Q. For the reaction $N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g);\ \ \ \ \ \mathrm{\Delta}H = - 93.6\ kJ\ mol^{- 1}$, the concentration of $NH_{3}$ at equilibrium can be increased by (I) lowering the temperature (II)low pressure (III)excess of $N_{2}$ (IV)excess of $H_{2}$
A) (II) and (IV) are correct
B) (II) only is correct
C) (I), (II) and (III) are correct
D) (III) and (IV) are correct
Explanation
(c)
$N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g);\ \$
$\ \ \ \mathrm{\Delta}H = - 93.6\ kJ\ mol^{- 1}$
This reaction is exothermic therefore the concentration of $NH_{3}$ can be increased by lowering the temperature, high pressure ($\mathrm{\Delta}n < 0$) excess $N_{2}\ and\ H_{2}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.