Kinetics: Rate of Appearance, Rate of Disappearance
Question
Consider the reaction below:
[latex]\begin{equation} \begin{gathered} 5\mathrm{Br}^{-}(\mathrm{aq})+\mathrm{BrO}_3^{-}(\mathrm{aq})+6\mathrm{H}^{+}(\mathrm{aq})\rightarrow3\mathrm{Br}_2(\mathrm{aq})+3\mathrm{H}_2\mathrm{O}(\ell) \end{gathered} \end{equation}[/latex]
If the rate of disappearance of Br– (aq) at a particular moment during the reaction is 3.5 × 10−4 M s−1, what is the rate of appearance of Br2 (aq) at that moment?
Show/Hide Answer
[latex]\begin{equation} \begin{gathered} \frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta\mathrm{t}}=2.1\times10^{-4}\mathrm{M}/\mathrm{s} \end{gathered} \end{equation}[/latex]
Refer to Section 4.2: Chemical Reaction Rates (1).
Strategy Map
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1. Write out the full chemical reaction rate expression.
Show/Hide HintFor the reaction: [latex]\begin{gathered} \mathrm{aA}+\mathrm{bB}\rightarrow\mathrm{cC}+\mathrm{dD} \end{gathered}[/latex] Reaction rate: [latex]\begin{equation} \begin{gathered} =-\frac{1}{\mathrm{a}}\dfrac{\Delta[\mathrm{~A}]}{\Delta\mathrm{t}}=-\frac{1}{\mathrm{~b}}\frac{\Delta[\mathrm{~B}]}{\Delta\mathrm{t}}=\frac{1}{\mathrm{c}}\frac{\Delta[\mathrm{~C}]}{\Delta\mathrm{t}}=\frac{1}{\mathrm{~d}}\frac{\Delta[\mathrm{D}]}{\Delta\mathrm{t}} \end{gathered} \end{equation}[/latex] |
2. Identify which 2 expressions represent what you have and what you need.
Show/Hide HintRemember that your rates of disappearance will be negative and your rates of appearance will be positive. Rate of disappearance of Br– (aq) is always negative, therefore: [latex]\begin{equation} \begin{gathered} \left(\frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}}\right)=-3.5\times 10^{-4}\mathrm{M}\mathrm{~s}^{-1} \end{gathered} \end{equation}[/latex] Rate of appearance Br2 (aq): [latex]\begin{equation} \begin{gathered} \frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta\mathrm{t}}=? \end{gathered} \end{equation}[/latex] |
3. Plug the rate given in the problem to the expression you created in your chemical reaction rate.
Show/Hide HintEnter the given value and units: [latex]\begin{equation} \begin{gathered} \left(\frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta\mathrm{t}}\right)=-3.5\times10^{-4}\mathrm{M}\mathrm{~s}^{-1} \end{gathered} \end{equation}[/latex] |
4. Solve for the rate of appearance of Br2.
Show/Hide HintMake sure to write out all steps as you rearrange the expressions and keep track of your signs and fractions. |
Solution
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[latex]\begin{aligned} -\frac{1}{5}\left(\frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta\mathrm{t}}\right)&=\frac{1}{3}\left(\frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta \mathrm{t}}\right)\\\\ -\frac{1}{5}\left(-3.5\times 10^{-4} \mathrm{M}/\mathrm{s}\right)&=\frac{1}{3}\left(\frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta\mathrm{t}}\right)\\\\ -\frac{1}{5} \times \frac{3}{1}\left(-3.5\times 10^{-4}\mathrm{M}/\mathrm{s}\right)&=\frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta\mathrm{t}}\\\\ \frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta\mathrm{t}}&=2.1\times 10^{-4}\mathrm{M}/\mathrm{s} \end{aligned}[/latex]
Answer: [latex]\begin{gathered}\frac{\Delta[\mathrm{Br}_2]}{\Delta\mathrm{t}}=2.1\times 10^{-4}\mathrm{M}/\mathrm{s}\end{gathered}[/latex]
Guided Solution
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The guided solution below will give you the reasoning for each step to get your answer, with reminders and hints.
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This question is a calculation type problem that requires you to set up the chemical reaction rate to calculate the appearance of one compound given the disappearance of another.
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If the rate of disappearance of Br–(aq) at a particular moment during the reaction is 3.5 × 10−4 M/s−1, what is the rate of appearance of Br2 (aq) at that moment?
Show/Hide Think About This!The problem gives that Br– is disappearing (and, therefore, is a reactant). The rate of its disappearance is -3.5 × 10-4 M/s. The problem also states that Br2 is appearing (and ,therefore, is a product). You are asked to find the rate of appearance of Br2. |
| The rate of a chemical reaction is the change in concentration over the change in time. Rate is always a positive value. Rate can be expressed in terms of the rate of disappearance of a reactant or in terms of rate of appearance of a product.
Rate = – Rate of disappearance of reactants: [latex]\begin{gathered} =-\frac{\Delta[\text {reactants}]}{\Delta\mathrm{t}} \end{gathered}[/latex] Rate = Rate of appearance of products: [latex]\begin{gathered} =\frac{\Delta[\text {products}]}{\Delta t} \end{gathered}[/latex] Show/Hide Don’t Forget!Because reactant concentrations decrease with time as the reaction is proceeding, Δ[reactants] will be negative. Since reactant concentrations are decreasing and Δ is final – initial, there must be a negative sign in front of the rate term for disappearance of reactants to relate it to the rate of the reaction. |
| Write out the chemical reaction rate expression in terms of rate of appearance of products and disappearance of reactants:
[latex]\begin{gathered} 5\mathrm{Br}^{-}(\mathrm{aq})+\mathrm{BrO}_3^{-}(\mathrm{aq})+6\mathrm{H}^{+}(\mathrm{aq})\rightarrow3\mathrm{Br}_2(\mathrm{aq})+3\mathrm{H}_2\mathrm{O}(\ell) \end{gathered}[/latex] Show/Hide Think About This!
