Problem 43 - Chemistry and Chemical Reactivity

Question

What is the rate law for each of the following elementary reactions? (a) \(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(g) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\) (b) \(\mathrm{Cl}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{H}(\mathrm{g})\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(\mathrm{aq}) \rightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\)

Step-by-Step Solution

Answer

(a) Rate = k [NO][NO3], (b) Rate = k [Cl][H2], (c) Rate = k [(CH3)3CBr].

Step 1: Understanding Rate Laws for Elementary Reactions

For an elementary reaction, the rate law can be written directly from the stoichiometry of the reaction. This means that the rate of the reaction is proportional to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.

Step 2: Write Rate Law for Reaction (a)

The reaction given is \( \mathrm{NO} (\mathrm{g}) + \mathrm{NO}_{3} (\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2} (\mathrm{g}) \). For this elementary reaction, the rate law is: \[ \text{Rate} = k [\mathrm{NO}][\mathrm{NO}_3] \]Here, \(k\) is the rate constant and each reactant concentration is raised to the power of 1 based on their coefficients in the reaction.

Step 3: Write Rate Law for Reaction (b)

The reaction given is \( \mathrm{Cl} (\mathrm{g}) + \mathrm{H}_{2} (\mathrm{g}) \rightarrow \mathrm{HCl} (\mathrm{g}) + \mathrm{H} (\mathrm{g}) \). For this elementary reaction, the rate law is: \[ \text{Rate} = k [\mathrm{Cl}][\mathrm{H}_2] \]Again, each reactant's concentration is raised to the power of 1 since they each have a stoichiometric coefficient of 1.

Step 4: Write Rate Law for Reaction (c)

The reaction given is \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr} (\mathrm{aq}) \rightarrow \left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+} (\mathrm{aq}) + \mathrm{Br}^{-} (\mathrm{aq}) \). For this elementary reaction, the rate law is: \[ \text{Rate} = k [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}] \]The concentration of the reactant \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) is raised to the power of 1 as it is the only reactant with a stoichiometric coefficient of 1.

Key Concepts

Understanding Elementary Reactions

In chemistry, an elementary reaction is a single step process where reactants convert into products in a single collision or simultaneous collisions. What makes elementary reactions special is their simplicity; they represent the fundamental movements in a chemical reaction. Because they occur in just one step, they have direct kinetic data that can be used to define rate laws easily.
Identifying whether a reaction is elementary is crucial. If it is, the rate law can be derived simply based on the stoichiometry of the reactants. This will not work with complex reactions, which may involve multiple steps. For example, given the elementary reaction: \( \mathrm{A} + \mathrm{B} \rightarrow \mathrm{C} \), it is possible to write the rate law directly from the equation, as \( \text{Rate} = k [\mathrm{A}][\mathrm{B}] \). Each reactant's concentration is raised to the power of its coefficient in the balanced equation.

Role of Stoichiometry in Rate Laws

Stoichiometry is the study of the quantitative relationships in chemical reactions. It's a bit like the recipe of a chemical reaction. In the context of rate laws for elementary reactions, stoichiometry directly influences how rate laws are written. Each reactant in an elementary reaction influences the rate to a degree proportional to its stoichiometric coefficient.
For elementary reactions, the stoichiometry tells us exactly which concentrations to include in the rate law and what power to raise them to. For example, if a balanced reaction is \( \mathrm{2A} + 3\mathrm{B} \rightarrow \mathrm{C} \), this would lead to a rate law: \[ \text{Rate} = k [\mathrm{A}]^2[\mathrm{B}]^3 \]. Here, 2 and 3 are the stoichiometric coefficients of \(\mathrm{A}\) and \(\mathrm{B}\), respectively. They indicate how changes in each reactant's concentration will affect the overall rate of the reaction.

Unpacking the Rate Constant

The rate constant, symbolized as \(k\), is a fundamental part of the rate law. It's a proportionality constant that links the rate of a reaction to the concentrations of its reactants. Unlike the concentrations in the rate law, which can change as the reaction proceeds, the rate constant is specific to the particular reaction at a given temperature and does not change during the reaction.
Understanding the units and magnitude of \(k\) can provide insights into the reaction's speed and mechanism. For example, if the rate law is \( \text{Rate} = k [\mathrm{A}][\mathrm{B}] \), the units of \(k\) would depend on the reaction order (sum of the powers in the rate law), ensuring that the rate always has units of concentration/time.

• The higher the value of \(k\), the faster the reaction.
• Temperature and catalysts are factors that can affect \(k\). Increasing temperature usually increases \(k\), making reactions proceed faster.