Chemistry - Chemical Kinetics Question with Solution | TestHub

$-\frac{\text{d}\left[\text{N}{\text{H}}_3\right]}{\text{d}\text{t}}=-\frac{\text{d}\left[{\text{O}}_2\right]}{\text{d}\text{t}}=-\frac{\text{d}\left[\text{N}\text{O}\right]}{\text{d}\text{t}}=-\frac{\text{d}\left[{\text{H}}_2\text{O}\right]}{\text{d}\text{t}}$

$\frac{\text{d}\left[\text{N}{\text{H}}_3\right]}{\text{d}\text{t}}=\frac{\text{d}\left[{\text{O}}_2\right]}{\text{d}\text{t}}=-\frac{1}{4}\frac{\text{d}\left[\text{N}\text{O}\right]}{\text{d}\text{t}}=-\frac{1}{6}\frac{\text{d}\left[{\text{H}}_2\text{O}\right]}{\text{d}\text{t}}$

$\frac{1}{4}\frac{\text{d}\left[\text{N}{\text{H}}_3\right]}{\text{d}\text{t}}=\frac{1}{5}\frac{\text{d}\left[{\text{O}}_2\right]}{\text{d}\text{t}}=\frac{1}{4}\frac{\text{d}\left[\text{N}\text{O}\right]}{\text{d}\text{t}}=\frac{1}{6}\frac{\text{d}\left[{\text{H}}_2\text{O}\right]}{\text{d}\text{t}}$

$-\frac{1}{4}\frac{\text{d}\left[\text{N}{\text{H}}_3\right]}{\text{d}\text{t}}=-\frac{1}{5}\frac{\text{d}\left[{\text{O}}_2\right]}{\text{d}\text{t}}=\frac{1}{4}\frac{\text{d}\left[\text{N}\text{O}\right]}{\text{d}\text{t}}=\frac{1}{6}\frac{\text{d}\left[{\text{H}}_2\text{O}\right]}{\text{d}\text{t}}$