Rate Law (Reaction Order)

With the method of initial rates: you measure v for several runs in which only one concentration is changed while all others stay constant. Comparing two measurements yields the exponent: if v doubles when [A] doubles, $m = 1$; if it quadruples, $m = 2$; if v stays the same, $m = 0$. Formally $m =\frac{ lg(\frac{v_{1}}{v_{2}})}{lg}(\frac{[A]_{1}}{[A]_{2}})$. Alternatively you test integrated rate laws: if ln[A] versus t is linear, the reaction is first order; if $\frac{1}{[A]}$ versus t is linear, second order. The order can never be read off the reaction equation, it is purely a measured result.

Because the reaction equation only describes the overall balance, not the pathway. Most reactions proceed through several elementary steps, and the rate is set by the slowest one. Its rate law contains only the particles involved in that step. A classic example is the reaction of hydrogen with bromine: the equation H₂ + Br₂ → 2 HBr suggests order 1 in each, but the measured rate law is fractional and complicated because a radical chain mechanism operates. Only for elementary reactions that really occur in a single collision do order and stoichiometric coefficients coincide. That is why, conversely, the measured rate law gives valuable clues about the mechanism.

The unit of k depends on the overall order of the reaction, because the product $k\cdot{}[A]^{m}\cdot{}[B]^{n}$ must always come out in $\frac{mol}{(L\cdot{}s)}$. For a zero-order reaction, k itself is in $\frac{mol}{(L\cdot{}s)}$. For first order one concentration cancels: k has the unit $s^{-1}$, which is convenient because k then relates directly to the half-life via $t\frac{1}{2} =\frac{ \ln{} 2}{k}$. Second order requires $\frac{L}{(mol\cdot{}s)}$, third order $\frac{L^{2}}{(mol^{2}\cdot{}s)}$. Conversely, the unit of a given k immediately reveals the overall order; this is a popular exam trick and a good self-check when calculating.

For zero order the rate is constant and independent of concentration, for example when a catalyst surface or an enzyme is saturated: material is processed as fast as possible. For first order, v is proportional to the concentration of a single substance; radioactive decay and many decompositions follow this pattern with a constant half-life. For second order two particles must collide, either two identical ones $(v = k\cdot{}[A]^{2})$ or two different ones $(v = k\cdot{}[A]\cdot{}[B])$; here the half-life grows as the reaction proceeds. The order thus describes how strongly the reaction responds to concentration changes and reveals something about the underlying mechanism.

The two equations split the description of the reaction rate between them. The rate law $v = k\cdot{}[A]^{m}\cdot{}[B]^{n}$ describes, at fixed temperature, how v depends on the concentrations. The entire temperature dependence sits in the rate constant k, and that is exactly what the Arrhenius equation $k = A\cdot{}e^{(-\frac{E_{A}}{(R\cdot{}T)})}$ describes. If you raise the temperature, the orders m and n stay the same, but k grows exponentially, roughly doubling per 10 K as a rule of thumb. For complete problems you combine both: first determine k at the desired temperature via Arrhenius, then insert the concentrations into the rate law. This is how chemists cleanly separate concentration and temperature effects.