Problem 67 - Chemistry: The Central Science

Question

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g)\) (b) \(\mathrm{OCl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HOCl}(a q)+\mathrm{OH}^{-}(a q)\) (c) \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}_{2}(g)\)

Step-by-Step Solution

Answer

(a) Molecularity: 1 (unimolecular). Rate law: Rate = k[\(Cl_{2}\)] (b) Molecularity: 2 (bimolecular). Rate law: Rate = k[\(OCl^-\)][\(H_2O\)] (c) Molecularity: 2 (bimolecular). Rate law: Rate = k[\(NO\)][\(Cl_{2}\)]

Step 1: (a) Determine the molecularity of reaction (a)

For reaction (a), \(\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g)\) There is only 1 reactant, which is Cl₂. The molecularity of this reaction is 1 since only one molecule is involved in the elementary reaction step. This reaction is also called a unimolecular reaction.

Step 2: (a) Write the rate law for reaction (a)

The rate law of any elementary reaction depends on the concentrations of the reactants involved. Since reaction (a) is unimolecular, the rate law is written as follows: Rate = k[\(Cl_{2}\)] where k is the rate constant and [\(Cl_{2}\)] represents the concentration of \(Cl_{2}\).

Step 3: (b) Determine the molecularity of reaction (b)

For reaction (b), \(\mathrm{OCl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HOCl}(a q)+\mathrm{OH}^{-}(a q)\) This reaction involves 2 reactant particles: \(OCl^-\) and \(H_2O\). The molecularity of this reaction is 2 since there are two molecules involved in the elementary reaction step. This reaction is also called a bimolecular reaction.

Step 4: (b) Write the rate law for reaction (b)

Since reaction (b) is bimolecular, the rate law is written as follows: Rate = k[\(OCl^-\)][\(H_2O\)] where k is the rate constant, [\(OCl^-\)] represents the concentration of \(OCl^-\), and [\(H_2O\)] represents the concentration of \(H_2O\).

Step 5: (c) Determine the molecularity of reaction (c)

For reaction (c), \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}_{2}(g)\) This reaction involves 2 reactant particles: \(NO\) and \(Cl_{2}\). The molecularity of this reaction is 2 since there are two molecules involved in the elementary reaction step. This reaction is also called a bimolecular reaction.

Step 6: (c) Write the rate law for reaction (c)

Since reaction (c) is bimolecular, the rate law is written as follows: Rate = k[\(NO\)][\(Cl_{2}\)] where k is the rate constant, [\(NO\)] represents the concentration of \(NO\), and [\(Cl_{2}\)] represents the concentration of \(Cl_{2}\).