Problem 1 - Chemistry and Chemical Reactivity

Question

Give the relative rates of disappearance of reactants and formation of products for each of the following reactions. (a) \(2 \mathrm{O}_{3}(\mathrm{g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{g})\) (b) \(2 \mathrm{HOF}(\mathrm{g}) \rightarrow 2 \mathrm{HF}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\)

Step-by-Step Solution

Answer

(a) Rate: \(-\frac{1}{2} \frac{d[\mathrm{O}_{3}]}{dt} = \frac{1}{3} \frac{d[\mathrm{O}_{2}]}{dt}\), (b) Rate: \(-\frac{1}{2} \frac{d[\mathrm{HOF}]}{dt} = \frac{1}{2} \frac{d[\mathrm{HF}]}{dt} = \frac{d[\mathrm{O}_{2}]}{dt}\).

Step 1: Understanding the Rate of Reaction

In a chemical reaction, the rate of reaction for a reactant is the rate at which the reactant is consumed. It is usually expressed as a negative value because the concentration of the reactant decreases over time. Conversely, for products, the rate is positive as their concentration increases over time. The general formula for rate is \( ext{Rate} = -\frac{1}{n} \frac{d[ ext{Reactant}]}{dt} = \frac{1}{m} \frac{d[ ext{Product}]}{dt} \), where \(n\) and \(m\) are the stoichiometric coefficients.

Step 2: Writing the Rate Expression for Reaction (a)

For the reaction \(2 \mathrm{O}_{3} \rightarrow 3 \mathrm{O}_{2}\), the rate of disappearance of \(\mathrm{O}_{3}\) is \(-\frac{1}{2} \frac{d[\mathrm{O}_{3}]}{dt}\), and the rate of formation of \(\mathrm{O}_{2}\) is \(\frac{1}{3} \frac{d[\mathrm{O}_{2}]}{dt}\). So, \(-\frac{1}{2} \frac{d[\mathrm{O}_{3}]}{dt} = \frac{1}{3} \frac{d[\mathrm{O}_{2}]}{dt}\).

Step 3: Writing the Rate Expression for Reaction (b)

For the reaction \(2 \mathrm{HOF} \rightarrow 2 \mathrm{HF} + \mathrm{O}_{2}\), the rate of disappearance of \(\mathrm{HOF}\) is \(-\frac{1}{2} \frac{d[\mathrm{HOF}]}{dt}\). The rate of formation of \(\mathrm{HF}\) is \(\frac{1}{2} \frac{d[\mathrm{HF}]}{dt}\), and the rate of formation of \(\mathrm{O}_{2}\) is \(\frac{d[\mathrm{O}_{2}]}{dt}\). These rates are equal; hence,\(-\frac{1}{2} \frac{d[\mathrm{HOF}]}{dt} = \frac{1}{2} \frac{d[\mathrm{HF}]}{dt} = \frac{d[\mathrm{O}_{2}]}{dt}\).

Key Concepts

Stoichiometric Coefficients

In chemical reactions, stoichiometric coefficients are the numbers placed before molecules in a chemical equation that indicate the proportion of each substance involved. These coefficients are crucial as they ensure the law of conservation of mass balances the equation. In the reactions given, for example:

• In reaction (a), the stoichiometric coefficients are 2 for \( \mathrm{O}_3 \) and 3 for \( \mathrm{O}_2 \). This shows that two moles of ozone yield three moles of oxygen.
• For reaction (b), the stoichiometric coefficients are 2 for \( \mathrm{HOF} \) and \( \mathrm{HF} \), while it's 1 for \( \mathrm{O}_2 \). Thus, two moles of \( \mathrm{HOF} \) produce two moles of \( \mathrm{HF} \) and one mole of \( \mathrm{O}_2 \).

The coefficients are critical for calculating rates, as they tell us how quickly each substance reacts or is formed relative to others in the equation. They directly influence the rate expressions, as seen in the way rates of disappearance and formation are calculated.

Rate of Disappearance

The rate of disappearance refers to how fast a reactant is consumed in a chemical reaction. Since reactants decrease in concentration over time, the rate of disappearance is expressed as a negative value. For accurate calculations, this rate is divided by the stoichiometric coefficient of the reactant.

• For reaction (a), the rate of disappearance for \( \mathrm{O}_3 \) is shown as \( -\frac{1}{2} \frac{d[\mathrm{O}_{3}]}{dt} \), illustrating that per mole of reaction progress, half a mole of \( \mathrm{O}_3 \) is used.
• In reaction (b), the rate of disappearance for \( \mathrm{HOF} \) is \( -\frac{1}{2} \frac{d[\mathrm{HOF}]}{dt} \), indicating that per mole of reaction progress, half a mole of \( \mathrm{HOF} \) disappears.

Understanding this rate helps predict how fast a reaction will proceed under given conditions.

Rate of Formation

The rate of formation describes the speed at which a product is produced in a reaction. Unlike the rate of disappearance, it is typically positive, as product concentrations increase over time. The rate is determined using stoichiometric coefficients, ensuring consistency with the reaction's balanced equation.

• In reaction (a), the rate of formation of \( \mathrm{O}_2 \) is \( \frac{1}{3} \frac{d[\mathrm{O}_2]}{dt} \). This fraction indicates that one molecule of \( \mathrm{O}_3 \) results in 1.5 molecules (or one and a half moles) of \( \mathrm{O}_2 \) being formed.
• For reaction (b), the rate of formation for \( \mathrm{HF} \) is \( \frac{1}{2} \frac{d[\mathrm{HF}]}{dt} \) and for \( \mathrm{O}_2 \), it is \( \frac{d[\mathrm{O}_2]}{dt} \). Thus, each mole of \( \mathrm{HOF} \) forms an equivalent of \( \mathrm{HF} \), showcasing balanced production rates based on the reaction stoichiometry.

These rates provide insight into the yield of a reaction and are essential for scaling reactions in lab and industrial settings.

Chemical Kinetics

Chemical kinetics is the study of the speed or rate of chemical reactions and the factors affecting these rates. It connects deeply with the rates of disappearance and formation, and stoichiometry of reactions.

• It considers concentrations of reactants and products, reaction temperature, catalysts, and other variables influencing reaction rates.
• Understanding kinetics helps in practical applications like optimizing reactions for industry, controlling reaction environments in laboratories, and predicting reaction behavior in different conditions.
• In our reactions, calculating the rates of disappearance and formation as per stoichiometric coefficients exemplifies how kinetics helps in understanding reaction dynamics.

Kinetics provides a theoretical framework for predicting and managing how quickly a reaction will occur, ensuring successful completion of processes in various scientific and industrial fields.