Thermo - Part II
Page 22
Question: Calculate the heat required to decompose limestone: CaCO3(s) → CaO(s) + CO2(g)
You are given:
ΔH°f [CaCO3(s)] = –1206.9 kJ/mol
ΔH°f [CaO(s)] = –635.1 kJ/mol
ΔH°f [CO2(g)] = –393.5 kJ/mol
1) The Long Way (Hess's Law):
(1)
Ca(s) + C(s) + 3/2O2(g) → CaCO3(s);
(2)
Ca(s) + ½O2(g) → CaO(s);
(3)
C(s) + O2(g) → CO2(g);
ΔH°f = –1206.9 kJ
ΔH°f = –635.1 kJ
ΔH°f = –393.5 kJ
Want: CaCO3(s) → CaO(s) + CO2(g)
Therefore, flip equation (1) and add (2) and (3):
CaCO3(s) → Ca(s) + C(s) + 3/2O2(g)
Ca(s) + ½O2(g) → CaO (s);
C(s) + O2(g) → CO2(g);
;
CaCO3(s) → CaO(s) + CO2(g) ;
ΔH°f = +1206.9 kJ
ΔH°f = –635.1 kJ
ΔH°f = –393.5 kJ
ΔH°Rxn = +178.3 kJ
2) The Short Way:
ΔH°rxn = ΣnΔHf°(products) – ΣnΔHf°(reactants)
ΔH°rxn = ΔH°f [CaO(s)] + ΔH°f [CO2(g)] – ΔH°f [CaCO3(s)]
ΔH°rxn = (–635.1 kJ/mol) + (–393.5 kJ/mol) – (–1206.9 kJ/mol)
ΔH°rxn = +178.3 kJ
Page 23
Problem 1:
Calculate ΔH°Rxn for the reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH°rxn = ΣnΔHf°(products) – ΣnΔHf°(reactants)
ΔH°rxn = ΔH°f [CO2(g)] + 2ΔH°f [H2O(l)] – ΔH°f [CH4(g)] – 2ΔH°f [O2(g)]
ΔH°rxn = (–393.5 kJ/mol) + 2(–285.8 kJ/mol) – (–74.87 kJ/mol) – 2(0 kJ/mol)
ΔH°Rxn = –890 kJ/mol
Problem 2:
Given
(a) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l); ΔH = –1169 kJ
(b) 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l); ΔH = –1530 kJ
What is ΔH°f for NO(g)?
Want:
½N2(g) + ½O2(g) → NO(g); ΔH°f[NO(g)]
Add (a) to the reverse of (b):
2
(a) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l); ΔH = –1169 kJ
(-b) 2N2(g) + 6H2O(l) → 4NH3(g) + 3O2(g) ; ΔH = +1530 kJ
2N2(g) + 2O2(g) → 4NO(g)
ΔHRxn = +361 kJ
Divide by 4:
½N2(g) + ½O2(g) → NO(g)
ΔH°f = +361/4 = 90.2 kJ