For the reaction ${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{NH}}_3$, the rate of disappearance of ${\mathrm{H}}_2$ is $0.01{\mathrm{molelit}}^{-1}{ \min }^{-1}$. The rate of appearance of ${\mathrm{NH}}_3$ would be

$$\begin{gathered} {\mathrm{N}}_2+3{\mathrm{H}}_2⟶2{\mathrm{NH}}_3 \\ \text{Rate of reaction}=-\frac{\mathrm{d}\left({\mathrm{N}}_2\right)}{\mathrm{dt}}=-\frac{1}{3}\frac{\mathrm{d}\left({\mathrm{H}}_2\right)}{\mathrm{dt}}=\frac{1}{2}\frac{\mathrm{d}\left({\mathrm{NH}}_3\right)}{\mathrm{dt}} \\ \text{From rate expression,} \\ \begin{array}{rl}=& \frac{\mathrm{d}\left({\mathrm{NH}}_3\right)}{\mathrm{dt}}=\frac{2}{3}\frac{\mathrm{d}\left({\mathrm{H}}_2\right)}{\mathrm{dt}}\\ =& (2/3)\times 0.01=0.07\\ =& 0.007\mathrm{mol}{\mathrm{lit}}^{-1}{\min }^{-1}\end{array} \end{gathered}$$