11 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.
Show/Hide Answer
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
Do you need a little help to get started?
Check out the strategy map.
Show/Hide Strategy Map
| Strategy Map Steps |
|---|
1. Identify stoichiometry.
Show/Hide HintMake sure the reaction is balanced. |
2. Identify what are reactants and what are products.
Show/Hide HintFor 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
Do you want to see the steps to reach the answer?
Check out this solution.
Show/Hide Solution
[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
Do you want more help?
The guided solution below will give you the reasoning for each step to get your answer, with reminders and hints.
Show/Hide Guided Solution
| Guided Solution Ideas |
|---|
This question is a theory problem that requires you to set up the reaction rates for the given reaction.
Show/Hide ResourceRefer 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] Show/Hide Think About This!
|
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?
Show/Hide Think About This!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?
Show/Hide Don’t Forget!
|
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.
Show/Hide Watch Out!
|
| 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?
Show/Hide Answer
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
- LibreTexts PASS Chemistry Book CHEM 1510/1520 (2).
- Question 4.E.2 from LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (3) is used under a CC BY 4.0 license.
- Question 4.E.2 is question 12.E.1.2: Q12.1.2 from LibreTexts Chemistry 1e (OpenSTAX) (4), which is under a CC BY 4.0 license.
- Question 12.E.1.2: Q12.1.2 is question 2 from OpenStax Chemistry 2e (5), which is under a CC BY 4.0 license. Access for free at https://openstax.org/books/chemistry-2e/pages/1-introduction
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.
