CHEMISTRY 161
Energy and Chemical Change
Chapter 7
Energy an Chemical Change
1.Forms of Energy
2.SI Unit of Energy
3.Energy in Atoms and Molecules
4.Thermodynamics
5.Calculation of Heat and Energy Changes
6.Measuring Heat and Energy Changes
1. Forms of Energy
1. Kinetic energy
energy of a moving microscopic or macroscopic object
E = ½ m v2
2. Radiant energy
energy in form of photons (‘light’) (solar energy)
E = h  (h = Planck’s constant) (Chapter 8)
3. Potential energy
energy by changing object’s position in height
E = m g h (h = height)
4. Thermal Energy
energy associated with random motion of atoms and molecules
Ekin = ½ M v2 = 3/2 R T (Chapter 7) M = m  Na
5. Chemical Energy
EXP1
energy stored in chemical bonds of substances (Chapter 7)
LAW OF CONSERVATION OF ENERGY
THE TOTAL ENERGY OF THE UNIVERSE
IS CONSTANT
2. SI Unit of Energy
1 Joule = 1 J
1 cal = 4.184 J
1 J = 1 Nm = 1 kg m2 s-2
Ekin = ½  m  v2
macroscopic versus microscopic
1 J vs. 1 kJ mol-1
3. Energy in Atoms and Molecules
Atoms – Kinetic and Thermal Energy
gases are constantly in motion
and hold a kinetic energy
Ekin = ½ M v2 = 3/2 R T
EXP2
Molecules – Kinetic, Thermal, & Potential Energy (N2)
molecules have different ‘internal’ (vibrational) energy
when bond distances are changed
EXP3/4
different bonds have different bond strength (stabilities) (H2 vs. N2)
4. Thermodynamics
reactants

products
(different energies)
THERMODYNAMICS
HEAT
CHANGE
study of the energy associated with change
THERMOCHEMISTRY
study of the energy associated with chemical change
2 H2(g) + O2(g) → 2 H2O(l) + energy
Hindenburg 1937
Challenger 1986
Surrounding
Surrounding
System
System
heat
heat
ENDOTHERMIC
EXOTHERMIC
EXP
2 HgO(s) → O2(g) + 2 Hg(l)
2 H2(g) + O2(g) → 2 H2O(l)
Energy
2 H2(g) + O2(g)
NH4NO3 (aq)
Exothermic
Endothermic
(heat given off by system)
(heat absorbed by system)
2 H2O(l)
NH4NO3(s) + H2O (l)
QUANTIFICATION
Enthalpy of Reaction
Enthalpy is the heat release at a constant pressure
(mostly atmospheric pressure)
DH = Hfinal - Hinitial
DH = Hproducts - Hreactants
board
Hfinal > Hinitial : DH > 0 ENDOTHERMIC
Hfinal < Hinitial : DH < 0 EXOTHERMIC
Energy
NH4NO3(aq)
2 H2(g) + O2(g)
Hfinal < Hinitial
Exothermic
2 H2O(l)
Hfinal > Hinitial
Endothermic
NH4NO3(s) + H2O(l)
Energy
H2O(l)
Hfinal > Hinitial
DH = Hfinal – Hinitial
Endothermic
H2O(s)
H2O(s) → H2O(l)
ΔH = + 6.01 kJ mol-1
Energy
H2O(l)
Hfinal < Hinitial
DH = Hfinal – Hinitial
Exothermic
H2O(s)
H2O(l) → H2O(s)
ΔH = - 6.01 kJ mol-1
THERMOCHEMICAL EQUATIONS
H2O(l) → H2O(s)
ΔH = - 6.01 kJ mol-1
CH4(g) + 2 O2(g) → 2 H2O(l) + CO2(g)
ΔH=-890.4 kJ mol-1
Calculate the heat evolved when combusting
24.0 g of methane gas.
