# 14.1: The Time Dependence of a Chemical Reaction Is Described by a Rate Law

The rate of a chemical reaction (or the reaction rate) can be defined by the time needed for a change in concentration to occur. But there is a problem in that this allows for the definition to be made based on concentration changes for either the reactants or the products. Plus, due to stoichiometric concerns, the rates at which the concentrations are generally different! Toward this end, the following convention is used.

For a general reaction

$a A + b B \rightarrow c C + d D$

the reaction rate can be defined by any of the ratios

$\text{rate} = - \dfrac{1}{a} \dfrac{\Delta [A]}{dt} = - \dfrac{1}{b} \dfrac{\Delta[B]}{dt} = + \dfrac{1}{c} \dfrac{\Delta [C]}{dt} = + \dfrac{1}{d} \dfrac{ \Delta [D]}{dt}$

Or for infinitesimal time intervals

$\text{rate} = - \dfrac{1}{a} \dfrac{d[A]}{dt} = - \dfrac{1}{b} \dfrac{d[C]}{dt} = + \dfrac{1}{c} \dfrac{d[C]}{dt} = + \dfrac{1}{d} \dfrac{d[D]}{dt}$

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 N2, H2, and NH3.

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.