be altered by changing the total pressure of the system.
${{N}_{2(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{(g)}}$

In the given equilibrium, the number of moles of gaseous products and reactants are equal and hence there is no pressure effect on the equilibrium that means $\Delta {{n}_{g}}=0$. Therefore, by increasing or decreasing the pressure, there will be no change in the equilibrium reaction as this will bring no change in the partial pressures of the gases involved hence concentration at equilibrium will remain the same.
Effect of change of concentration: At the conditions of equilibrium, the reactions mixture contains both the reactant and product molecules that is ${{N}_{2}}$​, ${{O}_{2}}$​ and NO molecules. The concentration of both the reactant and the product molecules are constant and remain the same at equilibrium state. If the concentration of NO is increased, the equilibrium gets disturbed. The excess amount of NO reacts to give back the reactants, 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 that is by increasing the concentration of ${{N}_{2}}$​ and ${{O}_{2}}$​, more amount of NO will be produced in the system.

Note: Effect of change temperature:
${{N}_{2(g)}}+{{O}_{2(g)}}\rightleftharpoons 2N{{O}_{(g)}}\text{ }\Delta H\text{ }=+ve$
In the given equilibrium the reaction of the product formation $\left( NO \right)$​ is endothermic and reverse reaction of reactant formation ${{N}_{2}}$and ${{O}_{2}}$​. is hence exothermic. If the above reaction mixture is heated its temperature will increase, the equilibrium will shift in the direction in which temperature of the system decreases. The equilibrium will shift towards the formation of NO​ as this uses up the energy of the system to give the product and subsequently dissociation of ${{N}_{2}}$and ${{O}_{2}}$​ increases. When the temperature is raised in a chemical equilibrium, endothermic reaction will be favoured and reaction will move in that direction. If the temperature is decreased exothermic reaction will be favoured over endothermic reaction.