Q. Consider the reaction N₂(g) + 3H₂(g) → 2NH₃(g). The equality relationship between d[NH₃]/dt and −d[H₂]/dt is :
A) d[NH₃]/dt = −(1/3) d[H₂]/dt
B) d[NH₃]/dt = −(2/3) d[H₂]/dt
C) d[NH₃]/dt = −(3/2) d[H₂]/dt
D) d[NH₃]/dt = −d[H₂]/dt
Explanation
[Solution from source] Rate = −⅓ d[H₂]/dt = ½ d[NH₃]/dt. Therefore d[NH₃]/dt = −(2/3) d[H₂]/dt: two moles of NH₃ form for every three moles of H₂ used. The ratio 3/2 inverts the stoichiometry. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.