Kinetics: Determine Reaction Order, Rate Constant, Rate Law
Question
Nitrogen(II) oxide reacts with chlorine according to the equation:
[latex]\begin{gathered} 2\mathrm{NO (g)}+\mathrm{Cl}_2\mathrm{(g)}\rightarrow2\mathrm{NOCl (g)} \end{gathered}[/latex]
The following initial rates of reaction have been observed for certain reactant concentrations:
| Experiment | [NO] (mol/L) | [Cl2] (mol/L) | Rate (M/h) |
|---|---|---|---|
| 1 | 0.50 | 0.50 | 1.14 |
| 2 | 1.00 | 0.50 | 4.56 |
| 3 | 1.00 | 1.00 | 9.12 |
- What are the orders with respect to each reactant?
- Determine the rate law and the rate constant for the reaction from the experimental data.
Show/Hide Answer
- The second order in NO.
The first order in Cl2. - [latex]\mathrm{k}=9.12\mathrm{M}^{-2}\mathrm{~h}^{-1}[/latex]
[latex]\text{Rate}=9.12\mathrm{M}^{-2}\mathrm{h}^{-1}[\mathrm{NO}]^2[\mathrm{Cl}_2][/latex]
Refer to Section 4.4: Rate Laws (1).
Strategy Map
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1. Find the order for NO.
Show/Hide HintSteps to find the order of NO:
These 2 rate laws will be used to create a ratio using the template below: [latex]\begin{gathered} \frac{\mathrm {Rate_x}}{\mathrm {Rate_y}}=\frac{\mathrm{k}[\mathrm{NO}]^{\mathrm{mx}}[\mathrm{Cl} 2]_2^{\mathrm{nx}}}{\mathrm{k}[\mathrm{NO}]^{\mathrm{my}}[\mathrm{Cl} 2]^{\mathrm{ny}}} \end{gathered}[/latex]
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2.Find the order for Cl2.
Show/Hide HintSteps to find the order of Cl2:
These 2 rate laws will be used to create a ratio using the template below: [latex]\begin{gathered} \frac{\mathrm {Rate_x}}{\mathrm {Rate_y}}=\frac{\mathrm{k}[\mathrm{NO}]^{\mathrm{mx}}[\mathrm{Cl} 2]_2^{\mathrm{nx}}}{\mathrm{k}[\mathrm{NO}]^{\mathrm{my}}[\mathrm{Cl} 2]^{\mathrm{ny}}} \end{gathered}[/latex]
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3.Choose 1 of the experimental examples to solve for k.
Show/Hide HintSteps to solve for k:
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Solution
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[latex]\begin{gathered} \text{Rate}=\mathrm{k}[\mathrm{NO}]^2[\mathrm{Cl}_2] \end{gathered}[/latex]
a. What are the orders with respect to each reactant?
Order of NO (m):
[latex]\begin{aligned} &\frac{\text {Rate }2}{\text {Rate }1}=\frac{4.56}{1.14}=4\\ &\begin{gathered} \frac{\text {Rate }2}{\text {Rate }1}=\frac{k_2[\mathrm{NO}]_2^m[\mathrm{Cl}]_2^n}{k_1[\mathrm{NO}]_1^m[\mathrm{Cl}]_1^n}=\frac{k_2(1.00)^m[\mathrm{Cl}]_2^n}{k_1(0.50)^m[\mathrm{Cl}]_1^n} \\ =\left(\frac{1.00}{0.50}\right)^m=2^m \\ 2^m=4 \end{gathered} \end{aligned}[/latex]
m = 2 is the order for NO
Answer: Order with respect to NO is 2.
Order of Cl2 (n):
[latex]\begin{aligned} &\begin{gathered} \frac{\text {Rate }3}{\text {Rate }2}=\frac{k_3\left[\mathrm{NO}_3^m[\mathrm{Cl}]_3^n\right.}{k_2[\mathrm{NO}]_2^m[\mathrm{Cl}]_2^n}=\frac{k_3[\mathrm{NO}]_3^m(1.00)^n}{k_2[\mathrm{NO}]_2^m(0.50)^n}\\ =\left(\frac{1.00}{0.50}\right)^m=\frac{\text {Rate }3}{\text {Rate }2}\\ \left(\frac{1.00}{0.50}\right)^m=2^n\\ 2^n=2 \\ \end{gathered} \end{aligned}[/latex]
n = 1 is the order for Cl2
Answer: Order with respect to Cl2 is 1.
b. Determine the rate law and the rate constant for the reaction from the experimental data.
