The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions for each. (a) 2Ag2 O (s) ⟶ 4Ag (s) + O2 (g) (b) SnO (s) + CO (g) ⟶ Sn (s) + CO2 (g) (c) Cr2 O3 (s) + 3H2 (g) ⟶ 2Cr (s) + 3H2 O (l) (d) 2Al (s) + Fe2 O3 (s) ⟶ Al2 O3 (s) + 2Fe (s)

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) - 851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O (s) → 4Ag (s) + O₂ (g)

ΔH°f, Ag₂O (s) = - 31.05 kJ/mol

ΔH°rxn = 0 - (2 * (-31.05)) = 62.1 kJ/mol

b) SnO (s) + CO (g) → Sn (s) + CO₂ (g)

ΔH°f, SnO (s) = - 285.8 kJ/mol

ΔH°f, CO (g) = - 110.53 kJ/mol

ΔH°f, CO₂ (g) = - 393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃ (s) + 3H₂ (g) → 2Cr (s) + 3H₂O (l)

ΔH°f, Cr₂O₃ (s) = - 1128.4 kJ/mol

ΔH°f, H₂O (l) = - 285.83 kJ/mol

ΔH°rxn = [3 * (-285.83) ] - [ (-1128.4) ] = 270.91 kJ/mol

d) 2Al (s) + Fe₂O₃ (s) → Al₂O₃ (s) + 2Fe (s)

ΔH°f, Fe₂O₃ (s) = - 824.2 kJ/mol

ΔH°f, Al₂O₃ (s) = - 1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = - 851.5 kJ/mol