Enthalpy Of Formation Practice Problems

Mastering Enthalpy of Formation: Practice Problems and Solutions

Understanding enthalpy of formation is crucial for mastering thermochemistry. This article provides a comprehensive guide, moving from foundational concepts to progressively challenging practice problems, complete with detailed solutions. We'll cover the definition of enthalpy of formation (ΔHf°), Hess's Law, and its application in calculating enthalpy changes for various reactions. By the end, you'll confidently tackle enthalpy of formation problems and gain a deeper understanding of this important thermodynamic concept.

Introduction to Enthalpy of Formation

Enthalpy of formation (ΔHf°) refers to the heat change that occurs when one mole of a compound is formed from its constituent elements in their standard states (usually at 298 K and 1 atm pressure). It's a fundamental quantity in thermochemistry, allowing us to predict the heat released or absorbed during chemical reactions. A negative ΔHf° indicates an exothermic reaction (heat is released), while a positive ΔHf° signifies an endothermic reaction (heat is absorbed).

Key Concepts and Equations

Before diving into the practice problems, let's revisit some essential concepts:

• Standard States: Elements in their most stable form at standard conditions (298 K and 1 atm). For example, the standard state of oxygen is O₂(g), not O(g).

• Hess's Law: The enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps. This law is fundamental for calculating enthalpy changes using enthalpy of formation data.

• Calculating Enthalpy Change (ΔHrxn°) using Enthalpy of Formation: The enthalpy change for a reaction can be calculated using the following equation:

  ΔHrxn° = Σ [ΔHf°(products)] - Σ [ΔHf°(reactants)]

  Where:

  • ΔHrxn° is the standard enthalpy change of reaction.
  • ΔHf°(products) represents the sum of the standard enthalpies of formation of the products, each multiplied by its stoichiometric coefficient.
  • ΔHf°(reactants) represents the sum of the standard enthalpies of formation of the reactants, each multiplied by its stoichiometric coefficient.

  Remember that the enthalpy of formation of an element in its standard state is zero (ΔHf° = 0).

Practice Problems: From Basic to Advanced

Now, let's tackle some practice problems of varying difficulty. Each problem will be followed by a detailed solution.

Problem 1: Basic Enthalpy of Formation Calculation

Calculate the enthalpy change for the following reaction using the given standard enthalpies of formation:

2CO(g) + O₂(g) → 2CO₂(g)

Given:

• ΔHf°[CO(g)] = -110.5 kJ/mol
• ΔHf°[CO₂(g)] = -393.5 kJ/mol
• ΔHf°[O₂(g)] = 0 kJ/mol (O₂ is in its standard state)

Solution 1:

Using the equation: ΔHrxn° = Σ [ΔHf°(products)] - Σ [ΔHf°(reactants)]

ΔHrxn° = [2 mol × ΔHf°(CO₂(g))] - [2 mol × ΔHf°(CO(g)) + 1 mol × ΔHf°(O₂(g))]

ΔHrxn° = [2 mol × (-393.5 kJ/mol)] - [2 mol × (-110.5 kJ/mol) + 1 mol × (0 kJ/mol)]

ΔHrxn° = -787 kJ - (-221 kJ) = -566 kJ

The enthalpy change for the reaction is -566 kJ. This indicates an exothermic reaction.

Problem 2: Incorporating Different Phases

Calculate the standard enthalpy change for the combustion of methane:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Given:

• ΔHf°[CH₄(g)] = -74.8 kJ/mol
• ΔHf°[CO₂(g)] = -393.5 kJ/mol
• ΔHf°[H₂O(l)] = -285.8 kJ/mol
• ΔHf°[O₂(g)] = 0 kJ/mol

Solution 2:

ΔHrxn° = [1 mol × ΔHf°(CO₂(g)) + 2 mol × ΔHf°(H₂O(l))] - [1 mol × ΔHf°(CH₄(g)) + 2 mol × ΔHf°(O₂(g))]

ΔHrxn° = [1 mol × (-393.5 kJ/mol) + 2 mol × (-285.8 kJ/mol)] - [1 mol × (-74.8 kJ/mol) + 2 mol × (0 kJ/mol)]

ΔHrxn° = (-393.5 kJ - 571.6 kJ) - (-74.8 kJ) = -890.3 kJ

The enthalpy change for the combustion of methane is -890.3 kJ. This is a highly exothermic reaction.

Problem 3: Using Hess's Law

Determine the enthalpy of formation of ethene (C₂H₄(g)) given the following information:

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

2. C(s) + O₂(g) → CO₂(g) ΔH° = -393.5 kJ

3. H₂(g) + ½O₂(g) → H₂O(l) ΔH° = -285.8 kJ

Solution 3:

We need to manipulate the given equations to obtain the formation reaction of ethene: 2C(s) + 2H₂(g) → C₂H₄(g).