Finding k:
[latex]\begin{aligned} &\begin{gathered} \text{Rate}=k[\mathrm{NO}]^{\mathrm{m}}\left[\mathrm{Cl}_2\right]^{\mathrm{n}}\\ k=\frac{\text {Rate}}{[\mathrm{NO}]^{\mathrm{m}}\left[\mathrm{Cl}_2\right]^{\mathrm{n}}}\\ k=\frac{1.14\mathrm{M}/\mathrm{h}}{(0.50\mathrm{M})^2(0.50\mathrm{M})^1}\\ k=9.12\mathrm{M}^{-2}\mathrm{~h}^{-1} \end{gathered} \end{aligned}[/latex]
Rate law is:
[latex]\begin{gathered} \text{Rate}=9.12\mathrm{M}^{-2}\mathrm{h}^{-1}[\mathrm{NO}]^2[\mathrm{Cl}_2] \end{gathered}[/latex]
Answers:
- [latex]\mathrm{k}=9.12\mathrm{M}^{-2}\mathrm{~h}^{-1}[/latex]
- [latex]\text{Rate}=9.12\mathrm{M}^{-2}\mathrm{h}^{-1}[\mathrm{NO}]^2[\mathrm{Cl}_2][/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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| Guided Solution Ideas |
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| This question is a calculation problem which asks you to determine the rate law and the rate constant for the reaction from the experimental data.
In order to determine the rate law, we must first find the reaction order with respect to each reactant. Show/Hide Resource
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To find the order of NO, first identify which 2 experiments from the given table have the same concentration of Cl2.
Think About This!Experiments 1 and 2 |
| The rate laws for experiment 1 and 2 will be used to create a ratio using the template below:
[latex]\begin{gathered} \frac{\mathrm{Rate_x}}{\mathrm{Rate_y}}=\frac{\mathrm{k}[\mathrm{NO}]^{\mathrm{mx}}[\mathrm{Cl} 2]_2^{\mathrm{nx}}}{\mathrm{k}[\mathrm{NO}]^{\mathrm{my}}[\mathrm{Cl} 2]^{\mathrm{ny}}} \end{gathered}[/latex] Identify which associated rate is larger (this one will be in the numerator of your fraction). Show/Hide Think About This!
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Cancel out the rates that are the same.
Show/Hide Think About This!
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| Solve for m |
To find the order of Cl2, follow the same process by first identifying which 2 experiments from the given table have the same concentration of NO.
Show/Hide Think About This!
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Then, create the rate law fraction using the template, putting the largest rate as the numerator.
Show/Hide Think About This![latex]\begin{gathered} \frac{\mathrm{Rate_x}}{\mathrm{Rate_y}}=\frac{\mathrm{k}[\mathrm{NO}]^{\mathrm{mx}}[\mathrm{Cl} 2]_2^{\mathrm{nx}}}{\mathrm{k}[\mathrm{NO}]^{\mathrm{my}}[\mathrm{Cl} 2]^{\mathrm{ny}}} \end{gathered}[/latex] Reaction 3 will be in the numerator. |
Cancel out the rates that are the same.
Show/Hide Don’t Forget!Cancel out [NO]m in the numerator and denominator. |
| Solve for n |
Then solve for k:
Show/Hide Reaction Equation![latex]\begin{gathered} \text{Rate}=\mathrm{k}\left[\mathrm{NO}^\mathrm{m}\left[\mathrm{Cl}_2\right]^\mathrm{n}]\right. \end{gathered}[/latex] |
Use your dimensional analysis approach by stating all your units in your rate expression to ensure you have the correct units for k.
