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CHEMISTRY 161
Chapter 6
www.chem.hawaii.edu/Bil301/welcome.html
REVISION
Measurement of Heat Changes
HHmm = = ΔΔQQmm = = ccmpmp ΔΔTT HHmm = = ΔΔQQmm = = ccmpmp ΔΔTT
ccmpmp(H(H22O) = 75.3 J molO) = 75.3 J mol-1-1 K K-1-1ccmpmp(H(H22O) = 75.3 J molO) = 75.3 J mol-1-1 K K-1-1
Calculation of Heat Changes
HHmm = H = Hm,productsm,products – H – Hm,reactantsm,reactantsHHmm = H = Hm,productsm,products – H – Hm,reactantsm,reactants
REFERENCE SYSTEM
molar volume at standard temperature and pressureVm = 22.4 l
oxidation numbers of elements are zero
Standard Enthalpy of Formation
HHffOOHHff
OO
heat change when 1 mole of a compound is formed from its elements at a pressure of 1 atm
(T = 298 K)
HHffO O (element) = 0 kJ/mol(element) = 0 kJ/molHHff
O O (element) = 0 kJ/mol(element) = 0 kJ/mol
HHffO O (graphite) = 0 kJ/mol(graphite) = 0 kJ/molHHff
O O (graphite) = 0 kJ/mol(graphite) = 0 kJ/mol HHffO O (diamond) = 1.9 kJ/mol(diamond) = 1.9 kJ/molHHff
O O (diamond) = 1.9 kJ/mol(diamond) = 1.9 kJ/mol
EN
TH
AL
PY
, H
HHreactantsreactantsHHreactantsreactants
HHproductsproductsHHproductsproducts
C(s, graphite) + O2(g)
CO2(g)
Hf0 = - 393.51 kJ mol-1
0
-393.51
C(s, graphite) + O2(g) CO2(g) Hf0 = - 393.51 kJ mol-1
Standard Enthalpy of Formation
C(s, graphite) + 2H2(g) CH4(g) Hf0 = - 74.81 kJ mol-1
½ N2(g) + 3/2 H2(g) NH3(g) Hf0 = - 46.11 kJ mol-1
(1/2) N2(g) + (1/2) O2(g) NO(g) Hf0 = + 33.18 kJ mol-1
EN
TH
AL
PY
, H
HHreactantsreactantsHHreactantsreactants
HHproductsproductsHHproductsproducts
a A + b B → c C + d D
Standard Enthalpy of Reaction
a A + b B
c C + d D
HHOOrxn rxn = = ΣΔΣΔHHff
00(prod) – (prod) – ΣΔΣΔHHff00(react)(react) HHOO
rxn rxn = = ΣΔΣΔHHff00(prod) – (prod) – ΣΔΣΔHHff
00(react)(react)
a a ××HHffO O (A) + b (A) + b × × ΔΔHHff
OO(B)(B)a a ××HHffO O (A) + b (A) + b × × ΔΔHHff
OO(B)(B)
c c ××HHffOO(C) + d (C) + d × × ΔΔHHff
OO(D)(D)c c ××HHffOO(C) + d (C) + d × × ΔΔHHff
OO(D)(D)
HHOOrxn rxn = = ΣΣnnΔΔHHff
00(prod) – (prod) – ΣΣmmΔΔHHff00(react)(react) HHOO
rxn rxn = = ΣΣnnΔΔHHff00(prod) – (prod) – ΣΣmmΔΔHHff
00(react)(react)
Standard Enthalpy of Reaction
CaO(s) + CO2(g) → CaCO3(s)
-635.6 -393.5 -1206.9 [kJ/mol]
HHOOrxn rxn = -177.8 kJ/mol= -177.8 kJ/molHHOO
rxn rxn = -177.8 kJ/mol= -177.8 kJ/mol
Standard Enthalpy of Reaction
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
HHOOrxn rxn = = ΣΣnnΔΔHHff
00(prod) – (prod) – ΣΣmmΔΔHHff00(react)(react) HHOO
rxn rxn = = ΣΣnnΔΔHHff00(prod) – (prod) – ΣΣmmΔΔHHff
00(react)(react)
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
2 H2O(g) → 2 H2O(l)
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
EN
TH
AL
PY
, H
Reactants
Products
CO2(g) + 2H2O(g)
- 802 kJ
- 88 kJ
- 890 kJ
Hess’s Law
The overall reaction enthalpy is the
sum of the reaction enthalpies
of the steps in which the reaction
can be divided
HHrxnrxn for S(s) + 3/2 O for S(s) + 3/2 O22(g) (g) SOSO33(g)(g)
S(s) + OS(s) + O22(g) (g) SOSO22(g) (g) HH11 = -320.5 kJ = -320.5 kJ
SOSO22(g) + 1/2 O(g) + 1/2 O22(g) (g) SOSO33(g) (g) HH22 = -75.2 kJ = -75.2 kJ
S solid
SO3 gas
direct path
+ 3/2 O2
H3 = -395.7 kJ
SO2 gas
+O2H1 = -320.5 kJ
+ 1/2 O2H2 = -75.2 kJ
Indirect Path
Enthalpy of Solution
NaCl(s) → Na+(aq) + Cl-(aq) HHOOsolution solution = ?= ?HHOO
solution solution = ?= ?
NaCl(s) → Na+(g) + Cl-(g)
Na+(g) + Cl-(g) → Na+(aq) + Cl-(aq)
HH11OO=+788 kJ/mol=+788 kJ/molHH11
OO=+788 kJ/mol=+788 kJ/mol
HH22OO=-784 kJ/mol=-784 kJ/molHH22
OO=-784 kJ/mol=-784 kJ/mol
HHOOsolution solution = 788 kJ/mol – 784 kJ/mol = + 4 kJ/mol = 788 kJ/mol – 784 kJ/mol = + 4 kJ/mol HHOO
solution solution = 788 kJ/mol – 784 kJ/mol = + 4 kJ/mol = 788 kJ/mol – 784 kJ/mol = + 4 kJ/mol
( solutebility tables)
EN
TH
AL
PY
, H
NaCl(s)
Na+(g) + Cl-(g)
Na+(aq) + Cl-(aq)
HH11OO=+788 kJ/mol=+788 kJ/molHH11
OO=+788 kJ/mol=+788 kJ/mol HH22OO=-784 kJ/mol=-784 kJ/molHH22
OO=-784 kJ/mol=-784 kJ/mol
SUMMARY
Standard Enthalpy of Formation
HHffOOHHff
OO HHffO O (element) = 0 kJ/mol(element) = 0 kJ/molHHff
O O (element) = 0 kJ/mol(element) = 0 kJ/mol
Standard Enthalpy of Reaction
Hess’s Law
Homework
Chapter 6, p. 217-222problems