Chapter 6: Enthalpy Changes for Chemical Reactions
Chapter Highlights
energy transfer, specific heat and heat transfer
heat of fusion & heat vapourization
exothermic & endothermic reactions
first law of thermodynamics
enthalpy changes
calorimetry
Hess’s Law
Chemistry-140 Lecture 13
Thermodynamics
Thermodynamics: energy and its transformations.
Thermochemistry: energy changes and chemical reactions
Force: "push" or "pull" exerted on an object.
Work: energy required to overcome a force:
Work = force x distance.
Heat: energy transferred from one object to another because of a temperature difference.
Chemistry-140 Lecture 13
Kinetic and Potential Energy
Kinetic energy (EK): energy of an object due to its motion.
EK = mv2 (m = mass, v = velocity)
Potential energy (EP): energy stored by an object
due to its relative position.
1
2
Chemistry-140 Lecture 13
Energy
Energy can be expressed in a wide variety of units.
The joule (J) is the metric unit of energy and is the energy
possessed by a 2 kg object moving at a velocity of 1 m/s.
1 J (joule) =1 kg-m2/s2.
The calorie (cal) is the energy required to raise the
temperature of 1 g of water by 1°C.
1 cal = 4.184 J
1 Cal = 1 kcal = 1000 cal = 4.184 kJ
Chemistry-140 Lecture 13
Energy Transfer
The environment of a chemical reaction is separated into system and surroundings.
System: reactants, products, solvents, etc. in the reaction.
Surroundings: the universe, including the vessel.
2 H2(g) + O2(g) 2 H2O(l) + energy
Processes that lower a system's internal energy are spontaneous. Processes that increase a system's internal energy are nonspontaneous.
Chemistry-140 Lecture 13
Heat capacity of an object, C, is the amount of heat energy required to raise its temperature by 1°C,
The specific heat of a substance, C, is the amount of heat required to raise the temperature of a 1 g sample of the substance by 1°C,
Energy Transfer & Specific Heat
quantity of heat supplied
mass of object temperature change
C =
heat transferred = q = C x Theat transferred = q = C x T
Chemistry-140 Lecture 13
Example 6.2:
A lake has a surface area of 2.6 x 106 m2 and an
average depth of 10 m. What quantity of heat (kJ) must be
transferred to the lake to raise the temperature by 1 oC?
(Assume a density of 1.0 g/cm3 the lake water)
Energy Transfer & Specific Heat
Chemistry-140 Lecture 13
Answer:
Determine the mass of water and calculate the energy
required using the concept of specific heat.
Volume(H2O) = (2.6 x 106 m2)(10 m) = 2.6 x 107 m3 =
Mass(H2O) = (2.6 x 1013 cm3)(1.0 g/cm3) =
Since:
q = m x C x T = (2.6 x 1013 g)(4.184 J/g-K)(1.0 K)
=
2.6 x 1013 cm32.6 x 1013 cm3
2.6 x 1013 g2.6 x 1013 g
quantity of heat supplied
mass of object temperature change
C =
1.1 x 1011 kJ1.1 x 1011 kJ
Chemistry-140 Lecture 13
Energy and Changes of State
Heat transfer: heat is
transferred to a
substance with no
structural change
Phase changes or phase
transitions: a
substance's structure is
altered as in melting,
freezing, vaporizing,
and condensing.
Chemistry-140 Lecture 13
500 kJ
500 kJ
Ice, 2.0 kg, 0oC
Iron, 1.0 kg, 0oC
0oC
0oC
1100oC
Energy and Changes of State
Heat of fusion: the enthalpy change associated with melting
a substance, Hfus (kJ/mol).
Heat of vapourization; the enthalpy associated with
vapourizing a substance, Hvap (kJ/mol).
Heating a substance is endothermic. BUT... during a phase
transition the temperature remains constant
Chemistry-140 Lecture 13
Energy and Changes of State
The energy required (q) for the five possible processes is determined by the amount of sample.
For warming:
q = C x mass x T,
(specific heat, C, is different for each physical state)
For the phase transitions: q = (# of moles) x H.
Chemistry-140 Lecture 13
Energy and Changes of State
Question (similar to example 6.4):
Calculate the enthalpy change upon converting 1.00
mol of ice at -25 oC to water vapour (steam) at 125 oC
under a constant pressure of 1 atm. The specific heat of
ice, water and steam are 2.09 J/g-K, 4.18 J/g-K and 1.84
J/g-K, respectively. Also for water; Hfus = 6.01 kJ/mol
and Hvap = 40.67 kJ/mol.
