Thermochemistry
Chapter 5 BLB 12th
Expectations
Heat & enthalpy – same or different?
Heat calculations:
– Temp. change
– Phase change (11.4, p. 438)
– Reactions
Enthalpy calculations
Read the chapter, study, and apply!
5.1 The Nature of Energy
Chemistry ⇐ ? ⇒ Energy
Energy – capacity to do work or transfer
heat
– Potential – stored energy; chemical
– Kinetic – released energy; energy of motion;
thermal
Electrostatic potential – interaction between
charged particles
Examples of Kinetic Energy
Energy, cont.
Units of energy:
Joule (J) – SI unit of energy;
kg  m
1J  1
2
s
2
calorie (cal)
– amount of energy required to raise the
temperature of exactly 1 gram of pure water
by 1°C (from 14.5°C to 15.5°C)
– 1 cal = 4.184 J (exactly)
– Calorie (dietary calorie),Cal
1 Cal = 1000 cal = 1 kcal
Energy, cont.
System and Surroundings
System – component(s) of interest
– Open – matter and energy can be exchanged between
system and surroundings
– Closed – can exchange energy but not matter
– Isolated – neither energy nor matter can be exchanged
Surroundings – everything outside of the system
Energy, cont.
Transferring Energy
Work (w) – energy used to move an object against
a force; w = F x d
Heat (q) – energy transferred from a hotter object to
a cooler one
Energy – capacity to do work or transfer heat;
ΔE = q + w
Combustion
heat & work
5.2 The First Law of Thermodynamics
Energy can be neither created nor destroyed.
Energy is conserved.
Internal energy, E – sum of all the kinetic and
potential energy of the system’s components
What kinds of energy are in here?
What changes could occur?
5.2 The First Law of Thermodynamics
More interested in the change in energy:
ΔE = Efinal – Einitial
Need to give number, units, and sign for all
thermodynamic quantities.
ΔE > 0 - Efinal > Einitial, system has gained
energy; endergonic
ΔE < 0 - Efinal < Einitial, system has lost energy;
exergonic
Note: Opposite change occurs with respect to
the surroundings.
Energy, heat & work
ΔE = q + w
Sign of ΔE depends upon sign and magnitude
of q and w.
Exothermic
Efinal < Einitial
Sample Exercise 5.2
A(g) + B(g) → C(s)
System loses 1150 J of heat to the
surroundings.
The piston move downwards doing
480 J of work on the system.
ΔE = ?
Calculate ΔE (in J); exothermic or endothermic?
a.
Balloon heating by adding 900 J of heat and
expands doing 422 J of work on atmosphere.
b.
50 g of H2O cooled from 30°C to 15°C losing
3140 J of heat.
c.
Reaction releases 8.65 kJ of heat, no work done.
Heat or Thermal Energy (q)
Exothermic: system → surroundings
– Heat energy released to surroundings
–q<0
– e.g. combustion reaction, crystallization
– Surroundings get warmer
Endothermic: system ← surroundings
– Heat energy flows into the system
–q>0
– e.g. melting, boiling, dissolution of NH4NO3
– Surroundings get colder
Heat, cont.
Evidenced by a change in temperature
Spontaneously transferred from the hotter
to the cooler object
Atoms or molecules with more energy
move faster
Temperature-dependent
Extensive property (depends on amount)
Total energy of system is the sum of the
individual energies of all the atoms and
molecules of the system.
Work (w)
Force acting over a distance
w = F x d = −PΔV
Compression: work ← surroundings
– Work is done on the system.
– ΔV < 0
–w>0
Expansion: work → surroundings
– Work is done on the surroundings.
– ΔV > 0
–w<0
Heat & Work, cont.
Work and heat are pathways by which energy
can be transferred.
State function – depends only on the system’s
present state; independent of the pathway;
internal energy, P, V, ΔE, ΔH, ΔS are state
functions
Energy is a state function, as is enthalpy.
