Kinetics — Writing Rate Expressions

Question

Ozone decomposes to oxygen according to the equation [latex]\mathrm{2O_3 (g)} \rightarrow \mathrm{3O_2 (g)}[/latex].

Write the equation that relates the rate expressions for this reaction in terms of the disappearance of [latex]\mathrm{O_3}[/latex] and the formation of oxygen.

 

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Rate Expression = [latex]-\dfrac{1}{2}\left(\dfrac{\Delta\left[\mathrm{O}_\mathrm{3}\right]}{\Delta \mathrm{t}}\right)=\dfrac{1}{3}\left(\dfrac{\Delta\left[\mathrm{O}_\mathrm{2}\right]}{\Delta \mathrm{t}}\right)[/latex]

Refer to Section 4.2: Chemical Reaction Rates (1).

Strategy Map

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Table 1: Strategy Map
Strategy Map Steps
1. Identify stoichiometry.

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Make sure the reaction is balanced.
2. Identify what are reactants and what are products.

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For general reaction: aA + bB → cC + dD

[latex]\begin{equation} \text { rate }=-\dfrac{1}{\mathrm{a}} \dfrac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}}=-\dfrac{1}{\mathrm{b}} \dfrac{\Delta[\mathrm{B}]}{\Delta \mathrm{t}}=\dfrac{1}{\mathrm{c}} \dfrac{\Delta[\mathrm{C}]}{\Delta \mathrm{t}}=\dfrac{1}{\mathrm{d}} \dfrac{\Delta[\mathrm{D}]}{\Delta \mathrm{t}} \end{equation}[/latex]

Solution

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[latex]\begin{equation} \text { Rate}=-\dfrac{1}{2}\left(\dfrac{\Delta\left[\mathrm{O}_\mathrm{3}\right]}{\Delta \mathrm{t}}\right)=\dfrac{1}{3}\left(\dfrac{\Delta\left[\mathrm{O}_\mathrm{2}\right]}{\Delta \mathrm{t}}\right) \end{equation}[/latex]

Ozone is a reactant, so it requires a negative. The stoichiometric coefficient from the balanced reaction is 2, so the rate is multiplied by one-half.

Oxygen is a product, so it is positive. The stoichiometric coefficient is 3, so the rate is multiplied by one-third.

Guided Solution

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Table 2: Guided Solution
Guided Solution Ideas
This question is a theory problem that requires you to set up the reaction rates for the given reaction.

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Refer to Section 4.2: Chemical Reaction Rates (1).
What is a chemical reaction rate?

Rate of formation/ decomposition = [latex]\dfrac{\text{(change in concentration)}}{\text{(time interval)}}[/latex]

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Change in concentration = [latex]\Delta{\text{[reactant or product]}={[\;]\mathrm{_f} – [\;]\mathrm{_i}}}[/latex]

[latex]\begin{equation} \text { time interval }=\Delta \mathrm{t=t_f-t_i} \end{equation}[/latex]
The reaction is written as [latex]\mathrm{2O_3 (g)} \rightarrow \mathrm{3O_2 (g)}[/latex], meaning for the breakdown of every two ozone molecules, three oxygen molecules are produced. How is this used in your expression?

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The stoichiometry in the reaction is used as coefficients in your rates. However, they are used in their reciprocal form.
Recall that the amount of reactants present decrease over time, and the amount of products increase. How is this information represented in your expression?

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Reaction rates are positive. In the reaction rate expression, reacting compounds/molecules will have a negative sign in front of their fraction to indicate that they are decreasing in concentration over reaction time, while product terms will be positive to indicate they are increasing in concentration over reaction time.

[latex]\begin{aligned} \text{For a reaction } \mathrm{A}\rightarrow\mathrm{B}\\ \end{aligned}[/latex]

[latex]\begin{aligned} \text{rate} & =-\dfrac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}}\\ \text{rate} & =\dfrac{\Delta[\mathrm{B}]}{\Delta \mathrm{t}} \end{aligned}[/latex]
Check Your Work: Reactants decrease in concentration as the reaction proceeds while products increase. The reactant term has a negative sign, and the product term has a positive sign.

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Make sure the reactant has a negative sign in front of the concentration, and you use the ‘Δ’ symbol to represent change in both the numerator and denominator. Square brackets in the numerator represent concentration in M (moles/litre).

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The number of products formed will always equal the number of reactants used. This is because reactants break to turn into products.

In a reaction rate, the reactants will need a negative sign. This is because the number of reactants will always decrease (negative) over time.

When a negative is multiplied by a negative, it becomes a positive, making the two rates equal.
Table 3: Complete Solution
Complete Solution
[latex]\begin{equation} \text { Rate Expression }=-\dfrac{1}{2}\left(\dfrac{\Delta\left[\mathrm{O_3}\right]}{\Delta \mathrm{t}}\right)=\dfrac{1}{3}\left(\dfrac{\Delta\left[\mathrm{O_2}\right]}{\Delta \mathrm{t}}\right) \end{equation}[/latex]Ozone is a reactant, so it requires a negative. Two are used, so the expression is multiplied by one-half.Oxygen is a product, so it is positive. Three are used, so the rate is multiplied by one-third.Square brackets representing concentration in molarity are used for the numerator because we are looking at the change in the concentrations through time. The products will increase over time, and the reactants will decrease.

Check Your Work

Make sure the reactant term has a negative sign in front of the concentration, and you use the ‘Δ’ symbol to represent change in both the numerator and denominator. Square brackets in the numerator represent concentration in M (moles/litre). Stoichiometric coefficients show up as appropriate fractions to relate the change in reactants and products.

Does your answer make chemical sense? 

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The rate expression formula reflects the balanced stoichiometry for the reaction equation incorporated as a fraction coefficient term. The reactant term is negative because the concentration is decreasing overtime as the reactant molecules are transformed into product molecules. The product terms are positive because the concentration of product molecules is increasing over time as the reaction proceeds.

PASS Attribution

References

1. OpenStax. 4.2: Chemical Reaction Rates. In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520). LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/04%3A_Kinetics/4.02%3A_Chemical_Reaction_Rates.

2. Blackstock, L.; Brewer, S.; Jensen, A. In PASS Chemistry Book CHEM 1510/1520; LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1510%2F%2F1520.

3. OpenStax. 4.E: Kinetics (Exercises). In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520). LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/04%3A_Kinetics/4.E%3A_Kinetics_(Exercises).

4. OpenStax. 12.E: Kinetics (Exercises). In Chemistry 1e (OpenSTAX). LibreTexts, 2023. https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/12%3A_Kinetics/12.E%3A_Kinetics_(Exercises).

5. Flowers, P.; Theopold, K.; Langley, R.; Robinson, W. R. (2019). Ch. 12 Exercises. In Chemistry 2e. OpenStax, 2019. https://openstax.org/books/chemistry-2e/pages/12-exercises.

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