Chemical Equilibrium: Calculate Equilibrium Constant from Concentrations at Equilibrium
Question
What is the value of the equilibrium constant at 500 °C for the formation of NH3 according to the following equation:
[latex]\mathrm{N}_2(\mathrm{g})+3 \mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{g})[/latex]
An equilibrium mixture of NH3(g), H2(g), and N2(g) at 500 °C was found to contain 1.35 M H2, 1.15 M N2, and 4.12 × 10−1 M NH3.
Show/Hide Answer
Kc = 6.00 × 10-2
Refer to Section 5.3: Equilibrium Constants (1).
Strategy Map
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| Strategy Map Steps |
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1. Recall the general format for the equilibrium constant (Kc) expression.
Show/Hide Hint[latex]\begin{gathered} \mathrm{K}_\mathrm{c}=\frac{[\text{Products}]}{[\text{Reactants}]} \end{gathered}[/latex] |
2. Write the equilibrium expression for the given reaction using the reactant and product species. Include stoichiometric coefficients as exponents.
Show/Hide HintBe sure that the reaction equation is balanced. Recall that only gaseous and aqueous species can be included in the expression. |
3. Plug in the provided information.
Show/Hide HintEnsure that the data provided is in units of concentration in molarity, moles per litre. |
| 4. Solve for the value of the equilibrium constant. |
Solution
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[latex]\begin{aligned} \mathrm{K}_\mathrm{c}&=\frac{[\text{Products}]}{[\text{Reactants}]}\\ \mathrm{K}_\mathrm{c}&=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^3}\\ \mathrm{K}_\mathrm{c}&=\frac{(0.412\mathrm{~M})^2}{(1.15\mathrm{~M})(1.35\mathrm{~M})^3}\\ \mathrm{K}_\mathrm{c}&=0.059992=6.00 \times 10^{-2} \end{aligned}[/latex]
Answer: Kc = 6.00 × 10-2
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 type problem that requires you to set up the equilibrium expression for a given reaction. You are provided with the equilibrium concentrations for a specific condition which will allow you to solve for the equilibrium constant.
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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. |
| 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{gathered} \mathrm{K}_{\mathrm{c}}=\frac{[\text{Products}]}{[\text{Reactants}]} \end{gathered}[/latex] [latex]\mathrm{N}_2(\mathrm{g})+3\mathrm{H}_2(\mathrm{g})\rightleftharpoons2\mathrm{NH}_3(\mathrm{g})[/latex] Show/Hide Think About This!Ammonia (NH3) is the product and nitrogen (N) and hydrogen (H) are the reactants, so the equilibrium expression will be: [latex]\begin{gathered} \mathrm{K}_\mathrm{c}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^3} \end{gathered}[/latex] |
You are given the equilibrium conditions for the reaction at 500°C.
These values will be plugged into your equilibrium expression. How do you use the temperature in your calculation? Show/Hide Don’t Forget!The temperature is not included in the calculation. The temperature only indicates the specific conditions of the reaction as the equilibrium constant is dependent on temperature. |
| Complete Solution |
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| Recall the Kc expression:
[latex]\begin{gathered} \mathrm{K}_\mathrm{c}=\frac{[\text{Products}]}{[\text{Reactants}]} \end{gathered}[/latex] |
| Recall that ammonia is the product and that nitrogen and hydrogen are the reactants:
[latex]\mathrm{N}_2(\mathrm{g})+3 \mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{g})[/latex] [latex]\begin{gathered} \mathrm{K}_\mathrm{c}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3} \end{gathered}[/latex] Show/Hide HintDo not forget to include all stoichiometric coefficients in your expression as exponents. |
| Plug the provided values into the equilibrium expression:
[latex]\begin{gathered} \mathrm{K}_\mathrm{c}=\frac{(0.412\mathrm{~M})^2}{(1.15\mathrm{~M})(1.35\mathrm{~M})^3} \end{gathered}[/latex] |
| Solve for Kc:
[latex]\mathrm{K}_\mathrm{c}=0.059992=6.00 \times 10^{-2}[/latex] Answer: Kc = 6.00 × 10-2 |
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 provided concentrations show significantly larger amounts of reactant species compared to the products. The product term is the numerator, and the reactant term is the denominator; therefore, we should expect that the equilibrium constant will be less than 1.
Does your answer make chemical sense?
Show/Hide Answer
An equilibrium constant is the value of the reaction quotient for a system at equilibrium. The equilibrium constant will vary depending on the conditions (such as temperature) and can be calculated by dividing the product concentrations by the reactant concentrations (raised to the power of their stoichiometric coefficient).
Note that the equilibrium constant can be less than or greater than 1 but will never be negative.
PASS Attribution
- LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (2).
- Question 5.E.11 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.11 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.4.4: Q13.4.3 is question 52 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. 5.3: Equilibrium Constants. 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/5.03%3A_Equilibrium_Constants.
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.
