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
The enthalpy of formation of reactants and products can be used to obtain the enthalpy of reaction.
We know that the enthalpy of reaction can be obtained using the relation;
ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reactants
The values of the enthalpies of formation are obtainable from standard 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
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