HCl(aq) + NaOH(s) → H2O + NaCl(aq)
Chapter 8:
Thermochemistry
thermometer
NaOH(s)
• How does the energy change
during this reaction?
• How does the change in energy
relate to change in temperature?
• Is this reaction “spontaneous”?
HCl(aq)
Chem 111
Dr. Gentry
Fundamental Kinds
of Energy
Conservation of Energy
Change in energy for the universe is zero
Potential energy
(mass * height * gravity)
Might change from one kind of energy to another…
Potential Energy → Kinetic Energy
Chemical Energy → Heat
Conservation
Of Energy
Or might transfer energy from one object to another…
But overall amount of energy must remain the same
Kinetic energy
(½ * mass * velocity2)
Kinetic Energy
Potential Energy
= Energy of Motion
= Stored Energy
= ½ m v2
m = mass of object
v = velocity of object
Examples:
Examples:
•
•
•
•
Skier at top of ski jump
Water behind a dam
Rock sitting on a hill
Electricity in a battery
•
•
•
•
Skier flying through air
Water falling over dam
Rock rolling downhill
Electric car driving on road
Convert potential energy into kinetic energy.
1
Conversion of Chemical Energy
Chemical Energy = energy involved in
bonds, intermolecular forces, etc.
Chemical Energy
CH4 + 2O2 → CO2 + 2H2O
(~potential energy)
TNT
Energy →
Conservation
Of Energy
Reactants
Change
in Energy,
∆E = 802kJ
Product
Heat or Work
Reaction Timeline →
(~kinetic energy)
Energy of system may change due to chemical rxn
Definitions Related
to Temperature
But conservation of energy requires transferring energy to
or from surroundings
Heat (q)
Work (w)
transfer of energy to raise temperature
all other transfers of energy
Heat + Work (Light)
Surroundings
Thermal Energy
= energy related to temperature (hot vs. cold)
= corresponds to increasing molecular motion
Temperature
= measure of how much thermal energy is present
System
Energy
of
Chemical
System
Heat
Heat
Work
Paraffin + O2 →
CO2 + H2O + energy
Chemical Energy
2 H2(g) + O2(g) → [ 2 H2O (g) ] → 2 H2O (l)
Reaction requires:
= FLOW of energy from hot object to cold object
… AN OBJECT DOES NOT HAVE HEAT !
Heat is what happens if object is hot in a cold room
Bond Dissociation Energy
• Energy required to break a covalent bond in an
isolated gaseous molecule.
1st: Break two H2 bonds and one O2 bond
(2*436 kJ) + (498 kJ)
2nd: Form four new O-H bonds in gas phase 2 H2O’s
(4*-460 kJ)
3d: Condensation of H2O from gas to liquid
(2*-44 kJ)
Total energy for reaction:
Add up all of the energies for each step in the process
(approximately - 558 kJ/mol )
actual number:
∆E= - 563kJ/mol
2
Work = – (External Pressure) * (Change in Volume)
One Small Problem
w = – Pext*∆V
• In open lab experiments
“∆” = change…
∆V = Vfinal - Vinitial
∆E (chemical energy) for 2 H2O formation is: – 563kJ/mol
But observe heat flow of:
– 570kJ/mol
• Must correct for work done by system
If system changes its volume
Then change in volume does work
by “pushing” against
surrounding air pressure
Energy
of
Chemical
System
Air Pressure
Heat
Water Pressure
Work
∆V
Work = – (External Pressure) * (Change in Volume)
w = – Pext*∆V
Due to weight of column of air overhead
Water Pressure
Atmosphere
height:
cross-section:
mass of air:
90 miles
1 in2
14.7 lb
∆V
14.7 pounds per square inch (psi) = 1.00 atm
Boiling
Water
Air Pressure
Volume change
due to water
changing to steam
Examples of Changes in Volume
Change in Enthalpy, ∆H
∆H = ∆E + P∆V
• Change in number of gas molecules
(Gases take up much for volume than solids or liquids)
Boiling water: H2O(l) → H2O(g)
• Dissolving salts in water
(volume of solid may be different than molecules in sol’n)
NaCl(s) → Na+(aq) + Cl-(aq)
Volume of NaCl(s)
≠
Volume of [ Na+(aq) + Cl-(aq) ]
(at constant pressure)
Enthalpy is what we observe in lab
∆E
= change in chemical energy
P∆V = additional work required if system
gains or loses volume during reaction
∆H
= total change in energy
Air Pressure, P
• Heating solids or liquids
(materials normally expand when get hot)
Cu(s, 25°C + heat → Cu(s, 85°C
(~ 14.7psi)
Change in
Volume,
∆V
∆H = ∆Ereaction + P∆V
Chem.