Chemistry-140 Lecture 13
Energy and Changes of State
Answer:
Divide the total change into calculable segments and
calculate the enthalpy change for each segment. Sum to get
the total enthalpy change (Hess's Law).
Segment 1: Heat ice from -25 oC to 0 oC.
Segment 2: Convert ice to water at 0 oC.
Segment 3: Heat water from 0 oC to 100 oC.
Segment 4: Convert water to steam at 100 oC.
Segment 5: Heat steam from 100 oC to 125 oC.
Chemistry-140 Lecture 13
3: Water
2: Melting
1: Ice
5: Steam4: Boiling
Heat absorbed
Heat liberated
Heat added (kJ)
Energy and Changes of StateT
empe
ratu
re o C
Chemistry-140 Lecture 13
Energy and Changes of State
Segment 1: Heating the ice from -25 oC to 0 oC.
H1 = (1.00 mol)(18.0 g/mol)(2.09 J/g-K)(25 K) =
Segment 2: Convert ice to water at 0 oC.
H2 = (1.00 mol)(6.01 kJ/mol) =
Segment 3: Heating the water from 0 oC to 100 oC.
H3 = (1.00 mol)(18.0 g/mol)(4.18 J/g-K)(100 K) =
940 J 940 J
6.01 kJ6.01 kJ
7520 J7520 J
Chemistry-140 Lecture 13
Energy and Changes of State
Segment 4: Convert water to steam at 100oC.
H4 = (1.00 mol)(40.67 kJ/mol) =
Segment 5: Heating the steam from 100oC to 125oC.
H5 = (1.00 mol)(18.0 g/mol)(1.84 J/g-K)(25 K) =
Total Enthalpy Change:
H = H1 + H2 + H3 + H4+ H5
= (0.94 kJ) + (6.01 kJ) + (7.52 kJ) + (40.67 kJ) + (0.83 kJ)
=
40.67 kJ40.67 kJ
830 J830 J
55.97 kJ55.97 kJ
Chemistry-140 Lecture 13
Energy is neither created nor destroyed; only exchanged between system and surroundings.
Internal energy is the total energy of a system. Only changes in internal energy, E, can be measured.
E = Efinal - Einitial
(+ve = gain by system & -ve = loss to surroundings)
E = q + w
(heat added is positive & heat withdrawn is negative)
(work done is positive & work done by is negative)
The First Law of Thermodynamics
heatadded
internalenergy
workdone
Chemistry-140 Lecture 13
The First Law of Thermodynamics
Energy transferredfrom surroundings
to system
Energy transferredfrom system
to surroundings
Chemistry-140 Lecture 13
Heat Transfer
Question:
The H2(g) and O2(g) in a cylinder are ignited and the
system loses 550 J to its surroundings. The expanding gas
does 240 J of work on its surroundings. What is the change
in internal energy of the system?
Chemistry-140 Lecture 13
Heat Transfer
Answer:
Energy flows from the system, so q = -550 J.
Work is done by the system, so w = -240 J
Therefore:
E = q + w = (-550 J) + (-240 J) = -790 J-790 J
Chemistry-140 Lecture 13
Chemistry-140 Lecture 13
Term Test #1
Friday October 12th, 2001
6:30 P.M.
Term Test #1
Friday October 12th, 2001
6:30 P.M.
Chemistry-140 Lecture 13
A to H Education Gym
I to Z Rm 104 Odette Bldg
A to H Education Gym
I to Z Rm 104 Odette Bldg
Chemistry-140 Lecture 13
Covers Material From Chapters 4 - 6Covers Material From Chapters 4 - 6
Duration: 75 minutes
Contents: SIX “Problems”!!
Duration: 75 minutes
Contents: SIX “Problems”!!
A Periodic Table & ALL Required
Constants will be Supplied
A Periodic Table & ALL Required
Constants will be Supplied
Chapter 6: Enthalpy Changes for Chemical Reactions
Chapter Highlights
energy transfer, specific heat and heat transfer
heat of fusion & heat vapourization
exothermic & endothermic reactions
first law of thermodynamics
enthalpy changes
calorimetry
Hess’s Law
Chemistry-140 Lecture 14
Energy is neither created nor destroyed; only exchanged between system and surroundings.
Internal energy is the total energy of a system. Only changes in internal energy, E, can be measured.