For the reaction: [latex]\begin{gathered} \mathrm{aA}+\mathrm{bB}\rightarrow\mathrm{cC}+\mathrm{dD} \end{gathered}[/latex] Reaction rate: [latex]\begin{gathered} =-\frac{1}{\mathrm{a}}\frac{\Delta[\mathrm{~A}]}{\Delta\mathrm{t}}=-\frac{1}{\mathrm{~b}}\frac{\Delta[\mathrm{~B}]}{\Delta\mathrm{t}}=\frac{1}{\mathrm{c}}\frac{\Delta[\mathrm{~C}]}{\Delta \mathrm{t}}=\frac{1}{\mathrm{~d}}\frac{\Delta[\mathrm{D}]}{\Delta\mathrm{t}} \end{gathered}[/latex] Square brackets [ ] represent concentration in moles per litre (M) Δ represents change, so Δ[ ] represents the change in concentration and Δt represents the time interval. |
Since the stoichiometric coefficients are different for our species, we must divide each change in concentration by the appropriate coefficient to relate their rate of appearance and disappearance to each other.
Show/Hide Don’t Forget![latex]\begin{array}{lcl} \textcolor{blue}{\mathbf{5}}\mathrm{Br}^{-}(\mathrm{aq})+\mathrm{BrO}_3^{-}(\mathrm{aq})+6\mathrm{H}^{+}(\mathrm{aq})&\rightarrow&\textcolor{blue}{\mathbf{3}}\mathrm{Br}_2(\mathrm{aq})+3\mathrm{H}_2\mathrm{O}(\ell)\\ \textcolor{blue}{\uparrow}&&\textcolor{blue}{\uparrow} \end{array}[/latex] |
What prediction can you make for the magnitude of the rates given the stoichiometry of the reactant and product in the question?
Show/Hide Don’t Forget!Look at the stoichiometric coefficients. For every 5 moles of Br– (aq) that disappears, 3 moles of Br2 (aq) appears. Since the stoichiometric coefficient for Br– is larger than for Br2, you can expect your numerical answer to be smaller than the rate you are given. Remember signs simply indicate appearance or disappearance. |
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| Set up the complete chemical reaction rate expression:
[latex]\begin{aligned} &\text {Rate }=-\frac{1}{5}\left(\frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}}\right)=-\left(\frac{\Delta\left[\mathrm{BrO}_8^{-}\right]}{\Delta\mathrm{t}}\right)=\\ & -\frac{1}{6}\left(\frac{\Delta\left[\mathrm{H}^{+}\right]}{\Delta\mathrm{t}}\right)=\frac{1}{3}\left(\frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta\mathrm{t}}\right) \end{aligned}[/latex] Show/Hide Don’t Forget!
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Check Your Work
Writing out the rate expression correctly is key to solving this problem. Make sure your sign for the rate of disappearance terms (reactants) is negative and your sign for the rate of appearance terms (products) is positive. Also, ensure you are using the appropriate stoichiometric coefficients for your fractional terms.
Refer to Section 4.2: Chemical Reaction Rates (1).
Show/Hide Think About This!
For the reaction:
[latex]\begin{gathered} \mathrm{aA}+\mathrm{bB}\rightarrow\mathrm{cC}+\mathrm{dD} \end{gathered}[/latex]
Reaction rate:
[latex]\begin{equation} \begin{gathered} =-\frac{1}{\mathrm{a}}\frac{\Delta[\mathrm{~A}]}{\Delta\mathrm{t}}=-\frac{1}{\mathrm{~b}}\frac{\Delta[\mathrm{~B}]}{\Delta\mathrm{t}}=\frac{1}{\mathrm{c}}\frac{\Delta[\mathrm{~C}]}{\Delta\mathrm{t}}=\frac{1}{\mathrm{~d}}\frac{\Delta[\mathrm{D}]}{\Delta\mathrm{t}} \end{gathered} \end{equation}[/latex]
Does your answer make chemical sense?
Show/Hide Answer
We were given the rate of disappearance of Br– (aq) to be 3.5 × 10−4 M s−1, and we expected the rate of appearance of Br2 (aq) at that time to be a smaller number as the mole ratio from the balanced chemical reaction for Br−(aq) : Br2(aq) is 5:3.
[latex]\begin{equation} \begin{gathered} 5\mathrm{Br}^{-}(\mathrm{aq})+\mathrm{BrO}_3^{-}(\mathrm{aq})+6\mathrm{H}^{+}(\mathrm{aq})\rightarrow3\mathrm{Br}_2(\mathrm{aq})+3\mathrm{H}_2\mathrm{O}(\ell) \end{gathered} \end{equation}[/latex]
Our answer of 2.1 × 10-4 M/s is a smaller value, which makes chemical sense as more Br–(aq) must decompose to make Br2(aq).
Show/Hide Don’t Forget!
For every 5 moles of Br– (aq) that disappears, 3 moles of Br2 (aq) appears.
Reactants break up to turn into products. For this reason, we can set up these chemical reaction rate expressions and identify the rate of disappearance or appearance of one from the other.
PASS Attribution
- LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (2).
- Question 4.6 from LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (3) is used under a CC BY 4.0 license.
- Question 4.6 is question 12.E.1.6: Q12.1.6 from LibreTexts Chemistry 1e (OpenSTAX) (4), which is under a CC BY 4.0 license.
- Question 12.E.1.6: Q12.1.6 is question 6 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. OpenStax. TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2024. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520).
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.
