Chemical Equilibrium: Calculate Q, Compare to K
Question
The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.
[latex]\begin{aligned} &2\mathrm{NH}_3(\mathrm{g})\rightleftharpoons\mathrm{N}_2(\mathrm{g})+3\mathrm{H}_2(\mathrm{g})\\ &\mathrm{K}_{\mathrm{c}}=17\\ &{\left[\mathrm{NH}_3\right]=0.20\mathrm{M},\left[\mathrm{N}_2\right]=1.00\mathrm{M},\left[\mathrm{H}_2\right]=1.00\mathrm{M}} \end{aligned}[/latex]
Show/Hide Answer
Qc = 25; the reaction proceeds to the left
Refer to Section 5.3: Equilibrium Constants (1).
Strategy Map
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Check out the strategy map.
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| Strategy Map Steps |
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| 1. Recall the Qc expression. |
| 2. Identify which compounds are products and which are reactants. |
| 3. Fill out the reaction quotient (Qc) expression and solve for the reaction quotient. |
| 4. Compare the provided Kc value and the Qc value you just solved for. Has there been an increase or decrease? |
| 5. Identify which direction the reaction would need to proceed to return to the equilibrium state |
Solution
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[latex]\begin{aligned} \mathrm{Q}_{\mathrm{c}} & =\frac{\text {[Products }]}{\lceil\text { Reactants }\rceil} \\\\ \mathrm{Q}_{\mathrm{c}} & =\frac{\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^3}{\left[\mathrm{NH}_3\right]^2} \end{aligned}[/latex]
[latex]\begin{gathered} \mathrm{Q}_\mathrm{c}=\frac{\left((1.00\mathrm{M})(1.00\mathrm{M})^3\right)}{(0.20\mathrm{M})^2}=\frac{(1.00\mathrm{M})(1.00\mathrm{M})}{0.04\mathrm{M}}=25 \end{gathered}[/latex]
[latex]\begin{equation} \begin{aligned} 25&>17\\ \mathrm{Q}_\mathrm{c}&>\mathrm{K}_\mathrm{c} \end{aligned} \end{equation}[/latex]
The Qc expression calculated a value larger than Kc. This means that that snapshot in time had more products and less reactants than at equilibrium. The reaction will proceed in the left direction, favouring the formation of reactants to return to equilibrium.
Answer: Qc = 25; the reaction proceeds to the left
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 where you must set up the reaction quotient expression and solve for it using the information provided. You are then tested on your knowledge of equilibrium shifts by comparing the reaction quotient and provided equilibrium constant to identify what shift must be made to return equilibrium.
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To solve this problem, you will need to set up the reaction quotient expression. It is created the same way as the equilibrium expression, with the concentration of the products divided by the concentration of the reactants.
Show/Hide Think About This![latex]\begin{gathered} \mathrm{Q}_{\mathrm{c}}=\frac{[\text{Products}]}{[\text{Reactants}]} \end{gathered}[/latex] |
| 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]2\mathrm{NH}_3(\mathrm{g})\rightleftharpoons\mathrm{N}_2(\mathrm{g})+3\mathrm{H}_2(\mathrm{g})[/latex] Show/Hide Think About This![latex]\begin{gathered} \mathrm{Q}_{\mathrm{c}}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]}{\left[\mathrm{NH}_2\right]} \end{gathered}[/latex] |
| Do not forget about the molar coefficients in the reaction equation.
You must include the stoichiometry into the expression. Recall how it is used. Show/Hide Don’t Forget!The stoichiometry is used as the exponent of its associated compound. [latex]\begin{gathered} \mathrm{Q}_{\mathrm{c}}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^3}{\left[\mathrm{NH}_3\right]^2} \end{gathered}[/latex] |
| The reaction quotient (Q) is used to measure the relative concentration of reactants and products in a reaction at a given time.
The equilibrium constant is used to measure the relative concentrations of reactants and products when a reaction is at equilibrium. The reaction quotient can be analyzed and compared to the equilibrium constant to determine how the reaction has shifted away from equilibrium. Show/Hide Think About This!
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| Complete Solution |
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| Recall the Qc expression:
[latex]\begin{gathered} \mathrm{Q}_{\mathrm{c}}=\frac{[\text{Products}]}{[\text{Reactants}]} \end{gathered}[/latex] |
| Fill out the template; recall that stoichiometry factors are used as the exponent to their associated concentration:
[latex]\begin{gathered} \mathrm{Q}_{\mathrm{c}}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^3}{\left[\mathrm{NH}_3\right]^2} \end{gathered}[/latex] |
| Plug in the concentrations and solve for Qc:
[latex]\begin{gathered} \mathrm{Q}_\mathrm{c}=\frac{\left((1.00\mathrm{M})(1.00 \mathrm{M})^3\right)}{(0.20\mathrm{M})^2}=\frac{(1.00\mathrm{M})(1.00\mathrm{M})}{0.04\mathrm{M}}=25 \end{gathered}[/latex] |
| Compare Qc and Kc:
[latex]\begin{aligned} 25&>17\\ \mathrm{Q}_\mathrm{c}&>\mathrm{K}_\mathrm{c} \end{aligned}[/latex] There are more products and less reactants in the Qc expression. To get back to equilibrium, the reaction will proceed in the left direction. This will use up some products and produce some reactants. Answer: Qc = 25; the reaction proceeds to the left |
Check Your Work
We found that the reaction is not at equilibrium, so it will be shifting to the reactants side to reach equilibrium. We used the provided concentrations to calculate the reaction quotient (Q) and then compared it to the equilibrium constant. To check your work, make sure you set up all terms correctly.
Show/Hide Think About This!
The reaction quotient (Qc) expression takes the same form as the equilibrium constant (Kc) expression. The formula has been set up correctly when the product species are in the numerator (top) and the reactant species are in the denominator (bottom). The stoichiometric coefficients are exponents to each species term.
Show/Hide Don’t Forget!
Make sure all values are in concentration units and that no solid species are included in the expression. Only materials in the gaseous and aqueous state can change concentration.
Show/Hide Think About This!
Because products are in the numerator and reactants in the denominator, a larger Q value means lots of products and a smaller Q value means lots of reactants.
Q is the ratio of reactants and products at a snapshot in time. K is the ratio at equilibrium.
Does your answer make chemical sense?
Show/Hide Answer
Reactions are constantly happening in the forward and reverse direction, even when the reaction is at equilibrium. When a reaction not at equilibrium, it will proceed in 1 direction to return to its equilibrium state.
Chemists can analyze the relative concentrations of a reaction before the change using the reaction quotient and compare it to the equilibrium constant. The difference between the 2 will tell you what direction the reaction must shift and by how much.
This makes sense mathematically when you look to the expression:
- If the numerator is larger, the value will be larger.
- If the denominator is larger the value will be smaller.
PASS Attribution
- LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (2).
- Question 5.E.9(a) 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.9(a) is question 13.E.2.18: Q13.2.12(a) in LibreTexts Chemistry 1e (4), which is under a CC BY 4.0 license.
- Question 13.E.2.18: Q13.2.12(a) is question 17(a) 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.
