Answer:
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