5.3 Enthalpy
Enthalpy – heat flow at constant pressure; from Gr.
enthalpien – to warm
Enthalpy change (ΔH) – energy transferred as heat
at constant pressure; ΔH = Hproducts – Hreactants
H = E + PV
For a change @ constant pressure:
ΔH = ΔE + PΔV
ΔH = ΔE − w = qP
5.3 Enthalpy
ΔH < 0 exothermic: reactants → products + heat
ΔH > 0 endothermic: reactants + heat → products
ΔH = q/mol
Enthalpy (or heat) of reaction, ΔHrxn – enthalpy
change that accompanies a reaction
5.4 Enthalpies of Reaction, ΔHrxn
1. ΔH is an extensive property; value depends
upon the BALANCED equation.
2 H2(g) + O2(g) → 2 H2O(g)
ΔH = +483.6 kJ per 2 moles of H2O
H2(g) + ½ O2(g) → H2O(g)
ΔH = −241.8 kJ per mole of H2O
5.4 Enthalpies of Reaction
2. For reverse reactions:
ΔH values are equal in magnitude, but
opposite in sign.
For water: ΔHvap = +44.0 kJ/mol
ΔHcond = −44.0 kJ/mol
For the combustion of methane:
5.4 Enthalpies of Reaction
3. ΔH is dependent upon physical state.
ΔHf values:
C6H6(g) = 82.9 kJ/mol
C6H6(l) = 49.0 kJ/mol
H2O(l) → H2O(g)
ΔH = +44 kJ
CH3OH(g) → CO(g) + 2 H2(g) ΔHrxn = +90.7 kJ
a. Exothermic or endothermic?
b. Heat transferred for 1.60 kg CH3OH?
c. If 64.7 kJ of heat were used, how many
grams of H2 would be produced?
CH3OH(g) → CO(g) + 2 H2(g) ΔHrxn = +90.7 kJ
d. ΔH of reverse reaction?
Heat (in kJ) released when 32.0 g of CO(g)
reacts completely?
5.5 Calorimetry
Calorimetry – science of
measuring heat flow
Calorimeter – a device
used to measure heat flow
Coffee-cup calorimeter ⇒
Heat Capacity and Specific Heat
Heat capacity (C) - amount of heat required for
a 1°C temperature change:
J/°C = J/K
extensive property
heat transferred
q
C

temperature change T
ΔT in K = ΔT in °C
Heat Capacity and Specific Heat
Specific heat capacity (Cs)
– heat capacity for 1 g;
J/g·°C or J/g·K
Molar heat capacity – heat
capacity for 1 mole;
J/mol·°C or J/mol·K
Heat Capacity and Specific Heat
Specific heat values (more on p. 176):
Fe 0.45 J/g·K
glass 0.84 J/g·K
water 4.18 J/g·K (highest of all liquids and solids except ammonia)
heat transferred
q
Cs 

grams  temperature change m  T
Calculating heat (q)
To calculate the quantity of heat transferred:
q = Cs x m x ΔT
q – heat (J)
Cs – specific heat (J/g∙K)
m – mass (g)
ΔT – change in temp. (K or °C)
Calculate the heat (in J) required to raise the
temperature of 62.0 g toluene from 16.3°C to 38.8°C.
The specific heat of toluene is 1.13 J/g·K.
Calculate the specific heat of lead if 78.2 J of heat
were required to raise the temperature of a 45.6-g
block of lead by 13.3°C.
Constant-Pressure Calorimetry
Constant-pressure, ΔH = qP and ΔE = qP + w
Assume no heat is lost to surroundings.
Usually exothermic (qrxn < 0)
Applications:
– Heat transfer between objects
– Reactions in aqueous solutions
Use specific heat of water (4.18 J/g·K).
Use mass (or moles) of solution.
Solution Calorimetry
heat lost by reaction = heat gained by solution
−qrxn = qsoln
qrxn = −(Cs,soln x msoln x ΔT)
Enthalpy of reaction (ΔHrxn) per mole
ΔHrxn = qrxn/mol of specified reactant
A 19.6-g piece of metal was heated to 61.67°C. When
the metal was placed into 26.7 g water, the temperature
of the water increased from 25.00 to 35.00°C. Calculate
specific heat of the metal.
A 15.0-g piece of nickel at 100.0°C is dropped into a
coffee-cup calorimeter containing 55.0 g H2O at 23.0°C.
What is the final temperature of the water and nickel
after reaching thermal equilibrium? The specific heat
capacity of nickel is 0.444 J/g·K and of water is 4.18
J/g·K.
