Enthalpy (H)
• The heat transferred sys ↔ surr during a
chemical rxn @ constant P
• Can’t measure H, only ΔH
• At constant P, ΔH = q = mCΔT, etc.
• Literally, ΔH = Hproducts - Hreactants
• ΔH = + (endothermic)
• Heat goes from surr into sys
• ΔH = - (exothermic)
• Heat leaves sys and goes into surr
In this example, the energy of the system (reactants and products) ↑, while
the energy of the surroundings ↓
Notice that the total energy does not change
Reactant + Energy
Product
Endothermic Reaction
Surroundings
Energy
Surroundings
System
System
Myers, Oldham, Tocci, Chemistry, 2004, page 41
Before
reaction
After
reaction
In this example, the energy of the system (reactants and products) ↓, while
the energy of the surroundings ↑
Notice again that the total energy does not change
Reactant
Product + Energy
Energy
Surroundings
Myers, Oldham, Tocci, Chemistry, 2004, page 41
System
Before
reaction
Exothermic Reaction
Surroundings
System
After
reaction
Reaction Coordinate Diagrams:
Endothermic Reaction
Activation Energy
Ea
D(PE)
PE
Products
Reactants
Progress of the Reaction
ΔHrxn= +
Reaction Coordinate Diagrams:
Exothermic Reaction
Activation Energy
Ea
ΔHrxn= -
PE
Reactants
D(PE)
Products
Progress of the Reaction
Reaction Coordinate Diagrams
Draw the reaction coordinate diagram for the following rxn:
C(s) + O2(g)  CO2 + 458.1kJ
Activation Energy
EXOTHERMIC
Ea
ΔHrxn= -458.1 kJ
PE
C + O2
D(PE)
CO2
Progress of the Reaction
Enthalpies of Reaction
• All reactions have some ΔH associated with it
H2(g) + ½ O2(g) → H2O(l)
ΔH = - 483.6 kJ
• How can we interpret this ΔH?
• Amount of energy released or absorbed per specific
reaction species
• Use balanced equation to find several definitions
- 483.6 kJ
1 mol H2
or
- 483.6 kJ
½ mol O2
or
- 483.6 kJ
1 mol H2O
Able to use like conversion factors in stoichiometry
Enthalpies of Reaction
• Formation of water
H2(g) + ½ O2(g) → H2O(l)
ΔH = - 483.6 kJ
• ΔH is proportional to amount used and will
change as amount changes
2H2(g) + O2(g) → 2H2O(l)
ΔH = - 967.2 kJ
• For reverse reactions, sign of ΔH changes
2H2O(l) → 2H2(g) + O2(g)
ΔH = + 967.2 kJ
• Treat ΔH like reactant or product
H2(g) + ½ O2(g) → H2O(l)
ΔH = - 483.6 kJ
H2(g) + ½ O2(g) → H2O(l) + 483.6 kJ (exo)
Enthalpies of Reaction Practice
Consider the following rxn:
C(s) + 1/2O2(g)  CO + 458.1kJ
Is the ΔH for this reaction positive or negative?
NEGATIVE (E released as a product)
What is the ΔH for 2.00 moles of carbon, if all the carbon
is used? 2.00 mol C - 458.1 kJ
1 mol C
= - 916 kJ
What is the ΔH if 50.0g of oxygen is used?
50.0 g O2
1 mol O2
32.0 g O2
- 458.1 kJ
= -1430 kJ
0.5 mol O2
1 mol CO
28.0 g CO
458.1 kJ
= 818 kJ
1 mol CO
What is the ΔH if 50.0 g of carbon monoxide decompose,
in the reverse reaction?
50.0 g CO
Hess’s Law
Reactants  Products
The change in enthalpy is the same whether the
reaction takes place in one step or a series of steps
Victor Hess
Why? Because enthalpy is a state function
To review:
1. If a reaction is reversed, ΔH is also reversed
2 CH4 + O2  2 CH3OH
ΔHrxn = -328 kJ
2 CH3OH  2 CH4 + O2
ΔHrxn = +328 kJ
2. If the coefficients of a reaction are multiplied by an integer,
ΔH is multiplied by that same integer
CH4 + 2 O2  CO2 + 2 H2O
2(CH4 + 2 O2  CO2 + 2 H2O)
ΔHrxn = -802.5 kJ
ΔHrxn = -1605 kJ
Example: Methanol-Powered Cars
2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(g)
ΔHrxn = ?