E = Efinal - Einitial
(+ve = gain by system & -ve = loss to surroundings)
E = q + w
(heat added is positive & heat withdrawn is negative)
(work done is positive & work done by is negative)
The First Law of Thermodynamics
heatadded
internalenergy
workdone
Chemistry-140 Lecture 14
The First Law of Thermodynamics
Energy transferredfrom surroundings
to system
Energy transferredfrom system
to surroundings
Chemistry-140 Lecture 14
Heat & Enthalpy Changes
Enthalpy: amount of heat energy possessed by a substance. Enthalpy change corresponds to the heat change of the
system at constant pressure:
Endothermic reaction: heat is added (positive value).
Exothermic reaction: heat is withdrawn (negative value).
H = qp
H = Hfinal - Hinitial
H = qp
H = Hfinal - Hinitial
Chemistry-140 Lecture 14
Enthalpies of Reaction
Enthalpy change: the sum of the absolute enthalpies of the products minus the absolute enthalpies of the reactants.
2 H2(g) + O2(g) 2 H2O(l) + energy
2 H2(g) + O2(g) 2 H2O(l) H = -483.6 kJ
H = H(products) - H(reactants)H = H(products) - H(reactants)
Chemistry-140 Lecture 14
Enthalpies of Reaction
Enthalpy is an extensive property. The magnitude of H is proportional to the amount of reactant consumed.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
H = -802 kJ
2 CH4(g) + 4 O2(g) 2 CO2(g) + 4 H2O(g)
H = -1604 kJ
Chemistry-140 Lecture 14
Enthalpies of Reaction
The enthalpy change is equal in magnitude but opposite in sign to H for the reverse reaction.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
H = -802 kJ
CO2(g) + 2 H2O(g) CH4(g) + 2 O2(g)H = +802 kJ
Chemistry-140 Lecture 14
Enthalpies of Reaction
The enthalpy change depends on the states of the reactants and the products.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
H = -802 kJ
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)H = -890 kJ
2 H2O(g) 2 H2O(l)
H = -88 kJ
Chemistry-140 Lecture 14
Enthalpies of Reaction
Question:
Ammonium nitrate can decompose by the reaction:
NH4NO3(s) N2O(g) + 2 H2O(g) H = -37.0 kJ
Calculate the quantity of heat produced when 2.50 g of NH4NO3 decomposes at constant pressure.
Chemistry-140 Lecture 14
Enthalpies of Reaction
Answer:
We know the heat produced from 1 mole of NH4NO3.Calculate moles in 2.50 g and determine the heat produced by that amount.
heat = (2.50 g)
=
1 mol NH NO
80.06 g NH NO 4 3
4 3
- 37.0 kJ
1 mol NH NO 4 3
-1.16 kJ-1.16 kJ
Chemistry-140 Lecture 14
Calorimetry
Calorimetry: The measurement of heat flow. Measurements are made using a calorimeter.
Recall that: The heat capacity of an object, C, is the amount of heat energy required to raise its temperature by 1°C,
In constant-pressure calorimetry, the pressure remains constant because the apparatus is open to the atmosphere. At constant pressure, the heat change of the reaction is the enthalpy change, H.
heat transferred = q = C x Theat transferred = q = C x T
Chemistry-140 Lecture 14
Calorimetry
Question:
When 50 mL of 1.0 M HCl(aq) and 50 mL of 1.0 M
NaOH(aq) are mixed in a constant-pressure calorimeter,
the temperature increases from 21.0 to 27.5 oC. Calculate
H for this reaction assuming a specific heat of 4.18 J/g-oC
and a density of 1.0 g/mL for the final solution.
Chemistry-140 Lecture 14
Calorimetry
Answer:
Recall, C =
q = C x m x T
Since the total solution is 100 mL and the density is 1.0 g/mL,
then m = (100 mL) (1.0 g/mL) = 100 g
T = (27.5 - 21.0 oC) = 6.5 oC
and C = 4.18 J/g-oC
q
m x T
Chemistry-140 Lecture 14
Calorimetry
Answer:
then qp = (4.18 J/g-oC) (100 g) (6.5 oC)
= -2700 J =
Since moles in solution were (0.050 L)(1.0 M ) = 0.050 mol
H = =
-2.7 kJ-2.7 kJ
- 2.7 kJ
0.050 L
-54 kJ/mol-54 kJ/mol
Chemistry-140 Lecture 14
Bomb Calorimetry
In bomb calorimetry, the apparatus is sealed and the
experiment is a constant-volume process. The heat
change of the reaction is the internal energy change.
E = qevolved = -Ccalorimeter x TE = qevolved = -Ccalorimeter x T
Chemistry-140 Lecture 14
Bomb Calorimetry
Chemistry-140 Lecture 14
A Student “Bomb” Calorimetry
Thermometer
Cardboard orStyrofoam Lid
NestedStyrofoam Cups
Exothermic ReactionOccurs in Solution
Chemistry-140 Lecture 14
Bomb Calorimetry
Question:
Hydrazine, N2H4, and its derivatives are widely used
as rocket fuels.
