For a general reaction $aA + bB \rightarrow cC + dD$, the rate of reaction is expressed as: $\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$

For the given reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$, the stoichiometric coefficients are 1 for $N_2$, 3 for $H_2$, and 2 for $NH_3$. Applying the rule: $\text{Rate} = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = +\frac{1}{2}\frac{d[NH_3]}{dt}$

To find the relationship between the rate of formation of ammonia ($\frac{d[NH_3]}{dt}$) and the rate of consumption of hydrogen ($-\frac{d[H_2]}{dt}$), we equate their terms: $+\frac{1}{2}\frac{d[NH_3]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt}$

Multiplying both sides by 2: $\frac{d[NH_3]}{dt} = -\frac{2}{3}\frac{d[H_2]}{dt}$