- ( ΔHrxn = -328 kJ)
- (2 CH4(g) + O2(g)  2 CH3OH(l) )
2 (CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)) 2 (ΔHrxn = -802.5 kJ )
2 CH3OH(l)  2 CH4(g) + O2(g)
ΔHrxn = +328 kJ
3
2 CH4(g) + 4 O2(g)  2 CO2(g) + 4 H2O(g) ΔHrxn = -1605 kJ
2 CH3OH + 3 O2  2 CO2 + 4 H2O
ΔHrxn = -1277 kJ
Tips for applying Hess’s Law…
Look at the final equation that you are
trying to create first…
• Find a molecule from that eq. that is only in
one of the given equations (i.e. CH3OH, CO2)
• Make whatever alterations are necessary to those
• Once you alter a given equation, you will not alter it again
• Continue to do this until there are no other
options
• Next, alter remaining equations to get things to
cancel that do not appear in the final equation
1. Given the following data:
S(s) + 3/2O2(g) → SO3(g)
ΔH = -395.2 kJ
- ½ ( 2SO2(g) + O2(g) → 2SO3(g) ) - ½ (ΔH = -198.2 kJ)
.
Calculate ΔH for the following reaction:
S(s) + O2(g) → SO2(g)
*
*
S(s) + 3/2O2(g) → SO3(g)
2SO3(g) → O2(g) + 2SO2(g)
SO3(g) → ½ O2(g) + SO2(g)
ΔH = -395.2 kJ
ΔH = +198.2 kJ
ΔH = +99.1 kJ
S(s) + O2(g) → SO2(g)
ΔH = -296.1 kJ
2. Given the following data:
-(C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = -1300. kJ)
2( C(s) + O2(g) → CO2(g) )
2(ΔH = -394 kJ)
H2(g) + 1/2O2(g) → H2O(l)
ΔH = -286 kJ
Calculate ΔH for the following reaction:
2C(s) + H2(g) → C2H2(g)
2C(s) + 2O2(g) → 2CO2(g)
ΔH = -788 kJ
2CO2(g) + H2O(l) → C2H2(g) + 5/2O2(g) ΔH = +1300 kJ
H2(g) + 1/2O2(g) → H2O(l)
ΔH = -286 kJ
*
*
*
2C(s) + H2(g) → C2H2(g)
ΔH = +226 kJ
3. Given the following data:
-½( 2O3(g) → 3O2(g))
-½(O2(g) → 2O(g) )
NO(g) + O3(g) → NO2(g) + O2(g)
-½( ΔH = - 427 kJ )
-½(ΔH = + 495 kJ)
ΔH = - 199 kJ
Calculate ΔH for the following reaction:
NO(g) + O(g) → NO2(g)
*
*
*
NO(g) + O3(g) → NO2(g) + O2(g)
O(g) → ½ O2(g)
3/2 O2(g) → O3(g)
NO(g) + O(g) → NO2(g)
ΔH = - 199 kJ
ΔH = - 247.5 kJ
ΔH = + 213.5 kJ
ΔH = - 233 kJ
4. Given the following data:
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
ΔH = -23 kJ
3Fe2O3(s) + CO(g) → 2Fe3O4(s) + CO2(g) ΔH = -39 kJ
Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g)
ΔH = 18 kJ
Calculate ΔH for the following reaction:
FeO(s) + CO(g) → Fe(s) + CO2(g)
Hess’s Law HW Questions
1.
A  B
C  B
∆H = + 30 kJ
∆H = - 60 kJ
Calculate ΔH for the following reaction: A  C
2. Suppose you are given the following reactions:
4X  2Y
DH = - 40 kJ
X  ½Z
DH = - 95 kJ
Calculate ΔH for the following reaction: Y  Z
3. From the following heats of reaction:
2 H2 (g) + O2 (g)  2 H2O (g)
3 O2 (g)  2 O3 (g)
DH = -483.6 kJ
DH = +284.6 kJ
Calculate the heat of the reaction (∆H):
3 H2 (g) + O3 (g)  3 H2O (g)
4. From the following enthalpies of reaction:
H2 (g) + F2 (g)  2 HF (g)
DH = - 537kJ
C (s) + 2 F2 (g)  CF4 (g)
DH = - 680 kJ
2 C (s) + 2 H2 (g)  C2H4 (g)
DH = + 52.3 kJ
Calculate the DH for the reaction of ethylene with F2.
C2H4 (g) + 6F2 (g)  2 CF4 (g) + 4 HF(g)
5. Given the following data:
N2 (g) + O2 (g)  2 NO (g)
2 NO (g) + O2 (g)  2 NO2 (g)
2 N2O (g)  2 N2 (g) + O2 (g)
Calculate DH for the reaction below:
N2O (g) + NO2 (g)  3 NO (g)
DH = + 180.7 kJ
DH = - 113.1 kJ
DH = - 162.3 kJ
DH = ?
Heats of Formation, ΔH°f
The enthalpy change when one mole of a compound is
formed from the elements in their standard states
° = standard conditions
• Gases at 1 atm pressure
• All solutes at 1 M concentration (remember M = mol/L)
• Pure solids and pure liquids
f
= a formation reaction
• 1 mole of product formed
• From the elements in their standard states (1 atm, 25°C)
For all elements in their standard states, ΔH°f = 0
What’s the formation reaction for adrenaline, C9H12NO3(s)?
