Hard Hess’s Law Practice Questions

Hess’s Law is a fundamental principle in thermochemistry stating that the total enthalpy change for a chemical reaction is independent of the route by which the chemical change occurs, provided the initial and final states are the same. This law is a direct application of the First Law of Thermodynamics and allows chemists to calculate enthalpy changes for reactions that are difficult to measure directly in a laboratory setting. By manipulating known thermochemical equations through reversing, scaling, and adding them together, we can determine the enthalpy of complex processes like the formation of unstable intermediates or high-temperature combustion.

Concept Explanation

Hess’s Law states that the enthalpy change of a reaction is constant regardless of whether the reaction occurs in one step or several steps. This concept relies on the fact that enthalpy (H) is a state function, meaning its value depends only on the current state of the system (pressure, temperature, and composition) and not on the history of how it reached that state. When solving Hard Hess’s Law Practice Questions, you must remember three key rules: if you reverse a reaction, the sign of ΔH must be flipped; if you multiply the coefficients of a reaction by a factor, you must multiply ΔH by that same factor; and when adding equations, the net enthalpy is the sum of the individual enthalpies. These principles are essential for mastering heat of reaction calculations and understanding the energy profiles of multi-step mechanisms.

According to Wikipedia's entry on Hess's Law, the law is particularly useful for measuring the heats of formation of compounds that cannot be synthesized directly from their elements. For instance, determining the enthalpy of formation for carbon monoxide (CO) is challenging because carbon usually combusts directly to carbon dioxide (CO2). By using Hess’s Law, we can subtract the enthalpy of combustion of CO from the enthalpy of combustion of graphite to find the missing value. This method is a cornerstone of thermochemical cycles taught in advanced chemistry courses.

Solved Examples

1. Calculate the enthalpy change for the reaction: C(graphite) + 2H₂(g) → CH₄(g)

   Given the following data:

   • (1) C(graphite) + O₂(g) → CO₂(g) ΔH = -393.5 kJ

   • (2) H₂(g) + ½O₂(g) → H₂O(l) ΔH = -285.8 kJ

   • (3) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = -890.3 kJ

   Solution:

   1. Keep equation (1) as is to get C(graphite) on the reactant side: ΔH = -393.5 kJ.

   2. Multiply equation (2) by 2 to get 2H₂: 2H₂(g) + O₂(g) → 2H₂O(l) ΔH = 2(-285.8) = -571.6 kJ.

   3. Reverse equation (3) to get CH₄ on the product side: CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH = +890.3 kJ.

   4. Sum the reactions: C + O₂ + 2H₂ + O₂ + CO₂ + 2H₂O → CO₂ + 2H₂O + CH₄ + 2O₂.

   5. Cancel intermediates: C(graphite) + 2H₂(g) → CH₄(g).

   6. Sum the enthalpies: -393.5 + (-571.6) + 890.3 = -74.8 kJ.

2. Find ΔH for: 2B(s) + 3H₂(g) → B₂H₆(g)

   Given:

   • (1) 2B(s) + 3/2O₂(g) → B₂O₃(s) ΔH = -1273 kJ

   • (2) B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(g) ΔH = -2035 kJ

   • (3) H₂(g) + ½O₂(g) → H₂O(g) ΔH = -242 kJ

   Solution:

   1. Use equation (1) for 2B: ΔH = -1273 kJ.

   2. Reverse equation (2) for B₂H₆: ΔH = +2035 kJ.

   3. Multiply equation (3) by 3 for 3H₂: ΔH = 3(-242) = -726 kJ.

   4. Total ΔH = -1273 + 2035 - 726 = +36 kJ.

3. Determine ΔH for: P₄O₆(s) + 2O₂(g) → P₄O₁₀(s)

   Given:

   • (1) P₄(s) + 3O₂(g) → P₄O₆(s) ΔH = -1640.1 kJ

   • (2) P₄(s) + 5O₂(g) → P₄O₁₀(s) ΔH = -2940.1 kJ

   Solution:

   1. Reverse equation (1) to put P₄O₆ on the reactant side: ΔH = +1640.1 kJ.

   2. Keep equation (2) to put P₄O₁₀ on the product side: ΔH = -2940.1 kJ.

   3. Sum: P₄O₆ → P₄ + 3O₂ and P₄ + 5O₂ → P₄O₁₀.

   4. Net: P₄O₆ + 2O₂ → P₄O₁₀.

   5. Total ΔH = 1640.1 - 2940.1 = -1300.0 kJ.

Practice Questions

Test your knowledge with these Hard Hess’s Law Practice Questions. Ensure you pay close attention to stoichiometry and phase changes.

