Let us consider the following mechanism:
\[\begin{align}
  & C{{H}_{3}}CN+{{H}^{+}}\rightleftharpoons C{{H}_{3}}CN{{H}^{+}}\text{ (fast)} \\
 & C{{H}_{3}}CN{{H}^{+}}+{{H}_{2}}O\to \text{Product (slow)} \\
\end{align}\]
What would be the rate law?

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Hint: In a reaction taking place in more than one step, the overall rate of the reaction is controlled by the slowest step of the reaction.

Complete answer:
The given reaction is a two step reaction taking place by the following mechanism:
     \[\begin{align}
  & C{{H}_{3}}CN+{{H}^{+}}\rightleftharpoons C{{H}_{3}}CN{{H}^{+}}\text{ (fast)} \\
 & C{{H}_{3}}CN{{H}^{+}}+{{H}_{2}}O\to \text{Product (slow)} \\
\end{align}\]
A reaction that takes place in several steps, its overall reaction rate is determined by the slowest step in the reaction mechanism. So, in the given reaction mechanism, second step is the slowest step. Hence, it is the rate determining step.
We know that the rate law expresses the rate of a reaction in terms of concentration terms of the reactants on which the reaction rate actually depends. Water being a solvent is present in large excess such that its concentration remains almost the same during the reaction. Therefore, its concentration will not affect the rate of the reaction. Therefore, the rate law for the given reaction will be given as:
     \[Rate=k\left[ C{{H}_{3}}CN{{H}^{+}} \right]\]
where k is rate constant.
Now, for the first step of the mechanism, we have
\[{{K}_{eq}}=\dfrac{\left[ C{{H}_{3}}CN{{H}^{+}} \right]}{\left[ {{H}^{+}} \right]\left[ C{{H}_{3}}CN \right]}\]
     \[\left[ C{{H}_{3}}CN{{H}^{+}} \right]={{K}_{eq}}\left[ {{H}^{+}} \right]\left[ C{{H}_{3}}CN \right]\]
Substituting equation (2) in (1), we get
     \[Rate=k\,{{k}_{eq}}\left[ {{H}^{+}} \right]\left[ C{{H}_{3}}CN \right]\]
     \[Rate=K\left[ C{{H}_{3}}CN \right]\]
where $K=k\,{{k}_{eq}}$ .
Hence, the rate law would be \[Rate=K\left[ C{{H}_{3}}CN \right]\].

Additional information: Order of a reaction is equal to the sum of the exponents of the concentration of the reactants in the rate law expression.

Note: Rate law for any reaction cannot always be determined from the balanced chemical equation. Therefore, do not simply write the rate law in terms of molar concentrations raised to the respective stoichiometric coefficients of all the reactants in the chemical equation.