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
Correct Answer: D
Solution :
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 in governed by Le-Chatelier's principle. According to Le-Chatelier's principle, equilibrium shifts in the opposite direction to undo the change. \[{{N}_{2}}(g)+3{{H}_{2}}(g)2N{{H}_{3}}(g)\] [a] Increasing the concentration of \[\mathbf{N}{{\mathbf{H}}_{\mathbf{3}}}(g)\] 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, pan (no. of moles), therefore, equilibrium shifts in the backward direction where number of moles are increasing. [c] Decreasing the concentration of \[{{\mathbf{N}}_{\mathbf{2}}}(g)\] and \[{{\mathbf{H}}_{\mathbf{2}}}(g)\] Equilibrium shifts in the backward direction when concentration of \[{{N}_{2}}(g)\] 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 while on decreasing temperature, it will move in forward direction where temperature increases.
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