Problem 74 - Chemistry The Central Science

Question

Using values from Appendix \(\mathrm{C},\) 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 of ΔH⁰ for the given reactions are as follows: (a) ΔH⁰ = -217.5 kJ/mol (b) ΔH⁰ = -560.4 kJ/mol (c) ΔH⁰ = 86.5 kJ/mol (d) ΔH⁰ = -286.8 kJ/mol

Step 1: (a) CaO(s) + 2 HCl(g) -> CaCl2(s) + H2O(g)

First, we need to find the ΔH⁰ values of the reactants and products from Appendix C. ΔH⁰(CaO(s)) = -635.1 kJ/mol ΔH⁰(HCl(g)) = -92.3 kJ/mol ΔH⁰(CaCl2(s)) = -795.4 kJ/mol ΔH⁰(H2O(g)) = -241.8 kJ/mol Next, we will apply the formula: ΔH⁰(reaction) = Σ ΔH⁰(products) - Σ ΔH⁰(reactants) ΔH⁰(reaction) = [ΔH⁰(CaCl2(s)) + ΔH⁰(H2O(g))] - [ΔH⁰(CaO(s)) + 2 × ΔH⁰(HCl(g))] ΔH⁰(reaction) = [(-795.4) + (-241.8)] - [(-635.1) + 2 × (-92.3)] ΔH⁰(reaction) = -1037.2 + 635.1 + 184.6 ΔH⁰(reaction) = -217.5 kJ/mol

Step 2: (b) 4 FeO(s) + O2(g) -> 2 Fe2O3(s)

First, we need to find the ΔH⁰ values of the reactants and products from Appendix C. ΔH⁰(FeO(s)) = -272.0 kJ/mol ΔH⁰(O2(g)) = 0 kJ/mol (as it is an element in its standard state) ΔH⁰(Fe2O3(s)) = -824.2 kJ/mol Next, we will apply the formula: ΔH⁰(reaction) = Σ ΔH⁰(products) - Σ ΔH⁰(reactants) ΔH⁰(reaction) = [2 × ΔH⁰(Fe2O3(s))] - [4 × ΔH⁰(FeO(s)) + ΔH⁰(O2(g))] ΔH⁰(reaction) = [2 × (-824.2)] - [4 × (-272.0) + 0] ΔH⁰(reaction) = -1648.4 + 1088.0 ΔH⁰(reaction) = -560.4 kJ/mol

Step 3: (c) 2 CuO(s) + NO(g) -> Cu2O(s) + NO2(g)

First, we need to find the ΔH⁰ values of the reactants and products from Appendix C. ΔH⁰(CuO(s)) = -156.1 kJ/mol ΔH⁰(NO(g)) = 90.3 kJ/mol ΔH⁰(Cu2O(s)) = -168.6 kJ/mol ΔH⁰(NO2(g)) = 33.2 kJ/mol Next, we will apply the formula: ΔH⁰(reaction) = Σ ΔH⁰(products) - Σ ΔH⁰(reactants) ΔH⁰(reaction) = [ΔH⁰(Cu2O(s)) + ΔH⁰(NO2(g))] - [2 × ΔH⁰(CuO(s)) + ΔH⁰(NO(g))] ΔH⁰(reaction) = [(-168.6) + 33.2] - [2 × (-156.1) + 90.3] ΔH⁰(reaction) = -135.4 + 312.2 - 90.3 ΔH⁰(reaction) = 86.5 kJ/mol

Step 4: (d) 4 NH3(g) + O2(g) -> 2 N2H4(g) + 2 H2O(l)

First, we need to find the ΔH⁰ values of the reactants and products from Appendix C. ΔH⁰(NH3(g)) = -45.9 kJ/mol ΔH⁰(O2(g)) = 0 kJ/mol (as it is an element in its standard state) ΔH⁰(N2H4(g)) = 50.6 kJ/mol ΔH⁰(H2O(l)) = -285.8 kJ/mol Next, we will apply the formula: ΔH⁰(reaction) = Σ ΔH⁰(products) - Σ ΔH⁰(reactants) ΔH⁰(reaction) = [2 × ΔH⁰(N2H4(g)) + 2 × ΔH⁰(H2O(l))] - [4 × ΔH⁰(NH3(g)) + ΔH⁰(O2(g))] ΔH⁰(reaction) = [2 × 50.6 + 2 × (-285.8)] - [4 × (-45.9) + 0] ΔH⁰(reaction) = 101.2 - 571.6 + 183.6 ΔH⁰(reaction) = -286.8 kJ/mol

Key Concepts

Chemical Reactions

Chemical reactions are processes where one or more substances, called reactants, are transformed into different substances, known as products. This transformation always involves the breaking and forming of bonds, which results in changes in the energy content of the system. For example, in a reaction like

• CaO(s) + 2HCl(g) -> CaCl₂(s) + H₂O(g)

The solid calcium oxide reacts with gaseous hydrogen chloride to form solid calcium chloride and gaseous water. In chemical reactions, the laws of conservation of mass and energy are always satisfied. Mass is neither created nor destroyed, and the total energy of the system remains constant. Chemical reactions can be exothermic, releasing energy, or endothermic, absorbing energy. The classification depends on the energy required to break existing bonds compared to the energy released when new bonds are formed. Understanding these concepts helps us grasp why figuring out energy changes, like enthalpy change, is so crucial.

Thermochemistry

Thermochemistry is the branch of chemistry concerned with the heat changes that occur during chemical reactions. It allows us to predict whether a reaction will release or absorb energy. This is crucial for industrial processes, environmental science, and even daily life activities like cooking. In thermochemistry, the primary focus is on:

• Heat (q): the energy transferred due to temperature differences.
• Work (w): energy transfer that is not due to temperature differences, often measured as pressure-volume work.

The study of these principles allows us to understand and calculate the enthalpy ( \(\Delta H\) ), which is the heat absorbed or released under constant pressure.If a reaction is described as endothermic, it means that the system absorbs heat from its surroundings, usually resulting in a positive \(\Delta H\) . Conversely, an exothermic reaction releases heat, giving a negative \(\Delta H\) . By using thermochemical data, such as those found in Appendix C of a chemistry textbook, we can determine the heat change for reactions, like those provided in our original exercise.

Enthalpy Calculations

Enthalpy calculations are essential for predicting the energy changes in chemical reactions. Enthalpy ( \(H\) ) is a measure of the total energy of a thermodynamic system, including both internal energy and energy associated with volume and pressure. In enthalpy calculations, we use the standard enthalpy of formation ( \(\Delta H^\circ_f\) ), which is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. The formula commonly applied is: \(\Delta H^{\circ} = \Sigma \Delta H^{\circ} (\text{products}) - \Sigma \Delta H^{\circ} (\text{reactants})\). By following these steps, you can effectively determine the enthalpy change ( \(\Delta H\)) for any given reaction:

• Find \(\Delta H^\circ_f\) values for all reactants and products from a trusted source, like Appendix C.
• Multiply these values by their respective coefficients in the balanced chemical equation.
• Apply the formula to find \(\Delta H^{\circ}\) by subtracting the sum of the reactants' enthalpies from the sum of the products' enthalpies.

Doing these calculations provides insight into whether a reaction is endothermic or exothermic, which in turn helps in understanding its energy profile and practical implications.