Q. The differential rate law for the reaction $4{NH}_{3}(g) + 5O_{2}(g) \rightarrow 4NO(g) + 6H_{2}O(g)\$is
A) $- \frac{d\left\lbrack {NH}_{3} \right\rbrack}{dt} = - \frac{d\left\lbrack O_{2} \right\rbrack}{dt} = - \frac{d\lbrack NO\rbrack}{dt} = - \frac{d\left\lbrack H_{2}O \right\rbrack}{dt}$
B) $\frac{d\left\lbrack {NH}_{3} \right\rbrack}{dt} = \frac{d\left\lbrack O_{2} \right\rbrack}{dt} = - \frac{1}{4}\frac{d\lbrack NO\rbrack}{dt} = - \frac{1}{6}\frac{d\left\lbrack H_{2}O \right\rbrack}{dt}$
C) $\frac{1}{4}\frac{d\left\lbrack {NH}_{3} \right\rbrack}{dt} = \frac{1}{5}\frac{d\left\lbrack O_{2} \right\rbrack}{dt} = \frac{1}{4}\frac{d\lbrack NO\rbrack}{dt} = \frac{1}{6}\frac{d\left\lbrack H_{2}O \right\rbrack}{dt}$
D) $- \frac{1}{4}\frac{d\left\lbrack {NH}_{3} \right\rbrack}{dt} = - \frac{1}{5}\frac{d\left\lbrack O_{2} \right\rbrack}{dt} = \frac{1}{4}\frac{d\lbrack NO\rbrack}{dt} = \frac{1}{6}\frac{d\left\lbrack H_{2}O \right\rbrack}{dt}$
Explanation
(d)
$4{NH}_{3}(g) + 5O_{2}(g) \rightarrow 4NO(g) + 6H_{2}O(g)$
$Rate = - \frac{1}{4}\frac{d\left\lbrack {NH}_{3} \right\rbrack}{dt} = - \frac{1}{5}\frac{d\left\lbrack O_{2} \right\rbrack}{dt} = + \frac{1}{4}\frac{d\lbrack NO\rbrack}{dt}$
$= + \frac{1}{6}\frac{d\left\lbrack H_{2}O \right\rbrack}{dt}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.