9 Cgr + 6 H2(g) + 1/2 N2(g) + 3/2 O2(g)  C9H12NO3(s)
Thermite Reaction
Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l)
ΔHrxn = ?
Welding railroad tracks
Thermite Reaction
Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(l)
Reactants
Elements
Products
(standard states)
Fe2O3(s)
2 Al(s)
2 Fe(s)
3/
2
O2(g)
2 Fe(l)
Al2O3(s)
2 Al(s)
ΔHrxn = 2ΔH°f(Fe(l)) + ΔH°f(Al2O3(s)) - ΔH°f(Fe2O3(s)) - 2ΔH°f(Al(s))
ΔHrxn = 2(15 kJ) + (-1676 kJ) - (-822 kJ) – 2(0)
ΔHrxn = - 824 kJ
ΔHrxn = nΔH°f(products) - nΔH°f(reactants)
ΔH°f Example Problems
∆Hrxn = Σ n∆Hof Products - Σn∆Hof Reactants
1. CH4(g) + 2 Cl2(g)  CCl4(g) + 2 H2(g)
ΔHrxn = ?
(- 106.7) 2 (0)
(- 74.8) 2 (0)
∆H = [(-106.7) + 0] – [(-74.8)+0]
= -106.7 + 74.8
= - 31.9 kJ
2. 2 KCl(s) + 3 O2(g)  2KClO3(s)
2 (- 435.9)
3 (0)
2 (- 391.2)
∆H = [(2)(- 391.2)] – [(2)(- 435.9) + (3)(0)]
= - 782.4 + 871.8
= 89.4 kJ
ΔHrxn = ?
ΔH°f Example Problems
∆Hrxn = Σ n∆Hof Products - Σn∆Hof Reactants
3. AgNO3(s) + NaCl (aq)  AgCl(s) + NaNO3(aq) ΔHrxn = ?
(-124.4)
(-407.1)
(-127.0)
(-446.2)
∆H = [(-127.0) + (-446.2)] – [(-124.4) + (-407.1)]
= -573.2 + 531.5
= - 41.7 kJ
4. C2H5OH(l) + 7/2 O2(g)  2CO2(g) + 3H2O(g) ΔHrxn = ?
(-277.7)
(7/2)(0)
(2)(-393.5) (3)(-241.8)
∆H = [(2)(-393.5) + (3)(-241.8)] – [(-277.7) + (7/2)(0)]
= -1512.4 + 277.7
= -1234.7 kJ
Enthalpy Review
#2. Calculate DH for the following reaction:
N2H4 (g) + O2 (g) → N2 (g) + 2H2O (g)
Given:
DH (kJ/mol)
2 NH3 (g) + 3 N2O (g) → 4 N2 (g) + 3 H2O (g)
-1010
N2O (g) + 3 H2 (g) → N2H4 (g) + H2O (g)
-317
2 NH3 (g) + ½ O2 (g) → N2H4 (g) + H2O (g)
-143
H2 (g) + ½ O2 (g) → H2O (g)
-286
Bond Energies
• Chemical reaction ⇔ Bond breakage & bond formation
• Bond energy = energy required to break a bond
• Bond breaking is endothermic (raises potential energy)
• Bond formation is exothermic (lowers PE)
• Average energy for one type of bond in different molecules
• Common Bond Energies
C-H : 413 kJ/mol
C=O : 799 kJ/mol
O=O : 495 kJ/mol
O-H : 467 kJ/mol
• ΔHrxn = (bonds broken) – (bonds formed)
Energy required
Energy released
Bond Energies
• ΔHrxn = (bonds broken) – (bonds formed)
ex. CH4 +
H
H C H +
H
2 O2  CO2 + 2 H2O
O O
O O
C-H : 413 kJ/mol
O=O : 495 kJ/mol
O
C +
O
H
H
O
O
H
H
C=O : 799 kJ/mol
O-H : 467 kJ/mol
ΔHrxn = [4(C-H) + 2(O=O)] – [2(C=O) + 4(O-H)]
ΔHrxn = [4(413 kJ) + 2(495 kJ)] – [2(799 kJ) + 4(467 kJ)]
ΔHrxn = -824 kJ
Compare to ΔHrxn = -802.5 kJ
Enthalpy Summary
• Enthalpy (ΔH) rxn = heat = q
• All rxns have some ΔH
• ΔH = +…endo
• ΔH = - …exo
• If given ΔH, can use stoich to quantify
• Three ways to estimate ΔH (if not given)
• Hess’s Law
• known eq’s manipulated into desired eq.  ΔHrxn
• Heats of Formation (ΔHf): values from appendix
• ΔHrxn = nΔH°f(products) - nΔH°f(reactants)
• Bond energies
• ΔHrxn = bonds broken – bonds formed