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| Complete Solution b) IF3 |
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Rate = k[NO]2[Cl2] |
| Order of NO (m)
A ratio is set up using the 2 rate laws where the experimental values of the Cl2 concentration are the same: [latex]\begin{gathered} \frac{\mathrm{k}[\mathrm{NO}]^\mathrm{m}\left[\mathrm{Cl}_2\right]^\mathrm{n}}{\mathrm{k}[\mathrm{NO}]^\mathrm{m}\left[\mathrm{Cl}_2\right]^\mathrm{n}}=\frac{\mathrm{k}(1.00)^\mathrm{m}(0.50)^\mathrm{n}}{\mathrm{k}(0.50)^\mathrm{m}(0.50)^\mathrm{n}}=\frac{4.56}{1.14}=\frac{\text {Rate } 2}{\text {Rate } 1} \end{gathered}[/latex] Cancel out the values that are the same in the numerator and denominator: [latex]\begin{gathered} \left(\frac{1.00}{0.50}\right)^\mathrm{m}=\frac{4.56}{1.14} \end{gathered}[/latex] Solve for m: [latex]\begin{aligned} (2)^\mathrm{m}&=4 \\ \mathrm{m}&=2 \end{aligned}[/latex] Answer: Order with respect to NO is 2. |
| Order of Cl2 (n)
A ratio is set up using the 2 rate laws where the experimental values of the NO concentration are the same: [latex]\begin{gathered} \frac{\mathrm{k}\left[\mathrm{NO}^\mathrm{m}\right]^\mathrm{m}\left[\mathrm{Cl}_2\right]^\mathrm{n}}{\mathrm{k}[\mathrm{NO}]^\mathrm{m}\left[\mathrm{Cl}_2\right]^\mathrm{m}}=\frac{\mathrm{k}(1.00)^\mathrm{m}(1.00)^\mathrm{n}}{\mathrm{k}(1.00)^\mathrm{m}(0.50)^\mathrm{n}}=\frac{9.12}{4.56}=\frac{\text{Rate } 3}{\text {Rate } 2} \end{gathered}[/latex] Cancel out the values that are the same in the numerator and denominator: [latex]\begin{gathered} \left(\frac{1.00}{0.50}\right)^\mathrm{m}=\frac{4.56}{1.14} \end{gathered}[/latex] Solve for n: [latex]\begin{aligned} (2)^\mathrm{n}&=2 \\ \mathrm{n}&=1 \end{aligned}[/latex] Answer: Order with respect to Cl2 is 1. |
| Finding k
Recall that rate laws are only the reactants of a reaction: [latex]\begin{gathered} \text{Rate}=\mathrm{k}[\mathrm{NO}]^{\mathrm{m}}\left[\mathrm{Cl}_2\right]^{\mathrm{n}} \end{gathered}[/latex] Isolate k to prepare to solve for it: [latex]\begin{gathered} \mathrm{k}=\frac{\text{Rate}}{[\mathrm{NO}]^{\mathrm{m}}\left[\mathrm{Cl}_2\right]^{\mathrm{n}}} \end{gathered}[/latex] Choose any of the 3 sets of experimental values from the table. In this case, experiment 1 was used: [latex]\begin{aligned} \mathrm{k}&=\frac{1.14\mathrm{M}/\mathrm{s}}{(0.50 \mathrm{M})^2(0.50\mathrm{M})^1}\\ \mathrm{k}&=9.12\mathrm{M}^{-2}\mathrm{~s}^{-1} \end{aligned}[/latex] Answers:
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Check Your Work
Summary of what we would expect based on the related chemistry theory.
Show/Hide Watch Out!
Verify that your rate law format matches below, that you have indicated the order as exponents, and that your units for k match the overall reaction order.
Show/Hide Think About This!
m + n = 3; the order for this reaction is third order. A third order rate constant will have the units of M-2s-1 .
Does your answer make chemical sense?
Show/Hide Answer
The rate shows that the rate of the reaction is first order with respect to Cl2 and second order with respect to NO. When looking back at the experimental data, the concentration of Cl2 doubles while NO stays the same; because of that, the rate doubles. We can also see that the rate increases by a factor of 4 when the concentration of NO doubles, so the calculated orders make sense.
Rate laws provide a mathematical description of how changes in the amount of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the amount of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible.
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Refer to Section 4.4: Rate Laws (1).
PASS Attribution
- LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (3).
- Question 4.25 from LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (4) is adapted under a CC BY 4.0 license.
- Question 4.25 is question 12.E.3.16: Q12.3.15 from LibreTexts Chemistry 1e (OpenSTAX) (5), which is under a CC BY 4.0 license.
- Question 12.E.3.16: Q12.3.15 is question 25 from OpenStax Chemistry 2e (6), 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. 4.4: Rate Laws. 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/4.04%3A_Rate_Laws.
2. OpenStax. 1.1.6: Mathematical Treatment of Measurement Results. 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)/01%3A_Background/1.01%3A_Essential_Ideas_of_Chemistry/1.1.06%3A_Mathematical_Treatment_of_Measurement_Results.
3. Thompson Rivers University. 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).
4. 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).
5. 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).
6. 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.
