Chemical Equilibrium: Given Starting Concentrations and K, Calculate Equilibrium Concentrations
Question
Calculate the equilibrium concentrations of N2O4 and NO2 in a 1.00 L vessel that was prepared starting with only 0.129 mol of N2O4.
[latex]\mathrm{N}_2\mathrm{O}_4(\mathrm{g})\rightleftharpoons2\mathrm{NO}_2(\mathrm{g});\,\mathrm{K}_{\mathrm{c}}={{1.07}\times 10^{-5}}[/latex]
Show/Hide Answer
- [NO2] = 1.17 × 10-3 M
- [N2O4] = 0.128 M
Refer to Section 5.4: Equilibrium Calculatioins (1).
Strategy Map
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1. Recognize we are starting with only N2O4 present
Show/Hide HintWe know the initial moles of N2O4 and that the vessel has a 1.00 L volume, so we know concentration in molarity. |
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2. Write the equilibrium constant expression for the given reaction using the reactant and product species. Include stoichiometric coefficients as exponents.
Show/Hide Hint[latex]\begin{aligned} \mathrm{K}_\mathrm{c}&=\frac{[\text{Product}]}{[\text{Reactants}]}\\ \mathrm{K}_{\mathrm{c}}&=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2 \mathrm{O}_4\right]} \end{aligned}[/latex] |
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3. Create an ICE table.
Show/Hide Hint
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| 4. Enter initial concentrations and use ‘x’ to represent the change.
Complete the table and enter equilibrium concentration terms in to your Kc expression. Solve for ‘x’ and calculate equilibrium concentrations. |
Solution
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[latex]\begin{gathered} \frac{0.129 \mathrm{~mol}\,\mathrm{N}_2 \mathrm{O}_4}{1.00 \mathrm{~L}}=0.129\mathrm{~M}\,\mathrm{N}_2\mathrm{O}_4 \end{gathered}[/latex]
[latex]\begin{gathered} \mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{N}_2\mathrm{O}_4\right]} \end{gathered}[/latex]
| Stage | [N2O4] | [NO2] |
|---|---|---|
| I | 0.129 | 0 |
| C | -x | +2x |
| E | 0.129 − x | 2x |
[latex]\begin{gathered} 1.07\times 10^{-5}=\frac{(2 \mathrm{x})^2}{(0.129-\mathrm{x})} \end{gathered}[/latex]
Since the equilibrium constant is much less than 1, the change is negligible in comparison to the initial reactant concentration.
[latex]\begin{aligned} 1.07\times 10^{-5}&=\frac{(2\mathrm{x})^2}{0.129}\\ 1.07\times 10^{-5}(0.129)&=(2\mathrm{x})^2 \\ 1.38\times10^{-6}&=(2\mathrm{x})^2 \\ \sqrt{1.38\times10^{-6}}&=\sqrt{(2\mathrm{x})^2} \\ 1.17\times10^{-3}&=2\mathrm{x} \\ 5.87\times10^{-4}&=\mathrm{x} \end{aligned}[/latex]
[latex]\begin{aligned} \,[\mathrm{NO}_2]&=2(5.87 \times 10^{-4})=1.17 \times 10^{-3}\mathrm{~M} \\ [\mathrm{N}_2\mathrm{O}_4]&=0.129\mathrm{~M} \end{aligned}[/latex]
Answer:
- [NO2] = 1.17 × 10-3 M
- [N2O4] = 0.128 M
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 where you must use an ICE table to solve for the equilibrium concentrations of a reaction.
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| Calculate the equilibrium concentrations of N2O4 and NO2 in a 1.00 L vessel that was prepared starting with only 0.129 mol of N2O4.
[latex]\mathrm{N}_2\mathrm{O}_4(\mathrm{g})\rightleftharpoons2\mathrm{NO}_2(\mathrm{g})\,|\,\mathrm{K}_\mathrm{c}=1.07\times 10^{-5}[/latex] |
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To solve this problem, you will need to set up the equilibrium constant expression.
Show/Hide Think About This!Recall that the equilibrium expression represents the proportion of products and reactants that results in an equal rate of forward and reverse reaction so that the concentrations of species does not change over time. The expression is the product of all product species concentrations divided by the product of all reactant concentrations. Each species concentration term is raised to the power of its stoichiometric coefficient. |
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| Write the equilibrium expression for the provided reaction. The compounds on the left side of the reaction are the reactants, and the compounds on the right side of the reaction are the products.
[latex]\begin{aligned} \mathrm{K}_{\mathrm{c}}&=\frac{[\text{Products}]}{[\text{Reactants}]}\\ \mathrm{K}_{\mathrm{c}}&=\frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2 \mathrm{O}_4]} \end{aligned}[/latex] Show/Hide Think About This!Since NO2 has a stoichiometric coefficient of 2 in the balanced reaction, the [NO2] term is raised to the power of 2 in the expression for Kc. |
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To solve this problem, you will need to create an ICE table. We know the initial concentrations and the equilibrium constant.
Complete the table using ‘x’ to represent change and writing the expressions for the equilibrium concentrations. Show/Hide Don’t Forget!Use the initial moles of N2O4 and the 1.00 L volume to calculate initial molarity of N2O4. We know we do not have any NO2 to start with as we were told that in the question. The mole ratio of the change will be the same as the mole ratio of reactant to product. Sine we start with N2O4, its concentration will decrease and the concentration of NO2 will increase as the reaction moves towards equilibrium. |
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Complete the ICE table and enter equilibrium concentration terms into your Kc expression. Look at the numerical value of Kc to determine appropriate assumptions.
