Problem 73 - Chemistry: The Central Science

Question

Using values from Appendix \(C\), calculate the standard enthalpy change for each of the following reactions: (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+4 \mathrm{HCl}(g)\)

Step-by-Step Solution

Answer

The standard enthalpy changes for the given reactions are: (a) -197.8 kJ/mol (b) 37.5 kJ/mol (c) -976.36 kJ/mol (d) -51.6 kJ/mol

Step 1: (a) Reaction: 2 SO₂(g) + O₂(g) → 2 SO₃(g)

To calculate the standard enthalpy change, we first look up the values of the standard enthalpy of formation (ΔH°f) for each species in the reaction. From Appendix C, we have: \(ΔH°f [SO₂(g)] = -296.8 \ \text{kJ/mol}\) Since there is no formation involved in the formation of elemental O₂, its value is: \(ΔH°f [O₂(g)] = 0\) \(ΔH°f [SO₃(g)] = -395.7 \ \text{kJ/mol}\) Now we apply the formula: ΔH°rxn = Σ n ΔH°f(products) - Σ n ΔH°f(reactants) For this reaction, it is: ΔH°rxn = [2 × (-395.7) ] - [ 2 × (-296.8) + 1 × 0] Calculate the value: ΔH°rxn = -791.4 + 593.6 = -197.8 kJ/mol

Step 2: (b) Reaction: Mg(OH)₂(s) → MgO(s) + H₂O(l)

From Appendix C, we have: \(ΔH°f [Mg(OH)₂(s)] = -924.9 \ \text{kJ/mol}\) \(ΔH°f [MgO(s)] = -601.6 \ \text{kJ/mol}\) \(ΔH°f [H₂O(l)] = -285.8 \ \text{kJ/mol}\) Apply the formula: ΔH°rxn = [(-601.6) + (-285.8) ] - [ (-924.9)] Calculate the value: ΔH°rxn = -887.4 + 924.9 = 37.5 kJ/mol

Step 3: (c) Reaction: N₂O₄(g) + 4 H₂(g) → N₂(g) + 4 H₂O(g)

From Appendix C, we have: \(ΔH°f [N₂O₄(g)] = 9.16 \ \text{kJ/mol}\) \(ΔH°f [H₂(g)] = 0\) \(ΔH°f [N₂(g)] = 0\) \(ΔH°f [H₂O(g)] = -241.8 \ \text{kJ/mol}\) Apply the formula: ΔH°rxn = [1 × 0 + 4 × (-241.8) ] - [ 1 × 9.16 + 4 × 0] Calculate the value: ΔH°rxn = -967.2 - 9.16 = -976.36 kJ/mol

Step 4: (d) Reaction: SiCl₄(l) + 2 H₂O(l) → SiO₂(s) + 4 HCl(g)

From Appendix C, we have: \(ΔH°f [SiCl₄(l)] = -657.5 \ \text{kJ/mol}\) \(ΔH°f [H₂O(l)] = -285.8 \ \text{kJ/mol}\) \(ΔH°f [SiO₂(s)] = -911.5 \ \text{kJ/mol}\) \(ΔH°f [HCl(g)] = -92.3 \ \text{kJ/mol}\) Apply the formula: ΔH°rxn = [(-911.5) + 4 × (-92.3) ] - [ (-657.5) + 2 × (-285.8) ] Calculate the value: ΔH°rxn = -911.5 - 369.2 + 657.5 + 571.6 = -51.6 kJ/mol So, the standard enthalpy changes for the given reactions are: (a) -197.8 kJ/mol (b) 37.5 kJ/mol (c) -976.36 kJ/mol (d) -51.6 kJ/mol