Example \(\PageIndex{1}\):

Under a certain set of conditions, the rate of the reaction

\[N_2 + 3 H_2 \rightarrow 2 NH_3\]

the reaction rate is \(6.0 \times 10^{-4}\, M/s\). Calculate the time-rate of change for the concentrations of N_{2}, H_{2}, and NH_{3}.

**Solution**:

Due to the stoichiometry of the reaction,

\[\text{rate} = - \dfrac{d[N_2]}{dt} = - \dfrac{1}{3} \dfrac{d[H_2]}{dt} = + \dfrac{1}{2} \dfrac{d[NH_3]}{dt} \]

so

\[\dfrac{d[N_2]}{dt} = -6.0 \times 10^{-4} \,M/s\]

\[\dfrac{d[H_2]}{dt} = -2.0 \times 10^{-4} \,M/s\]

\[\dfrac{d[NH_3]}{dt} = 3.0 \times 10^{-4} \,M/s\]

*Note*: The time derivatives for the reactants are negative because the reactant concentrations are decreasing, and those of products are positive since the concentrations of products increase as the reaction progresses.