In a coffee-cup calorimeter, 2.50 g of MgO was reacted
with 125 mL of 1.0 M HCl. The temperature increased
by 9.6°C. Calculate the enthalpy of reaction per mole of
MgO for the following reaction.
Mg2+(aq) + H2O(l) → MgO(s) + 2 H+(aq)
Constant-Volume Calorimetry (p. 178)
Bomb calorimetry
No work is done (ΔV = 0), so ΔE = qV
Used for combustion reactions
The bomb components absorb the heat lost
by the reaction.
Heat capacity of the bomb (Ccal) needed to
calculate the heat of combustion (reaction)
qrxn = −(Ccal x ΔT)
Bomb Calorimeter
A 1.320-g sample of a new organic substance is
combusted in a bomb calorimeter (Ccal = 8.74 kJ/K).
The temperature of the bomb increased from 22.14°C
to 26.82°C. What is the heat of combustion per gram
of the substance?
Phase Changes
(Fig. 11.20, p. 439)
(or crystallization)
Phase Changes
(p. 440)
Endothermic →
Cs = 1.84 J/g·K
ΔHvap = 40.67 kJ/mol
← Exothermic
Cs = 4.18 J/g·K
ΔHfus = 6.01 kJ/mol
Cs = 2.03 J/g·K
Heat transfer – phase changes
To calculate the quantity of heat transferred
during a change of state:
q = ΔHprocess x m
or
q = ΔHprocess x mol
No change in temperature, so no ΔT.
For a complete process, add together the
heat transferred for each segment.
See Sample Exercise 11.3, p. 441.
11.46 Calculate the heat transferred for the conversion
of 35.0 g of the fluorocarbon, C2Cl3F3, from a liquid at
10.00°C to a gas at 105.00°C.
Data: b.p. 47.6°C, ΔHvap= 27.49 kJ/mol, Cs(liquid) =
0.91 J/g·K, Cs(gas) = 0.67 J/g·K
5.6 Hess’s Law
If a reaction is carried out in a series of
steps, ΔHrxn will equal the sum of the ΔH
values of the individual steps.
Hess’s Law works because ΔH is a state
function, i.e. it only depends upon the
initial reactant and the final product states.
Hess’s Law Example
Hess’s Law Example
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
The sum of these
equations is:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Based on the following reactions:
N2(g) + O2(g) → 2 NO(g)
2 NO(g) + O2(g) → 2 NO2(g)
2 N2O(g) → 2 N2(g) + O2(g)
ΔH, kJ
180.7
−113.1
−163.2
Calculate the ΔHrxn for the following reaction:
N2O(g) + NO2(g) → 3 NO(g)
Based on the following reactions:
ΔH, kJ
C2H2(g) + 5/2 O2(g) → 2 CO2(g) + H2O(l) −1300.
C(s) + O2(g) → CO2(g)
−394
H2(g) + ½ O2(g) → H2O(l)
−286
Calculate the ΔHrxn for the following reaction:
2 C(s) + H2(g) → C2H2(g)
5.7 Enthalpies of Formation
Standard state of a substance – pure form
at atmospheric pressure (1 atm) and
temperature of interest (usually 25°C).
Standard enthalpy - ΔH°
Standard enthalpy of formation, ΔHf°
- the change in enthalpy for the formation
of 1 mole of a substance from its elements
in their standard states, kJ/mol of product.
ΔHf° of an element in its most stable form
= 0 kJ/mol
Standard Enthalpies of Formation
(see Appendix C, p. 1059)
Reactions for ΔHf°
Calculating Enthalpies of Reaction
ΔH°rxn = Σn ΔHf°(products) − Σ n ΔHf°(reactants)
5.73 (c) Calculate ΔHrxn for the following reaction:
N2O4(g) + 4 H2(g) → N2(g) + 4 H2O(g)
Calculate ΔHrxn for the following reaction:
2 KOH(s) + CO2(g) → K2CO3(s) + H2O(g)
5.8 Foods and Fuels
Glucose is our body’s fuel source.
Carbs and fats are metabolized into glucose.
Excess fat is stored.
What’s the big deal? Take a look at the “fuel
value”.
Nonbiological Fuel