Rxn.
3
HCl + NaOH
→ H2O + NaCl + Energy
Reactants
(exothermic)
Release energy,
∆H = negative
(Energy)
Enthalpy →
Exothermic Reaction
If reaction molecules (system)
change energy by ∆H,
Products
NaOH + HCl
∆Hreleased
heat (q)
Products
(Energy)
Enthalpy →
Endothermic Reaction
Absorb energy,
∆H = positive
Reactants
Heat Transfer and Temperature
qreaction
Energy
of
Chemical
System
thermometer
Treat as Two
Different
Phases
q system = ∆H
Heat
Work
(at constant P)
where ∆H = ∆E + P∆V work
qreaction
= – qsurroundings
qsurroundings = m * C * ∆T
Surroundings = solvent water
m = amount of solvent
C = heat capacity of solvent
∆T = change in temp. of solvent
(Tfinal – Tinitial)
Water
Water
+
Reaction
…then must transfer heat to
surrounding water and air
to compensate for ∆H.
Heat Transfer and Temperature
= – qsurroundings
Reactants
Dissolved in
Water
Transfer of Energy
to Surroundings
‒ qsurrounding
∆Hrxn qrxn
∆Hrxn qrxn
Chem. Rxn.
CH2O = 1.0 cal / (gm*ºC)
= 4.18 J / (gm*ºC)
Heat Capacity
• Heat capacity (C) is the amount of heat required to raise
the temperature of an object or substance.
q
C=
m*∆T
a conversion factor
between heat and
temperature
Specific Heat: Amount of heat required to raise the
temperature of 1.00 g of substance by 1.00°C.
Molar Heat: Amount of heat required to raise the
temperature of 1.00 mole of substance by 1.00°C.
4
Calorimetry and Heat Capacity
Standard Enthalpy of Reaction
∆H°r
CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g)
• What is the specific heat of lead if it takes 96 J to raise the
temperature of a 75 g block by 10.0°C?
• When 25.0 mL of 1.0 M H2SO4 is added to 50.0 mL of 1.0 M
NaOH at 25.0°C in a calorimeter, the temperature of the
solution increases to 33.9°C. Assume specific heat of
solution is 4.184 J/(g·°C), and the density is 1.00 g/mL,
calculate ∆H for the reaction.
How much heat will be released if 0.35 moles of Cl2 are
reacted with excess methane (CH4) to create methylene
chloride (CH2Cl2)?
∆H°r values assume kJ per 1 mole of reaction
Need to look to stoichiometry.
How many moles Cl2 per reaction?
 rxn 
 −202kJ 
0.35molCl2 × 
 × 
 = − 35.4kJ
 rxn 
 2molCl2 
Hess’s Law
Hess’s Law
Energy and Enthalpy are
Properties of State
If environmental conditions are stable, then material
has a given level of energy and enthalpy
• Hess’s Law: Overall enthalpy change for reaction is same as
sum of individual enthalpy changes for smaller steps in rxn.
Because… Energy and Enthalpy are Properties of State
H2 + N2H4 (intermediate)
Doesn’t matter how a material got to where it is,
Only matters on where it is now
The road doesn’t matter, only the destination
Elevation in Philadelphia
= 39 ft
Elevation in Denver
= 5280 ft
Change from P. to D.
= 5141 ft
1
2
Hess’s Law
2 H2
(2)
H2
(1+2)
+
N2
+ N2H4
→
N2H4
∆H°1 = + 95 kJ
→
2 NH3
∆H°2 = – 187 kJ
3 H2 + N2 + N2H4 → N2H4 + 2 NH3
If rxn’s can be added, then ∆H’s can be added
(net)
3 H2(g) + N2(g)
→
2 NH3(g)
– 187 kJ
3 H2(g) + N2(g) → 2 NH3(g)
∆H° = – 92 kJ
net
– 92 kJ
1) 2 H2(g) + N2(g)
→
N2H4(g)
∆H°= + 95 kJ
H2(g) + N2H4(g →
2NH3(g)
∆H°= ‒187 kJ
Hess’s Law
• Reactants and products in individual steps can
be added and subtracted to determine the overall
net equation.
