Chapter 6: Thermochemistry
Energy is the capacity to do work.
Work: force X distance
Potential Energy: available by virtue of an object’s position
Kinetic Energy: energy of motion
Radiant Energy: in electromagnetic waves, (light) (kinetic)
Thermal Energy: random motion of atoms and molecules (kinetic)
Chemical Energy: store within chemical bonds (potential)
Nuclear Energy: stored with atomic nuclei (potential)
Energy changes in chemical reactions:
Heat: transfer of thermal energy between two bodies that are at
Temperature: measure of the thermal energy
Temperature IS NOT the same as thermal energy
Bathtub=lower temperature, greater thermal energy
Coffee=higher temperature, lower thermal energy
Thermal energy in an object depends on…
Temperature vs. heat
Analogy: bank account How much money is in my account?
How many transfers are in my account? Doesn’t make sense.
Consider a solution
What is the temperature?
How much heat is in it? Doesn’t make sense
How much energy?
Thermochemistry: study of heat change in chemical reactions
System: the part of the universe being studied
Exchanged=mass and energy
Exothermic process: gives off heatthermal energy goes from
system to surroundings
2H2 (g) + O2 (g) ▯ 2H2O (l) + penergy
H2O (g) ▯H2O (l) + energy
Endothermic process: heat has to be suppliedthermal energy goes
from surroundings to system
energy + 2HgO (s) ▯ 2Hg (l) + O2 (g)
energy + H2O (s) ▯H2O (l)
Exothermic or exothermic?
Making ice cubes from water=exo
Conversion of frost to water vapor=endo
Forming a He+ from He (g)=endo
Cl2 ▯ 2Cl=endo
2Na + S ▯Na2S=exo
Units of energy:
1 kg x m^2 x s^2 ~1 human heartbeat
Lift 1 kg by 10 cm
Raise temperature of 1g of H2O by 1°C
Calorie=1000 calories = 1 kcal
Label on food information
37.2 J to cal
37.2 J 1 cal
x = 8.81cal
892.1 cal to kJ
892.1 cal 4.184 J 1 kJ
x x = 3.732 kJ
1 cal 1000 J
104.6 kJ to kcal
104.6 kJ 1000J 1 cal 1kcal
x x x = 25.00 kcal
1kJ 4.184 J 1000 cal
29.4 cal to J
29.4 cal 4.184 J
x =123 J 1 cal
What is energy?
The ability to do work or produce heat
Heat is microscopic
Work is macroscopic
Analogy: coffee in a fridge vs. pushing a desk
Heat and work: the only ways to transfer energy
State functions: properties that are determined by the state of the
system, regardless of how that condition was achieved.
ΔU=U finU initial
ΔV=V finV initial
ΔT=T finT initial
1 law of thermodynamics: energy can’t be created or destroyed
(can change form).
ΔU syst+ ΔU surround = 0 or ΔUsyste= ΔU surrounding
C3H3 + 5O2 ▯ 3CO2 + 4H2O
Exothermic chemical reactions!
Chemical energy lost by combustion=energy gained by the
ΔU= q + w
ΔU is the change in internal energy of a system q is the heat exchange between the system and surroundings
w is the work done (or by) the system
w=PΔV when a gas expands against a constant external pressure
Work done by the system:
PxV=F x d^3=Fd=w
w=F x Δd
F=P x A
w= P x A x Δd
A x Δd =ΔV
*Work is not state function
(x)J=L x atm
1 atm=101.3x10^3 Pa=101.3x10^3N/m^2
N=kg x M
101.3J=1L x atm
A gas expands from 1.6 L to 5.4 L (constant T). What is the work
(in joules) (a) against a vacuum (b) against a pressure of 3.7 atm?
(a) ΔV=5.4L1.6L=3.8L P=0atm
w=0atm x 3.8 L = 0Latm=0joules
(b) ΔV=5.4L1.6L=3.8L P=3.7atm
w=3.7atm x 3.8 L =14.1 Latm
w=14.1Latm x 101.3 J
50g of water is cooled from 30°C to 15°C, losing 3140 J of heat.
What is the energy change?
ΔU= q + w
ΔU= 3140 J
A balloon is heated by adding 900 J of heat. Expansion does 422 J
of work on the atm. What is the energy change?
ΔU= q + w
ΔU= 900J – 422J=478J
42.6L of gas expands 48.2L following addition of 1060 J of heat at
1.0 atm. What is the energy change?
ΔU= q +w
w=5.6Latm 101.3 J
x = 570J
30.2 L of gas expands to 84.2 L following addition of 512 J of heat
at 1.2 atm. What is the energy change?
ΔU=q + w
w=64.8Latm 101.3 J
x = 6600J
ΔU=512J6600J=6100J or 6.1kJ
612 J of heat is removed from 37.2 L of gas at 0.80 atm and the
volume becomes 20.0L. What is the energy change?
ΔU=q + w
w=(0.80atm)(17.2L) w=14Latm 101.3 J
x = 1400J
Enthalpy (ΔH) and the first law of thermodynamics
At constant pressure: q=ΔH and w= PΔV
ΔU=ΔHPΔV or ΔH=ΔU+PΔV
Definition: Thermodynamic potential
Measure of a system’s potential energy
Mathematically: H=U + PV
▯ ΔH=ΔU + PΔV (at constant pressure)
Units: J or cal
Recall: ΔU=q + w = q – PΔV
ΔH=qPΔV + PΔV
*But only at constant P!
Differences between q and ΔH
Imagine: P1V1 P2V2
Lots of q/w combinations
Heat and work are NOT state functions
ΔH: Same regardless of path
Enthalpy IS a state function
Defined in terms of U, P, and V ΔH=H fHi
ΔH is intrinsic
Enthalpy and the first law of thermodynamics
ΔU= q + w
At constant pressure:
q=ΔH and w=PΔV
Constant P: often good enough for chemistry
Gas expands, 18 KJ work (done on surroundings)
79 kJ heat released
ΔH? = 79kJ
ΔU? = ΔHPΔV
ΔU = 97kJ
418 J heat added
ΔH? = 418J
ΔU? = ΔHPΔV
= 418J(1.12atm)(0.928L) 101.3 J
x = 105J
=418J105J ΔU =313J
Exothermic: heat given off by the system to the surroundings
ΔH<0 products0 products>reactants
6.01 kJ are absorbed for every 1 mole of ice that melts at 0°C and 1
atm. Endothermic reactions, positive, system absorbs heat.
890.4 kJ are released for every 1 mole of methane that is
combusted at 25°C and 1 atm. Exothermic reaction, negative,
system gives off heat.
Coefficients: always of moles of a substance
H2O (s) ▯H2O (l) ΔH=6.01kJ
Reverse the reactions, change the sign of ΔH
H2O (l) ▯H2O (s) ΔH—6.01kJ
Multiply both sides of the equation: ΔH changes bi same factor
2H2O (s) ▯ 2H2O (l) ΔH=2 x 6.01= 12.0kJ
Note: change in state ▯change in enthalpy!
The physical states of all reactants and products must be specified
in the thermochemical equations
Heat evolved during combustion of 266 g of P 4?
P4 (s) + 5O2 (g) ▯P4O10 (s) ΔH=