Question

For the reversible reaction,${N_2 (g) + 3H_2 (g) <=> 2NH_3 (g) + heat}$ the equilibrium shifts in forward direction

• A. by increasing the concentration of $NH_3(g)$
• 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)

by increasing pressure and decreasing temperature

Answer 2

The correct option is(D): by increasing pressure and decreasing temperature,

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.
${ N_2 (g) + 3H_2 (g) <=> 2Nh_3 (g) + Heat}$
(a) Increasing the concentration of $NH_3 (g):$
On increasing the concentration of $NH_3 (g)$. the equilibrium shifts in the backward direction where concentration of $NH_3 (g)$ decreases
(b) Decreasing the pressure: Since, $p ? n$ (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