1. Calculate ΔH for the reaction: 2C(s) + H₂(g) → C₂H₂(g), given:

   • C₂H₂(g) + 5/2 O₂(g) → 2CO₂(g) + H₂O(l) ΔH = -1299.5 kJ

   • C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ

   • H₂(g) + 1/2 O₂(g) → H₂O(l) ΔH = -285.8 kJ

2. Find ΔH for the reaction: N₂H₄(l) + CH₄O(l) → CH₂O(g) + N₂(g) + 3H₂(g), given:

   • 2NH₃(g) → N₂H₄(l) + H₂(g) ΔH = +22.5 kJ

   • 2NH₃(g) → N₂(g) + 3H₂(g) ΔH = +57.5 kJ

   • CH₂O(g) + H₂(g) → CH₄O(l) ΔH = -81.2 kJ

3. Calculate ΔH for: ClF(g) + F₂(g) → ClF₃(g), given:

   • 2ClF(g) + O₂(g) → Cl₂O(g) + F₂O(g) ΔH = +167.4 kJ

   • 2ClF₃(g) + 2O₂(g) → Cl₂O(g) + 3F₂O(g) ΔH = +341.4 kJ

   • 2F₂(g) + O₂(g) → 2F₂O(g) ΔH = -43.4 kJ

4. Determine the enthalpy change for the conversion of diamond to graphite: C(diamond) → C(graphite), given:

   • C(diamond) + O₂(g) → CO₂(g) ΔH = -395.4 kJ

   • C(graphite) + O₂(g) → CO₂(g) ΔH = -393.5 kJ

5. Calculate ΔH for: 2Fe(s) + 3/2 O₂(g) → Fe₂O₃(s), given:

   • 2Fe(s) + 6H₂O(l) → 2Fe(OH)₃(s) + 3H₂(g) ΔH = +321.8 kJ

   • 2H₂(g) + O₂(g) → 2H₂O(l) ΔH = -571.6 kJ

   • Fe₂O₃(s) + 3H₂O(l) → 2Fe(OH)₃(s) ΔH = +288.6 kJ

6. Find the enthalpy change for: WO₃(s) + 3H₂(g) → W(s) + 3H₂O(g), given:

   • 2W(s) + 3O₂(g) → 2WO₃(s) ΔH = -1685.4 kJ

   • 2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -477.84 kJ

7. Given the following equations, calculate ΔH for 4NH₃(g) + 3O₂(g) → 2N₂(g) + 6H₂O(g):

   • N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = -92.2 kJ

   • 2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -483.6 kJ

8. Determine ΔH for: C₂H₄(g) + H₂(g) → C₂H₆(g), using the enthalpies of combustion:

   • C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l) ΔH = -1411 kJ

   • C₂H₆(g) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH = -1560 kJ

   • H₂(g) + 1/2 O₂(g) → H₂O(l) ΔH = -286 kJ

9. Calculate ΔH for the reaction: 2Al(s) + 3Cl₂(g) → 2AlCl₃(s), given:

   • 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g) ΔH = -1049 kJ

   • HCl(g) → HCl(aq) ΔH = -74.8 kJ

   • H₂(g) + Cl₂(g) → 2HCl(g) ΔH = -184.5 kJ

   • AlCl₃(s) → AlCl₃(aq) ΔH = -323 kJ

10. Find ΔH for: NO(g) + ½ O₂(g) → NO₂(g), given:

    • ½ N₂(g) + ½ O₂(g) → NO(g) ΔH = +90.3 kJ

    • ½ N₂(g) + O₂(g) → NO₂(g) ΔH = +33.2 kJ

Answers & Explanations

1. Answer: +226.7 kJ. Explanation: Reverse the combustion of C₂H₂ (+1299.5 kJ). Multiply C combustion by 2 (2 * -393.5 = -787 kJ). Keep H₂ combustion (-285.8 kJ). Sum: 1299.5 - 787 - 285.8 = +226.7 kJ.

