Question

In the formation of the $N{H}_3$ by the Haber`s process which condition will give maximum yield the equation is as
$N_2\left(g\right)$+$3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$$\Delta H=-Qkcal$, where Q is positive.

• A. Low temperature, high pressure and high concentrations of the reactants
• B. High temperature, low pressure and low concentration of the reactants
• C. High temperature, high pressure
• D. Low temperature, low pressure and low concentration of $H_2$

Answer 1

The correct option is A Low temperature, high pressure and high concentrations of the reactants

The Haber`s process have equilibrium reaction as
$N_2\left(g\right)$ + $3H_2\left(g\right)$$\rightleftharpoons$ $2N{H}_3\left(g\right)+\Delta$
according to the Le chatelier`s principle if any changes happen in any component then equilibirium will move that direction where the effect of that change neutralize.

This is exothermic reaction so low temperature gives more product.

This reaction is having less number of moles in product side so according to Le chatelier's principle more product will be produced at high pressure.

If we add the reactants (high concentration) then also more product will be formed.