Question

For the reversible reaction,$N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)+heat$ the equilibrium shifts in forward direction

• A. by increasing the concentration of $N{H}_3(g)$
• B. by decreasing the pressure
• C. by decreasing the concentrations of $N_2(g)$ and $h_2(g)$
• D. by increasing pressure and decreasing temperature

Answer 1

Answer: (D)

Any change in the concentration, pressure and temperature of the reaction results in change in the direction of equilibrium. This change in the direction of equilibrium is governed by
Le-Chatelier's principle. According to
Le-Chatelier's principle, equilibrium shifts in die opposite direction to undo the change.
$N_2(g)+3H_2(g)\rightleftharpoons 2N{h}_3(g)+Heat$
(a) Increasing the concentration of $N{H}_3(g):$
On increasing the concentration of $N{H}_3(g)$. the equilibrium shifts in the backward direction where concentration of $N{H}_3(g)$ decreases
(b) Decreasing the pressure: Since, $p\propto n$ (number of moles), therefore, equilibrium shifts in the backward direction where number of moles are increasing.
(c) Decreasing the concentration of $N_2(g)$ and $H_2(g)$.
Equilibrium shifts in the backward direction when concentration of $H_2(g)$ and $H_2(g)$ decreases.
(d) Increasing pressure and decreasing temperature: On increasing pressure, equilibrium shifts in the forward direction where number of moles decreases. It is an example of exothermic reaction therefore decreasing temperature favours the forward direction