2. Answer: +116.2 kJ. Explanation: Reverse the first equation (-22.5 kJ). Keep the second (+57.5 kJ). Reverse the third (+81.2 kJ). Sum: -22.5 + 57.5 + 81.2 = +116.2 kJ.

3. Answer: -108.7 kJ. Explanation: Divide eq 1 by 2 (+83.7 kJ). Divide eq 2 by -2 (-170.7 kJ). Divide eq 3 by 2 (-21.7 kJ). Sum: 83.7 - 170.7 - 21.7 = -108.7 kJ.

4. Answer: -1.9 kJ. Explanation: C(diamond) → CO₂ (-395.4). Reverse C(graphite) → CO₂ to get CO₂ → C(graphite) (+393.5). Sum: -395.4 + 393.5 = -1.9 kJ.

5. Answer: -824.2 kJ. Explanation: Eq 1 (+321.8). Multiply eq 2 by 1.5 to get 3H₂O (1.5 * -571.6 = -857.4). Reverse eq 3 (-288.6). Sum: 321.8 - 857.4 - 288.6 = -824.2 kJ.

6. Answer: +125.94 kJ. Explanation: Reverse first eq and divide by 2 (+842.7). Multiply second eq by 1.5 (-716.76). Sum: 842.7 - 716.76 = +125.94 kJ.

7. Answer: -1266.4 kJ. Explanation: Reverse first eq and multiply by 2 (2 * +92.2 = +184.4). Multiply second eq by 3 (3 * -483.6 = -1450.8). Sum: 184.4 - 1450.8 = -1266.4 kJ.

8. Answer: -137 kJ. Explanation: Keep C₂H₄ combustion (-1411). Keep H₂ combustion (-286). Reverse C₂H₆ combustion (+1560). Sum: -1411 - 286 + 1560 = -137 kJ.

9. Answer: -1407.9 kJ. Explanation: Eq 1 (-1049). Eq 2: 6 * -74.8 = -448.8. Eq 3: 3 * -184.5 = -553.5. Reverse Eq 4 and multiply by 2: 2 * +323 = +646. Sum: -1049 - 448.8 - 553.5 + 646 = -1405.3 kJ (rounding differences may occur).

10. Answer: -57.1 kJ. Explanation: Reverse first eq (-90.3). Keep second (+33.2). Sum: -90.3 + 33.2 = -57.1 kJ.

Quick Quiz

Frequently Asked Questions

What is the main purpose of using Hess’s Law?

The main purpose of Hess’s Law is to determine the enthalpy change of a reaction that cannot be easily measured experimentally. By combining several simpler reactions with known enthalpies, scientists can calculate the energy change of a complex process via a mathematical summation.

Does Hess’s Law apply to entropy and Gibbs free energy?

Yes, because entropy (S) and Gibbs free energy (G) are also state functions, they follow the same additive principles as enthalpy. You can use similar cycles to calculate ΔS and ΔG for reactions where direct measurement is difficult or dangerous.

Why must we pay attention to the state of matter (s, l, g) in Hess’s Law?

The state of matter is crucial because phase changes, such as vaporization or melting, involve their own enthalpy changes. For example, the combustion of hydrogen to form liquid water releases more energy than forming water vapor because the heat of condensation is also released.

Can Hess’s Law be used for reactions that are not at standard conditions?

Hess’s Law applies at any condition as long as all the component reactions are measured at the same temperature and pressure. However, it is most commonly used with standard enthalpies of formation (ΔH°f) at 298 K and 1 atm.

Is Hess's Law related to bond energies?

Yes, Hess’s Law provides the theoretical framework for using bond energies to estimate reaction enthalpies. By considering a reaction as the breaking of all reactant bonds and the forming of all product bonds, you are essentially creating a multi-step Hess’s Law cycle. For more practice on this, see our bond energy practice questions.

What happens if I forget to multiply the ΔH value when I multiply the reaction coefficients?

If you fail to scale the ΔH value, your final result will be incorrect because enthalpy is an extensive property, meaning it depends on the amount of substance present. Doubling the amount of fuel burned will double the total heat released, so the ΔH must reflect that change.

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