# Why are the stoichiometric coefficients the powers in the rate law?

Generally, for a reaction of the form $$\ce{n_A A + n_B B + \dots -> products},$$ the rate law is given by the following: $$\text{rate} = k [\ce{A}]^a [\ce{B}]^b.$$

Do the stoichiometric coefficients equal the powers in the rate law? \begin{align} a &= n_{\ce{A}},& b &= n_{\ce{B}} \end{align}

If that is true, why is it so?

In an elementary reaction, then, what actually takes place at a microscopic levels fits the reaction formula we write (if we express the reaction as $\ce{A + B -> C}$, a molecule of $\ce{A}$ and a molecule of $\ce{B}$ actually collide and produce a molecule of $\ce{C}$ as a single step). But, in general, that's not the case; reactions in general have mechanisms comprising one or more elementary reaction steps.
A related concept is the molecularity of a reaction step, that is, the number of molecules that intervene in a certain step. Since elementary reactions match their chemical expression, if a reaction is elementary, the order of a reagent will match its stoichiometric coefficient (which is your question: that means that $a=n_a$ and $b=n_b$). This is not the case for multi-step mechanisms - kinetic order is not related to stoichiometry in general (and you can have the reaction rate depend on concentrations that don't even appear in the formula, such as in homogeneous catalysis).