$\frac{d[NH_3]}{dt} = 2 \times 10^{-4} \, mol \,L^{-1} \, s^{-1}$ હોય, તો $\frac{-d[H_2]}{dt}$ ની કિંમત ............. $mol \,L^{-1} \, s^{-1}$ થશે.
Rate $=\frac{1}{2} \frac{d\left[\mathrm{NH}_{3}\right]}{d t}=-\frac{1}{3} \frac{d\left[\mathrm{H}_{2}\right]}{d t}=-\frac{d\left[\mathrm{N}_{2}\right]}{d t}$
$\frac{d\left[\mathrm{NH}_{3}\right]}{d t}=2 \times 10^{-4} \,\mathrm{mol} \,\mathrm{L}^{-1} \,\mathrm{s}^{-1}$
$\therefore-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=\frac{3}{2} \frac{d\left[\mathrm{NH}_{3}\right]}{d t}=\frac{3}{2} \times 2 \times 10^{-4}$
$\Rightarrow-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=3 \times 10^{-4}\, \mathrm{mol} \,\mathrm{L}^{-1} \,\mathrm{s}^{-1}$
| No | $[NH_4^+]$ | $[NO_2^-]$ | rate of reaction |
| $1.$ | $0.24\, M$ | $0.10\, M$ | $7.2 \times {10^{ - 6}}$ |
| $2.$ | $0.12\, M$ | $0.10\, M$ | $3.6 \times {10^{ - 6}}$ |
| $3.$ | $0.12\, M$ | $0.15\, M$ | $5.4 \times {10^{ - 6}}$ |