5. Calculation of Heat and Enthalpy Changes
DHm = Hm,products – Hm,reactants
molar
REFERENCE SYSTEM
e.g. oxidation numbers of elements are zero
Standard Enthalpy of Formation
DHfO
heat change when 1 mole of a compound is formed from
its elements at a pressure of 1 atm
(T = 298 K)
DHfO (element) = 0 kJ/mol
DHfO (graphite) = 0 kJ/mol
DHfO (diamond) = 1.9 kJ/mol
ENTHALPY, H
0
C(s, graphite) + O2(g)
Hreactants
DHf0 = - 393.51 kJ mol-1
Hproducts
-393.51
CO2(g)
Standard Enthalpy of Formation
C(s, graphite) + O2(g)
CO2(g) DHf0 = - 393.51 kJ mol-1
C(s, graphite) + 2H2(g)
CH4(g) DHf0 = - 74.81 kJ mol-1
½ N2(g) + 3/2 H2(g)
0
-1
NH3(g) DHf = - 46.11 kJ mol
(1/2) N2(g) + (1/2) O2(g)
NO(g)
DHf0 = + 33.18 kJ mol-1
Standard Enthalpy of Reaction
aA+bB→cC+dD
ENTHALPY, H
aA+bB
a × DHfO (A) + b × ΔHfO(B)
Hreactants
DHOrxn = ΣΔHf0(prod) – ΣΔHf0(react)
Hproducts
c × DHfO(C) + d × ΔHfO(D)
cC+dD
Standard Enthalpy of Reaction
DHOrxn = ΣnΔHf0(prod) – ΣmΔHf0(react)
CaO(s) + CO2(g) → CaCO3(s)
-635.6
-393.5
-1206.9
DHOrxn = -177.8 kJ/mol
[kJ/mol]
Standard Enthalpy of Reaction
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
DHOrxn = ΣnΔHf0(prod) – ΣmΔHf0(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)
Hess’s Law
The overall reaction enthalpy is the
sum of the reaction enthalpies
of the steps in which the reaction
can be divided
CH4(g) + 2O2(g)
ENTHALPY, H
- 802 kJ
Reactants
- 890 kJ
CO2(g) + 2H2O(g)
- 88 kJ
CO2(g) + 2H2O(l)
Products
DHrxn for S(s) + 3/2 O2(g) SO3(g)
S(s) + O2(g)  SO2(g)
DH1 = -320.5 kJ
SO2(g) + 1/2 O2(g) SO3(g) DH2 = -75.2 kJ
S solid
direct path
+ 3/2 O2
DH3 =
-395.7 kJ
SO3 gas
+O2
Indirect Path
DH1 =
-320.5 kJ
SO2 gas
+ 1/2 O 2
DH2 = -75.2 kJ
6. State Functions
THERMODYNAMICS
quantitative study of heat and energy changes of a system
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
the state (condition) of a system is defined by
T, p, n, V, E
the state (condition) of a system is defined by
T, p, n, V, E
STATE FUNCTIONS
properties which depend only on the initial
and final state, but not on the way how this
condition was achieved
Hess Law
ΔV = Vfinal – Vinitial
Δp = pfinal – pinitial
ΔT = Tfinal – Tinitial
ΔE = Efinal – Einitial
Energy is a STATE FUNCTION
ΔE = m g Δh
IT DOES NOT MATTER WHICH PATH YOU TAKE
Hess Law
ENTHALPY, H
CH4(g) + 2O2(g)
- 802 kJ
Reactants
- 890 kJ
CO2(g) + 2H2O(g)
- 88 kJ
CO2(g) + 2H2O(l)
Products
Applications
Zeroth Law of Thermodynamics
a system at thermodynamical equilibrium
has a constant temperature
heat is spontaneous transfer of thermal energy two bodies
at different temperatures T1 > T2
spontaneous
T2
T1
EXP LN2/Metal/H2O
First Law of Thermodynamics
energy can be converted from one form to another,
but cannot be created or destroyed
CONSERVATION OF ENERGY
SURROUNDINGS
+
-
SYSTEM
THE TOTAL ENERGY OF THE UNIVERSE IS CONSTANT
First Law of Thermodynamics
ΔEsystem = ΔQ + ΔW
ΔQ heat change
ΔW work done
DQ > 0 ENDOTHERMIC
DQ < 0 EXOTHERMIC
?
mechanical work
ΔW = - p ΔV
M
ΔV < 0
the energy of gas goes up
M
ΔV > 0
the energy of gas goes down
6. Measurement of Heat Changes
Surrounding
System
heat
temperature
increase
DH = ΔQ ∞ ΔT
(pressure is constant)
Where does the ‘heat’ go?
DH = ΔQ ∞ ΔT
DH = ΔQ = const × ΔT
DH =ΔQ = C  ΔT
temperature
change
enthalpy
change
heat capacity
C=ms
s = specific heat capacity
DH = ΔQ = m s ΔT
EXP
specific heat capacity
capability of substances to store heat and energy
s = J g-1 K-1
the J necessary to increase the temperature of
1 g of a compound by 1 K
DH = ΔQ = m s ΔT
1.prepare two styrofoam cups
2. carry out chemical reaction
in a compound with known s
s (H2O) = 4.184 J g-1 K-1
3. measure temperature change
4. determine ΔH
calorimeter
100 ml of 0.5 M HCl is mixed with 100 ml 0.5 M NaOH in a
constant pressure calorimeter (scup = 335 J K-1). The initial
temperature of the HCl and NaOH solutions are 22.5C, and
the final temperature of the solution is 24.9C. Calculate the
molar heat of neutralization assuming the specific heat of
the solution is the same as for water.
DH = ΔQ = C ΔT
DH = ΔQ = (c1 + c2) ΔT
1.Neutralization reactions
2.Redox reactions
3. Precipitation reactions
Constant Volume Calorimeter
ΔQ = (m s(H2O) + cbomb) ΔT
Energy an Chemical Change
1.Forms of Energy
2.SI Unit of Energy
3.Energy in Atoms and Molecules
4.Thermodynamics
5.Calculation of Heat and Energy Changes
6.Measuring Heat and Energy Changes