Thermochemistry: Hess’s Law Calculation
Question
The following sequence of reactions occurs in the commercial production of aqueous nitric acid:
[latex]\begin{aligned} & 4\mathrm{NH}_3(\mathrm{g})+5\mathrm{O}_2(\mathrm{g})\rightarrow4\mathrm{NO}(\mathrm{g})+6\mathrm{H}_2\mathrm{O}\,\mathrm{(l)}\Delta\mathrm{H}=-907\mathrm{~kJ}\\ & 2\mathrm{NO}\,\mathrm{(g)}+\mathrm{O}_2(\mathrm{g})\rightarrow2\mathrm{NO}_2(\mathrm{g})\Delta\mathrm{H}=-113\mathrm{~kJ}\\ & 3\mathrm{NO}_2(\mathrm{g})+\mathrm{H}_2\mathrm{O}\,\mathrm{(l)}\rightarrow 2\mathrm{HNO}_3(\mathrm{aq})+\mathrm{NO}\,\mathrm{(g)}\Delta\mathrm{H}=-139\mathrm{~kJ} \end{aligned}[/latex]
The overall reaction is:
[latex]\begin{gathered} 4\mathrm{NH}_3(\mathrm{g})+6\mathrm{O}_2(\mathrm{g})+\mathrm{NO}_2(\mathrm{g})\rightarrow2\mathrm{HNO}_3(\mathrm{aq})+3\mathrm{NO}\,\mathrm{(g)}+5\mathrm{H}_2\mathrm{O}\,\mathrm{(l)} \end{gathered}[/latex]
Determine the total enthalpy change for the production of 1 mole of aqueous nitric acid by this overall reaction.
Show/Hide Answer
-580.0 kJ/mol
Refer to Section 3.6: Hess’s Law (1).
Strategy Map
Do you need a little help to get started?
Check out the strategy map.
Show/Hide Strategy Map
| Strategy Map Steps |
|---|
| 1. The reactions given are steps that give the overall reaction when added together. Since adding all 3 reactions as is gives the overall reaction, adding their corresponding enthalpy change values together gives the overall enthalpy change. |
| 2. Recognize that nitric acid (HNO3) is a product. |
3. Look at the overall reaction and identify if it has the correct stoichiometry. We want to produce 1 mole HNO3 in order to answer the question.
Show/Hide HintIf the reaction produces more than 1 mole, you must manipulate the equation to produce only 1 and manipulate the ΔH in the same way. |
Solution
Do you want to see the steps to reach the answer?
Check out this solution.
Show/Hide Solution
To produce 2 moles of HNO3 (aq):
[latex]\begin{equation} \begin{gathered} -907\mathrm{~kJ}-113\mathrm{~kJ}-139\mathrm{~kJ}=-1159\mathrm{~kJ} \end{gathered} \end{equation}[/latex]
To produce 1 mole of HNO3 (aq):
[latex]\begin{equation} \begin{gathered} \frac{-1159 \mathrm{~kJ}}{2}=-579.5\mathrm{~kJ}=-580.0\mathrm{~kJ} \end{gathered} \end{equation}[/latex]
Answer: -580.0 kJ/mol
Guided Solution
Do you want more help?
The guided solution below will give you the reasoning for each step to get your answer, with reminders and hints.
Show/Hide Guided Solution
| Guided Solution Ideas |
|---|
|
This question is a calculation problem where you must use Hess’s Law to manipulate and add the 3 elementary reaction steps (and their associated enthalpies) to find the overall reaction and the total enthalpy change to produce 1 mole of nitric acid (HNO3). Show/Hide Resource
|
Add the reactions in the steps to determine the overall reaction.
Show/Hide Think About This!These 3 reactions combine to produce 2 moles of nitric acid. |
Add the ΔH of the steps to determine the overall ΔH.
