Transcript Enthalpy

Heat transfer experiments
150 mL
Tf
50 mL
100o
100 mL
25o
q = C x T x mass
q1 = (4.184 J/oC g) x (Tf - 100) x (50 g)
q2 = (4.184 J/oC g) x (Tf - 25) x (100 g)
q1 = - q2
(Tf - 100) x (50) = - (Tf - 25) x (100)
Tf = 50o C
Enthalpy of reaction
Hrxn = qrxn
coffee cup calorimeter
10.5 g KBr in 125g water at 24o
KBr(s)  K+ (aq) + Br-(aq)
Tf = 21o
Calculate Hrxn
qsystem = - qsurroundings = Hrxn
Hrxn
qsystem = - qsurroundings = Hrxn
qsurroundings = C x T x mass
qsurroundings = (4.184 J/goC) (21- 24oC) (10.5 g + 125 g)
= -1756 J
qsystem = - qsurroundings = +1756 J = Hrxn
a) endothermic b) exothermic H is extensive
= 167 J/g = 19873 J/mol
Hrxn = 1756 J
10.5 g KBr
E = q + w
E = q - PV
At constant V,
E = qv
Bomb calorimeter
qrxn = qsystem = -qcalorimeter
qcalorimeter = C (J / oC) x T (oC)
Constant Volume calorimetry
2Fe (s) + 3/2 O2 (g)  Fe2O3 (s)
11.2 g Fe(s), 1 atm O2
Ccalorimeter = 2.58 kJ/oC
Tcalorimeter = + 31.9 oC
qrxn = -qcalorimeter = Erxn
= - (2.58 kJ/oC) (31.9oC)
Erxn = - 82.2 kJ / 0.1 mol Fe2O3
= - 822 kJ/mol Fe2O3
Thermite reaction
2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l)
H is an extensive, State function
Hess’ Law
2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l)
2Al(s) + 3/2 O2(g)  Al2O3(s)
Fe
O3(s) +3/22Fe(s)
2 2Fe(s)
O2(g)+3/2FeO22O(g)
_______________________________
3(s)
H = -1676 kJ/mol
H = +-__________
822 kJ/mol
2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(s) -854 kJ/mol
2 (+15 kJ/mol )
2
2
Fe(s)

Fe(l)
_______________________________
__________
2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l)
Hrxn = -824 kJ/mol
Hess’ Law
• Always end up with exactly the same reactants
and products
• If you reverse a reaction, reverse the sign of
H
• If you change the stoichiometry, change H
Heats of formation, Hof
H = heat lost or gained by a reaction
“o” = standard conditions:
all solutes 1M
all gases 1 atm
“f” = formation reaction:
1mol product
from elements
in standard states
for elements in standard states, Hof = 0
2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l)
reactants
2 Al(s)
products
elements
2 Al(s)
Hof
Al2O3(s)
2 Fe(s)
Fe2O3
3/2 O2(g)
Hof Al2O3(s) + 2 Hof Fe (l)
2 Fe (l)
Fe2O3
Al(s)
2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l)
reactants
2 Al(s)
products
elements
2 Al(s)
Hof
Al2O3(s)
2 Fe(s)
Fe2O3
3/2 O2(g)
2 Fe (l)
Hof Al2O3(s) + 2 Hof Fe (l) - Hof Fe2O3 - Hof Al(s)
Hrxn = nHof products - nHof reactants
2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l)
Hrxn = nHof products - nHof reactants
Hrxn = [Hof Al2O3(s) + 2 Hof Fe(l)]
- [Hof Fe2O3(s) + 2 Hof Al(s)]
Hrxn = [(-1676) + 2 (15)]- [(-822) + 0]kJ
= -824 kJ
Bond Energies
chemical reactions = bond breakage and
bond formation
bond energies positive
energy required to break bond
bond breakage a) endothermic (raise P.E.)
exothermic (lower P.E.)
bond formation b)
exothermic
Bond energies
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)
C-H
O=O
C=O
O-H
413 kJ
495 kJ
799 kJ
467 kJ
Hrxn = bonds broken
- bonds formed
Hrxn =[ 4 (C-H)+ 2 (O=O)]
- [ 2 (C=O)+ 4 (O-H)]
= -824 kJ
Hrxn= Hof products- Hof reactants =- 802 kJ
qv v.s. qp
qv = E
qp = H
H = E + PV
H = E + PV = E + nRT
if n = 0
H = E
2Fe (s) + 3/2 O2 (g)  Fe2O3 (s)
n = (0 - 3/2) = - 3/2
H = - 822 kJ/mol + (- 3/2)(8.314 x 10-3 kJ)(298)
H = -826 kJ/mol