Problem 74 - Chemistry: The Central Science

Question

Using values from Appendix C, calculate the value of \(\Delta H^{\circ}\) for each of the following reactions: (a) \(\mathrm{CaO}(s)+2 \mathrm{HCl}(g) \longrightarrow \mathrm{CaCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(4 \mathrm{FeO}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) (c) \(2 \mathrm{CuO}(s)+\mathrm{NO}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{NO}_{2}(g)\) (d) \(4 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-Step Solution

Answer

The calculated values for the standard enthalpy change (ΔH°) for each reaction are as follows: (a) ΔH°(reaction) = -217.9 kJ/mol (b) ΔH°(reaction) = -560.4 kJ/mol (c) ΔH°(reaction) = 84.8 kJ/mol (d) ΔH°(reaction) = -287.6 kJ/mol

Step 1: Identify reactants and products

The reactants are CaO(s) and 2 HCl(g), and the products are CaCl₂(s) and H₂O(g).

Step 2: Find the values of ΔH° from Appendix C

From Appendix C, we have: ΔH°(CaO) = -635.1 kJ/mol ΔH°(HCl) = -92.3 kJ/mol ΔH°(CaCl₂) = -795.8 kJ/mol ΔH°(H₂O) = -241.8 kJ/mol

Step 3: Calculate the standard enthalpy change of the reaction

Using Hess's Law formula: ΔH°(reaction) = ((-795.8) + (-241.8)) - ((-635.1) + 2 × (-92.3)) ΔH°(reaction) = -1037.6 - (-819.7) = -217.9 kJ/mol (b) 4 FeO(s) + O₂(g) → 2 Fe₂O₃(s)

Step 1: Identify reactants and products

The reactants are 4 FeO(s) and O₂(g), and the products are 2 Fe₂O₃(s).

Step 2: Find the values of ΔH° from Appendix C

From Appendix C, we have: ΔH°(FeO) = -272 kJ/mol ΔH°(O₂) = 0 kJ/mol (standard element in its standard state) ΔH°(Fe₂O₃) = -824.2 kJ/mol

Step 3: Calculate the standard enthalpy change of the reaction

Using Hess's Law formula: ΔH°(reaction) = (2 × (-824.2)) - (4 × (-272) + 0) ΔH°(reaction) = -1648.4 - (-1088) = -560.4 kJ/mol (c) 2 CuO(s) + NO(g) → Cu₂O(s) + NO₂(g)

Step 1: Identify reactants and products

The reactants are 2 CuO(s) and NO(g), and the products are Cu₂O(s) and NO₂(g).

Step 2: Find the values of ΔH° from Appendix C

From Appendix C, we have: ΔH°(CuO) = -156 kJ/mol ΔH°(NO) = 90.3 kJ/mol ΔH°(Cu₂O) = -170 kJ/mol ΔH°(NO₂) = 33.1 kJ/mol

Step 3: Calculate the standard enthalpy change of the reaction

Using Hess's Law formula: ΔH°(reaction) = ((-170) + 33.1) - (2 × (-156) + 90.3) ΔH°(reaction) = -136.9 - (-221.7) = 84.8 kJ/mol (d) 4 NH₃(g) + O₂(g) → 2 N₂H₄(g) + 2 H₂O(l)

Step 1: Identify reactants and products

The reactants are 4 NH₃(g) and O₂(g), and the products are 2 N₂H₄(g) and 2 H₂O(l).

Step 2: Find the values of ΔH° from Appendix C

From Appendix C, we have: ΔH°(NH₃) = -45.9 kJ/mol ΔH°(O₂) = 0 kJ/mol (standard element in its standard state) ΔH°(N₂H₄) = 50.6 kJ/mol ΔH°(H₂O) = -285.8 kJ/mol

Step 3: Calculate the standard enthalpy change of the reaction

Using Hess's Law formula: ΔH°(reaction) = (2 × 50.6 + 2 × (-285.8)) - (4 × (-45.9) + 0) ΔH°(reaction) = (-471.2) - (-183.6) = -287.6 kJ/mol

Key Concepts

Hess's Law

Understanding Hess's Law is crucial for analyzing energy changes during chemical reactions. Hess's Law asserts that the total enthalpy change during a chemical reaction is the same, regardless of the number of intermediate steps involved. Enthalpy, symbolized as H, is the measure of the total heat content in a thermodynamic system at constant pressure.

To apply Hess's Law, one must know the standard enthalpy changes of individual reactions which can be summed to derive the enthalpy change of the overall reaction. This law is invaluable because it allows us to calculate the enthalpy changes of reactions for which direct measurements are complex or unfeasible. In simple terms, it lets us 'build' the desired reaction from known reactions, in much the same way one might use building blocks to create a structure.

Thermochemistry

The study of heat energy and its transformations is known as thermochemistry. This branch of chemistry focuses on the energy changes that occur during chemical reactions and changes of state. An essential concept in thermochemistry is the first law of thermodynamics, which states that energy can be transformed from one form to another, but cannot be created or destroyed.

Thermochemistry commonly deals with the enthalpy changes associated with chemical reactions, phase transitions, and the formation of compounds from their elements. Enthalpy changes can be measured or calculated from tables of standard enthalpy of formation values and are an important indicator of whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).

Enthalpy of Reaction

The term enthalpy of reaction, often denoted as \( \Delta H_{reaction} \), refers to the amount of heat released or absorbed during a chemical reaction at constant pressure. This value can be either negative or positive. A negative \( \Delta H_{reaction} \) indicates an exothermic reaction, where energy is given off to the surroundings, commonly experienced as warmth. Conversely, a positive \( \Delta H_{reaction} \) signifies an endothermic reaction, one that absorbs energy, and often causes cooling.

By calculating the enthalpy of reaction, scientists and engineers can predict how a reaction will behave thermodynamically, helping them to control processes and design reactions for a wide variety of applications, including energy production and material synthesis.

Standard Enthalpy of Formation

The standard enthalpy of formation, symbolized as \( \Delta H^{\circ}_{f} \), represents the heat change that occurs when one mole of a substance is formed from its elements at their standard states (most stable form under standard conditions of 1 bar and 25°C).

For an elementary substance in its standard state, the standard enthalpy of formation is zero by definition. Compounds have positive or negative \( \Delta H^{\circ}_{f} \) values, which can be found in tables like Appendix C used in our exercise. These values are fundamental in calculating the enthalpy changes for chemical reactions using Hess's Law. By understanding these concepts, students can better grasp the energetics of chemical processes and apply this knowledge to solve complex problems.