Show/Hide Don’t Forget!Recall that the total enthalpy change for the overall reaction is the sum of the ΔH of all the steps (Hess’s Law). |
| Complete Solution |
|---|
| Add the reactions to get the overall reaction:
[latex]\begin{equation} \begin{aligned} &4\mathrm{NH}_{3}(\mathrm{g})+5\mathrm{O}_2(\mathrm{g})\rightarrow4\mathrm{NO}\,\mathrm{(g)}+6\mathrm{H}_2\mathrm{O}\,\mathrm{(l)}\\ &2\mathrm{NO}\,\mathrm{(g)}+\mathrm{O}_2(\mathrm{g})\rightarrow2\mathrm{NO}_2(\mathrm{g})\\ & 3 \mathrm{NO}_2(\mathrm{g})+\mathrm{H}_2\mathrm{O}\,\mathrm{(l)}\rightarrow2\mathrm{HNO}_3(\mathrm{aq})+\mathrm{NO}\,\mathrm{(g)}\\ &\hline4\mathrm{NH}_3(\mathrm{g})+6\mathrm{O}_2(\mathrm{g})+\mathrm{NO}_2(\mathrm{g})\rightarrow2\mathrm{HNO}_3(\mathrm{aq})+3\mathrm{NO}\,\mathrm{(g)}+5\mathrm{H}_2\mathrm{O}\,\mathrm{(l)} \end{aligned} \end{equation}[/latex] |
| Add the enthalpy changes provided for each individual step.
To produce 2 moles of HNO3 (aq): [latex]\begin{equation} \begin{gathered} -907\mathrm{~kJ}-113\mathrm{~kJ}-139\mathrm{~kJ}=-1159\mathrm{~kJ} \end{gathered} \end{equation}[/latex] This will give the total enthalpy change for the overall reaction. However, it produces 2 moles of nitric acid. Since we want to know the total enthalpy to produce 1 mole, we divide the entire reaction and its corresponding enthalpy change by a factor of 2: [latex]\begin{equation} \begin{gathered} 2\mathrm{NH}_3(\mathrm{g})+3\mathrm{O}_2(\mathrm{g})+1/2\mathrm{NO}_2(\mathrm{g})\rightarrow\mathrm{HNO}_3(\mathrm{aq})+3/2\mathrm{NO}\,\mathrm{(g)}+5/2\mathrm{H}_2\mathrm{O}\,\mathrm{(l)} \end{gathered} \end{equation}[/latex] To produce 1 mole of HNO3 (aq): [latex]\begin{equation} \begin{gathered} \frac{-1159\mathrm{~kJ}}{2}=-579.5\mathrm{~kJ}=-580.0\mathrm{~kJ} \end{gathered} \end{equation}[/latex] Answer: −580.0 kJ/mol |
Check Your Work
Summary of what we would expect based on the related chemistry theory.
Show/Hide Think About This!
Make sure you keep the appropriate signs of ΔH. All the reactions we are working with are exothermic and have negative ΔH values. When we add them together, we get a larger negative number.
We were told these 3 reactions were steps in the overall process to produce nitric acid (HNO3). When we add the reactions together, we get the overall reaction, which produces 2 moles of HNO3. So, we divide by 2 to get the ΔH for the production of 1 mole of HNO3.
Does your answer make chemical sense?
Show/Hide Answer
Enthalpy change is an extensive property as it depends on the amount of a substance. To answer this question, we must recognize that when we add the enthalpy changes of the steps, we get the ΔH of the overall reaction, which produces 2 moles of HNO3. Since we are asked for the ΔH for the production of 1 mole, we must divide by 2.
Enthalpy is a state function, which means that what matters are the amounts and physical states of the reactants and products, not how many steps we take to go from reactant to products. The overall ΔH is the same if we do the reaction in 1 step or 3 steps. Hess’s Law states that we can add the ΔH of steps taken to do a reaction to get the overall ΔH of a reaction.
PASS Attribution
- LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (2).
- Question 3.E.21 from TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (3) is used under a CC BY-NC-SA 4.0 license.
- Question 3.E.21 is question Q5.3.20 from LibreTexts Chemistry 1e (OpenSTAX) (4), which is under a CC BY 4.0 license.
- Question Q5.3.20 is question 59 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. Thompson Rivers University. 3.6: Hess’s Law. 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)/03%3A_Thermochemistry/3.06%3A_Hess’s_Law.
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. Thompson Rivers University. 3.E: Thermochemistry (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)/03%3A_Thermochemistry/3.E%3A_Thermochemistry_(Exercises).
4. OpenStax. 5.E: Thermochemistry (Exercises). In Chemistry 1e (OpenSTAX); LibreTexts, 2022. https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/05%3A_Thermochemistry/5.E%3A_Thermochemistry_(Exercises).
5. Flowers, P.; Robinson, W. R.; Langley, R.; Theopold, K. Ch. 5 Exercises. In Chemistry 2e; OpenStax, 2019. https://openstax.org/books/chemistry-2e/pages/5-exercises.
