Thermochemistry
Types of Energy
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Radiant
Thermal
Chemical
Potential
The 1st Law
ΔE = q + w
Sign Convention for q and w..
Work done by sys. (expansion) = Work done on sys. (contraction) = +
Heat absorbed by sys. (endo) = +
Heat released by sys. (exo) = 1
Work and Energy
ΔE = q + w
P
P
ΔV
wgas = ­PΔV
+ΔV
q + w
ΔE =
Constant Pressure Conditions (typical)
wgas = ­PΔV
ΔH = ΔE + PΔV
ΔH = qP
2
Exothermic and Endothermic
+ΔH
­ΔH
H2O(s) + HEAT H2O(l)
ΔH = +6.01 kJ/mol
CH4 + 2O2 CO2 + 2H2O + HEAT
ΔH = ­ 890 kJ/mol
Heat Capacity
2,000,000 g H2O
90 oC
30 oC
200 g H2O
q
Heat Capacity = q = (H.C.)ΔT
ΔT
Specific Heat Capacity = C = q
mΔT
q = mCΔT
WARM AIR
COOL
COOL AIR
WARM
An 150 g iron bar absorbs 3500 J of heat.
• What will be its temperature change?
• If it was 30oC initially what will be its final temperature?
3
Calorimetry
­qsys = qwater
­mCΔT = m(4.184)ΔT
sample
water
50 g of a metal at 150oC is placed into 200 g of water at 30oC. The final temperature of the system is 34.5oC. What is the specific heat of the metal?
J
[0.652 ]
o
gC
Bomb Calorimtery
q = Ccal x ΔT
4
Determining ΔH from Reactions
target
Hess's Law
2 C (s)+ H2 (g)
C2H2 (g)+ 52 O2 (g)
C2H2 (g)
ΔH = ??
2 CO2 (g) + H2O (l)
ΔH = ­1300 J
C (s) + O2 (g)
CO2 (g)
H2 (g) + 12 O2 (g)
H2O (l)
ΔH = ­394 J
ΔH = ­286 J
5
Determining ΔH from Reaction Tables
NH4Cl (g)
NH3 (g) + HCl (g)
o
o
o
ΔHrxn = ΣaΔHprods ­ ΣbΔHreacts
C2H5OH (l) + 3 O2 (g)
2 CO2 (g)+ 3 H2O (g)
6
Heat of Solution (ΔHsol)
NH4NO3 (s)
100 g H2O
ΔHsol = +26.2 kJ/mol
CaCl2 (s)
100 g H2O
ΔHsol = ­82.8 kJ/mol
7