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