Question

Favorable conditions for the \[N{{H}_{3}}\] formation by Haber's process are high pressure and low temperature.
A.True
B.False

Answer 1

Hint:The reaction in the Haber’s process follows Le Chatelier's principle. According to this principle if a chemical system at equilibrium experiences a change in concentration, temperature, volume, or pressure, then the equilibrium shifts to counteract the imposed change and a new equilibrium is established.

Complete step by step answer:
The reaction involved in Haber's process is:
${{N}_{2}}\,+\,3{{H}_{2}}\,\,-2N{{H}_{3}}$
This reaction is exothermic (heat released) and has a negative \[(2-(3-1))\]
$\vartriangle n\,=\,2$.
If the temperature is lowered then, the reaction will go in the direction so that the temperature is retained to initial, therefore reaction will go in the forward direction because the reaction is exothermic.
When pressure is increased, then the reaction will shift such that the net pressure decreases, since the forward reaction decreases the pressure, thus forward reaction is favourable.
Hence, by LeChatelier's principle, high pressure and low temperature is favoured

So, the answer will be A.

Additional information:
 The Haber Process combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia. The reaction is a reversible reaction. The production of ammonia is exothermic i.e. heat is evolved.
In LeChatelier principle on changing the concentration of a chemical will shift the equilibrium to the side that would counter that change in concentration. The chemical system will oppose the change affected to the original state of equilibrium.

Note:
The condition which affects the LeChatellier principle should be known to solve this question. Adding a noble gas into a gas-phase equilibrium at constant volume doesn't end in a shift. This is because the addition of a non-reactive gas doesn't change the equilibrium equation, because the noble gas appears on each side of the reaction equation.