The rate of a chemical reaction can be expressed in terms of any of the reactants or products. For a reaction $aA + bB \rightarrow cC + dD$, the rate is related to the stoichiometric coefficients 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 specific reaction $N_2 + 3H_2 \rightarrow 2NH_3$: $\text{Rate} = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = +\frac{1}{2}\frac{d[NH_3]}{dt}$

We are given the rate of appearance of ammonia, $\frac{d[NH_3]}{dt} = 2 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$. We need to find the rate of disappearance of hydrogen, $-\frac{d[H_2]}{dt}$.

Using the equality: $-\frac{1}{3}\frac{d[H_2]}{dt} = +\frac{1}{2}\frac{d[NH_3]}{dt}$ $-\frac{d[H_2]}{dt} = \frac{3}{2} \times \frac{d[NH_3]}{dt}$ $-\frac{d[H_2]}{dt} = \frac{3}{2} \times (2 \times 10^{-4})$ $-\frac{d[H_2]}{dt} = 3 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$