Show/Hide Don’t Forget!When dealing with a small equilibrium constant (Kc << 1) the change (‘x’) will be negligible compared to the initial reactant concentration (x << 1.07 × 10-5). |
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Solve for ‘x’ and calculate equilibrium concentrations.
Show/Hide Don’t Forget!Plug your value for ‘x’ back in to the equilibrium concentration expressions from the last line of your table to calculate the concentrations. |
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| Recall the Kc expression, and write out the expression for Kc for this reaction:
[latex]\begin{aligned} \mathrm{K}_\mathrm{c}&=\frac{[\text{Products}]}{[\text{Reactants}]}\\ \mathrm{K}_{\mathrm{c}}&=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2 \mathrm{O}_4\right]} \end{aligned}[/latex] |
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| Divide the initial moles of N2O4 by the volume of its container to calculate the initial molarity:
[latex]\begin{gathered} \frac{0.129 \mathrm{~mol} \mathrm{~N}_2 \mathrm{O}_4}{1.00 \mathrm{~L}}=0.129 \mathrm{~M} \mathrm{~N}_2 \mathrm{O}_4 \end{gathered}[/latex] Show/Hide HintDo not forget to include all stoichiometric coefficients in your expression as exponents. |
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Create an ICE table for the reaction. Use ‘x’ to represent the change in concentration.
[latex]\begin{gathered} 1.07 \times 10^{-5}=\frac{(2 \mathrm{x})^2}{(0.129-\mathrm{x})} \end{gathered}[/latex] Since the equilibrium constant is much less than 1, the change is negligible in comparison to the initial reactant concentration. [latex]\begin{gathered} 1.07 \times 10^{-5}=\frac{(2 \mathrm{x})^2}{0.129} \end{gathered}[/latex] Solve for ‘x’ algebraically: [latex]\begin{aligned} 1.07 \times 10^{-5}(0.129)&=(2 \mathrm{x})^2 \\ 1.38 \times 10^{-6}&=(2 \mathrm{x})^2 \\ \sqrt{1.38 \times 10^{-6}}&=\sqrt{(2 \mathrm{x})^2} \\ 1.17 \times 10^{-3}&=2 \mathrm{x} \\ 5.87 \times 10^{-4}&=\mathrm{x} \end{aligned}[/latex] |
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| Solve for equilibrium concentrations:
[latex]\begin{aligned} \,[\mathrm{NO}_2]&=2(\mathrm{x})=2(5.87\times 10^{-4})=1.17\times 10^{-3}\mathrm{M} \\ [\mathrm{N}_2\mathrm{O}_4]&=0.129-\mathrm{x}=0.129-5.87 \times 10^{-4}=0.129\mathrm{~M} \end{aligned}[/latex] Show/Hide Think About This!The small value for ‘x’ shows that we were correct to neglect it compared to the initial concentration of N2O4. Answer:
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Check Your Work
The equilibrium constant describes the relative proportions of reactant and product species once the system has reached equilibrium (i.e., the reactant and product concentrations do not change).
The stoichiometric coefficient for NO2 is 2, and it shows up as the exponent in the Kc expression and in the mole ratio of the change in concentration. For every mole of N2O4 that is reacted, 2 moles of NO2 are formed. We started with only N2O4, so the concentration of NO2 increases while the reaction comes to equilibrium.
Does your answer make chemical sense?
Show/Hide Answer
The sign of the change in our ICE table is positive when the concentration increases and negative when it decreases. When given the equilibrium constant and some of the concentrations involved, we can solve for the missing concentrations.
For this reaction, the value for Kc is small (1.07 × 10-5) which tells us that the reactant should be favoured at equilibrium; our calculated concentrations confirm this.
PASS Attribution
- LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (2).
- Question 5.E.18 from LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (3) is used under a CC BY 4.0 license.
- Question 5.E.18 is question 13.E.4.4: Q13.4.3 in LibreTexts Chemistry 1e (4), which is under a CC BY 4.0 license.
- Question 13.E.4.35: Q13.4.16(a) is question 74 from OpenStax Chemistry (5), which is under a CC BY 4.0 license. Access for free at https://openstax.org/books/chemistry/pages/1-introduction
References
1. OpenStax. 5.4: Equilibrium Calculations. 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)/05%3A_Chemical_Equilibrium/5.04%3A_Equilibrium_Calculations.
2. 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)
3. OpenStax. 5.E: Fundamental Equilibrium Concepts (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)/05%3A_Chemical_Equilibrium/5.E%3A_Fundamental_Equilibrium_Concepts_(Exercises).
4. OpenStax. 13.E: Fundamental Equilibrium Concepts (Exercises). In Chemistry 1e (OpenSTAX); LibreTexts, 2022. https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/13%3A_Fundamental_Equilibrium_Concepts/13.E%3A_Fundamental_Equilibrium_Concepts_(Exercises).
5. Flowers, P.; Robinson, W. R.; Langley, R.; Theopold, K. Ch. 13 Exercises. In Chemistry; OpenStax, 2015. https://openstax.org/books/chemistry/pages/13-exercises.