• Equation 1 (reversed): 2CO₂(g) + 2H₂O(l) → C₂H₄(g) + 3O₂(g) ΔH° = +1411 kJ (sign changes when reversed)
• Equation 2 (multiplied by 2): 2C(s) + 2O₂(g) → 2CO₂(g) ΔH° = -787 kJ (multiplied by the stoichiometric coefficient)
• Equation 3 (multiplied by 2): 2H₂(g) + O₂(g) → 2H₂O(l) ΔH° = -571.6 kJ (multiplied by the stoichiometric coefficient)

Adding these three modified equations:

2C(s) + 2O₂(g) + 2H₂(g) + O₂(g) + 2CO₂(g) + 2H₂O(l) → 2CO₂(g) + 2H₂O(l) + C₂H₄(g) + 3O₂(g)

Simplifying (cancelling out common species on both sides):

2C(s) + 2H₂(g) → C₂H₄(g)

The enthalpy change for this reaction is the sum of the enthalpy changes of the modified equations:

ΔHf°(C₂H₄(g)) = +1411 kJ - 787 kJ - 571.6 kJ = +52.4 kJ

The enthalpy of formation of ethene is +52.4 kJ/mol. This is an endothermic reaction.

Problem 4: Dealing with Ionic Compounds

Calculate the enthalpy change for the following reaction:

Mg(s) + ½O₂(g) → MgO(s)

Given:

• ΔHf°[MgO(s)] = -601.7 kJ/mol
• ΔHf°[Mg(s)] = 0 kJ/mol
• ΔHf°[O₂(g)] = 0 kJ/mol

Solution 4:

This is a straightforward enthalpy of formation calculation since the reaction represents the formation of MgO from its elements in their standard states.

ΔHrxn° = ΔHf°[MgO(s)] - [ΔHf°[Mg(s)] + ½ΔHf°[O₂(g)]]

ΔHrxn° = -601.7 kJ/mol - (0 kJ/mol + 0 kJ/mol) = -601.7 kJ/mol

The enthalpy change for the reaction is -601.7 kJ/mol.

Problem 5: A More Complex Reaction

Calculate the enthalpy change for the reaction:

2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l)

Given:

• ΔHf°[C₂H₆(g)] = -84.7 kJ/mol
• ΔHf°[CO₂(g)] = -393.5 kJ/mol
• ΔHf°[H₂O(l)] = -285.8 kJ/mol
• ΔHf°[O₂(g)] = 0 kJ/mol

Solution 5:

ΔHrxn° = [4 mol × ΔHf°(CO₂(g)) + 6 mol × ΔHf°(H₂O(l))] - [2 mol × ΔHf°(C₂H₆(g)) + 7 mol × ΔHf°(O₂(g))]

ΔHrxn° = [4 mol × (-393.5 kJ/mol) + 6 mol × (-285.8 kJ/mol)] - [2 mol × (-84.7 kJ/mol) + 7 mol × (0 kJ/mol)]

ΔHrxn° = (-1574 kJ - 1714.8 kJ) - (-169.4 kJ) = -3119.4 kJ

The enthalpy change for this reaction is -3119.4 kJ.

Frequently Asked Questions (FAQs)

• Q: What does the "°" symbol mean in ΔHf°?

  • A: The "°" symbol indicates that the enthalpy of formation is measured under standard conditions (298 K and 1 atm pressure).

• Q: Can I use enthalpy of formation data for reactions at temperatures other than 298 K?

  • A: Strictly speaking, the values are only accurate at 298 K. However, small deviations from this temperature will often result in only minor inaccuracies. For larger temperature changes, more sophisticated thermodynamic calculations are required.

• Q: What if I don't have the enthalpy of formation for all the compounds in a reaction?

  • A: You can often use Hess's Law to combine known enthalpy changes for other reactions to calculate the unknown enthalpy of formation.

• Q: Why is the enthalpy of formation of an element in its standard state zero?

  • A: By definition, the enthalpy of formation refers to the heat change involved in forming a compound from its elements. Since an element in its standard state is already in its most stable form, no energy is required or released in "forming" it from itself.

• Q: How accurate are enthalpy of formation values?

  • A: The accuracy of enthalpy of formation data varies depending on the compound and the method used to determine it. They are typically reported with some uncertainty.

Conclusion

Mastering enthalpy of formation calculations requires a firm grasp of the fundamental concepts and the ability to apply Hess's Law. Through practice, you'll become more proficient in using enthalpy of formation data to predict the enthalpy changes for a wide variety of chemical reactions. Remember to always pay attention to stoichiometric coefficients and the standard states of elements when performing these calculations. This article provides a solid foundation; continued practice with more diverse problems will further solidify your understanding of this vital thermodynamic principle. Keep practicing, and you'll be well on your way to mastering thermochemistry!