(1)
2
1
+ 95 kJ
2)
add
rxn’s
∆H°r = ‒202kJ
∆H°1+2 = +95 ‒ 187kJ
= – 92 kJ
∆H°net = – 92 kJ
• The industrial degreasing solvent methylene chloride (CH2Cl2,
dichloromethane) is prepared from methane by reaction with
chlorine:
CH4(g) + 2 Cl2(g)
→
CH2Cl2(g) + 2 HCl(g)
∆H°r = ?
• Use the following data to calculate ∆H°(in kilojoules) for the
above reaction:
(a) CH4(g) + Cl2(g)
→
CH3Cl(g) + HCl(g)
(b) CH2Cl2(g) + HCl(g) → CH3Cl(g) + Cl2(g)
∆H°r = – 98 kJ
∆H°r = +104 kJ
5
Alternative Pathways for Hess’s Law
Hess’s Law
The beauty of Hess’s Law = Pathway is your choice
Methylene Chloride
H2 + •OH → H2O + •H
Keep (a) as is:
(a) CH4(g) + Cl2(g)
CH3Cl(g) + HCl(g)
→
∆H°r = – 98 kJ
Actual
Multi-step
mechanism
• H + •OH → H2O
H2 + O2 → 2 •OH
Reverse (b): (need to multiply ∆H°by ‒1)
(b) CH3Cl(g) + Cl2(g) →
CH2Cl2(g) + HCl(g)
∆H°r = –104 kJ
Overall
Observed
Reaction
2 H2(g)
+ O2(g)
∆Hrxn
2 H2O (g)
→
Then cancel common items and add together:
(c) CH4(g) + 2 Cl2(g) + CH3Cl
→ CH3Cl(g) + CH2Cl2 + 2 HCl(g)
∆H°r = –202 kJ
Bond Dissociation Energy
Using
Bond
Energies
2H2 → 4 H
O2 → 2 O
4 H + 2 O → 2 H 2O
Standard Heats of Formation, ∆H°f
• Standard Heats of Formation (∆
∆H°f):
The enthalpy change for … the formation … of 1 mole of
substance in its standard state … from its elements in their
standard states.
• Std. Heat of Formation for CuSO4(s)
Cu(s) + S(s) + 2O2(g) → CuSO4(s)
2 H2(g)
H
H
H
H
break
+ O2(g)
O
• Stable forms of elements in their standard state at 25ºC
2 H2O (g)
→
O
O
H
break
- Most elements are single atoms, e.g. H, B, Na, etc.
O
H
∆Hºf = -771 kJ/mol
H
- But diatomic molecule for H2, N2, O2, F2, Cl2, other halogens
H
form
Using Standard Heats of Formation
and Hess’s Law
To Determine Heat of Rxn
Standard Heats of Formation
∆Hf°(kJ/mol)
*** NOT ON EXAM EVEN THOUGH ON HOMEWORK ***
CO(g)
- 111
C2H2(g)
+ 227
NO(g)
+ 90.
CO2(g)
- 394
C2H4(g)
+ 52
NH3(g)
- 46
H2O(l)
- 286
C2H6(g)
- 85
H2O(g)
- 242
CH3OH(g)
- 201
HF(g)
- 271
N2H4(g)
+ 95
C2H5OH(g)
- 235
AgCl(s)
- 127
HCl(g)
- 92
C6H6(l)
+ 49
C6H12O6(s)
Na2CO3(s)
∆H°rxn = ?
Reactants
tear apart
- 1260
–(∆H°f)reactants
“–” because
destroying reactants,
not forming them
Products
rebuild
(∆H°f)products
Collection of Elements
- 1131
∆H°rxn = – (∆H°f)reactants + (∆H°f)products
6
Standard Heats of Formation
• Calculate ∆H°(in kilojoules) for the reaction of ammonia
with O2 to yield nitric oxide (NO) and H2O(l), a step in the
Ostwald process for the commercial production of nitric
acid.
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(l)
–4*(–46)
–5*(0)
4*(+90.)
6*(–286)
Standard Heats of Formation
• Calculate ∆H°(in kilojoules) for the production of
glucose (C6H12O6) and oxygen from CO2 and liquid
water, a reaction carried out by all green plants in
photosynthesis.
6 CO2 (g) + 6 H2O (l)
C6H12O6 (s) + 6 O2 (g)
– 6*(–394) – 6*(–286) + 1*(–1260) + 6*(0) = + 2820 kJ/mol
# moles
Elements ( collection of N, H2, O2)
– 4*(–46) – 5*(0) + 4*(+90) + 6*(–286) = –1172 kJ/mol
7