Given the following reaction at equilibrium $N_2\left ( g \right )+3H_2\left ( g \right )\rightleftharpoons 2NH_3$. Some inert gas at constant pressure is added to the system. Predict which of the following facts will be affected:
A.more ${\rm{N}}{{\rm{H}}_{\rm{3}}}\left( g \right)$ is produced
B.less ${\rm{N}}{{\rm{H}}_{\rm{3}}}\left( g \right)$ is produced
C.no effect on the equilibrium
D.${K_P}$ of the reaction is decreased
Answer
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Hint: We know that in a state of equilibrium both the forward and backward processes precede at the same speed. Also the state of a system can be altered by changing the conditions of pressure, temperature and concentration of the different species which results in disturbance of the equilibrium.
Complete step by step answer:
Let’s first understand Le Chatelier’s principle. According to this principle if a system under equilibrium is subjected to a change in temperature, pressure or concentration, then the equilibrium shifts itself in such a way as to undo or neutralize the effect of the change.
Now, come to the question. Here, the equilibrium reaction of formation of ammonia from nitrogen and hydrogen is given. And some inert gas at constant pressure is added to the reaction mixture. The addition of inert gas increases the volume of the reaction mixture, so the concentration of all reacting species decreases. That means the equilibrium will shift in the direction in which more number of moles of gas. Now, in the reaction
$N_2\left ( g \right )+3H_2\left ( g \right )\rightleftharpoons 2NH_3$.
The mole of reactant is 4 and the mole of the product is 2. So, the shift of equilibrium takes place towards reactants. That means, ammonia undergoes dissociation to result in nitrogen and hydrogen. So, the concentration of ammonia $\left( {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right)$ decreases.
So, the correct answer is Option B.
Note: Always remember that increase of concentration of any reactant causes shifting of equilibrium towards forward direction and the increase in concentration of any products causes shifting of equilibrium towards backward direction.
Complete step by step answer:
Let’s first understand Le Chatelier’s principle. According to this principle if a system under equilibrium is subjected to a change in temperature, pressure or concentration, then the equilibrium shifts itself in such a way as to undo or neutralize the effect of the change.
Now, come to the question. Here, the equilibrium reaction of formation of ammonia from nitrogen and hydrogen is given. And some inert gas at constant pressure is added to the reaction mixture. The addition of inert gas increases the volume of the reaction mixture, so the concentration of all reacting species decreases. That means the equilibrium will shift in the direction in which more number of moles of gas. Now, in the reaction
$N_2\left ( g \right )+3H_2\left ( g \right )\rightleftharpoons 2NH_3$.
The mole of reactant is 4 and the mole of the product is 2. So, the shift of equilibrium takes place towards reactants. That means, ammonia undergoes dissociation to result in nitrogen and hydrogen. So, the concentration of ammonia $\left( {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right)$ decreases.
So, the correct answer is Option B.
Note: Always remember that increase of concentration of any reactant causes shifting of equilibrium towards forward direction and the increase in concentration of any products causes shifting of equilibrium towards backward direction.
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