Q. In the Synthesis of ammonia by Haber process, if 60 moles of ammonia is obtained in one hour, then the rate of disappearance of nitrogen is
A) 30 mol/min
B) 6 mol/min
C) 0.5 mol/min
D) 60 mol/min
Explanation
(c)
In Haber’s process, ammonia is synthesized as
$N_{2}$+3$H_{2}\frac{Fe,Mo,300℃}{High\ Pressure} \rightarrow \ 2NH_{3}$
Rate of synthesis of ammonia=$\frac{60}{60}$
=1 mol/min
$\therefore$ Rate of disappearance of nitrogen, i.e.,
$\frac{- d\lbrack N_{2}\rbrack}{dt} = \frac{1}{2}\frac{d\lbrack NH_{3}\rbrack}{dt}$
=$\frac{1}{2} \times = 0.5\ mol/min$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.