N2H4(l) + O2(g) N2(g) + 2 H2O(g)
When 1.00 g of hydrazine is burned in a bomb
calorimeter, the temperature of the calorimeter increases
by 3.51 oC. If the calorimeter has a heat capacity of
5.510 kJ/oC, what is the quantity of heat evolved?
Chemistry-140 Lecture 14
Bomb Calorimetry
Answer:
Recall that:
qevolved = (-5.510 kJ/oC) x (3.51 oC)
qevolved = -19.3 kJ
Calculate this on a per mole basis:
qevolved = =
E = qevolved = -Ccalorimeter x TE = qevolved = -Ccalorimeter x T
-618 kJ/mol-618 kJ/mol- 19.3 kJ
1.00 g N H2 4
32.0 g N H
1 mol N H2 4
2 4
Chemistry-140 Lecture 14
Term Test #1
Friday October 12th, 2001
6:30 P.M.
Term Test #1
Friday October 12th, 2001
6:30 P.M.
Chemistry-140 Lecture 14
A to H Education Gym
I to Z Rm 104 Odette Bldg
A to H Education Gym
I to Z Rm 104 Odette Bldg
Chemistry-140 Lecture 14
Covers Material From Chapters 4 - 6Covers Material From Chapters 4 - 6
Duration: 75 minutes
Contents: SIX “Problems”!!
Duration: 75 minutes
Contents: SIX “Problems”!!
A Simple Periodic Table & ALL
Required Constants will be Supplied
A Simple Periodic Table & ALL
Required Constants will be Supplied
Chemistry-140 Lecture 14
Chapter 6: Enthalpy Changes for Chemical Reactions
Chapter Highlights
energy transfer, specific heat and heat transfer
heat of fusion & heat vapourization
exothermic & endothermic reactions
first law of thermodynamics
enthalpy changes
calorimetry
Hess’s Law
Chemistry-140 Lecture 16
Hess’s Law
If a reaction is carried out in a series of steps, H for the reaction will be equal to the sum of the enthalpy changes
for each step.
For example:
(1) CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)H = -802 kJ
(2) 2 H2O(g) 2 H2O(l)
H = -88 kJ
Chemistry-140 Lecture 16
Hess’s Law
Add: (1) + (2)
(1) + (2) CH4(g) + 2 O2(g) + 2 H2O(g)
CO2(g) + 2 H2O(l) + 2 H2O(g)
H = (-802 - 88) kJ = -890 kJ
Net Equation:
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)
H = -890 kJ
Chemistry-140 Lecture 16
Hess’s Law
Question:
Calculate H for the reaction:
2 C(s) + H2(g) C2H2(g)
Chemistry-140 Lecture 16
Hess’s LawQuestion:
Given the following:
(1) C2H2(g) + O2(g) 2 CO2(g) + H2O(l)
H = -1299.6 kJ
(2) C(s) + O2(g) CO2(g)
H = -393.5 kJ
(3) H2(g) + O2(g) H2O(l)
H = -285.9 kJ
1
2
5
2
Chemistry-140 Lecture 16
Hess’s Law
Answer:
Step 1: Target equation has C2H2(g) as a product so we reverse the first equation and change the sign on H:
-(1) 2 CO2(g) + H2O(l) C2H2(g) + O2(g)
H = +1299.6 kJ
5
2
Chemistry-140 Lecture 16
Hess’s Law
Step 2: Target equation has 2 C(s) as a reactant so multiply the second equation x 2:
2 x (2) 2 C(s) + 2 O2(g) 2 CO2(g)
H = -787.0 kJ
Chemistry-140 Lecture 16
Hess’s Law
Step 3: Target equation has H2(g) as a reactant so we keep the third equation as is:
(3) H2(g) + O2(g) H2O(l)
H = -285.9 kJ
1
2
Chemistry-140 Lecture 16
Step 4: Add the three equations from steps 1,2 and 3:
-(1) 2 CO2(g) + H2O(l) C2H2(g) + O2(g)
2 x (2) 2 C(s) + 2 O2(g) 2 CO2(g) (3) H2(g) + O2(g) H2O(l)
Total
2 CO2(g) + H2O(l) + 2 C(s) + 2 O2(g) + H2(g) + O2(g)
C2H2(g) + O2(g) + 2 CO2(g) + H2O(l)
5
2
1
2
1
25
2
Hess’s Law
Chemistry-140 Lecture 16
Hess’s Law
Net
2 C(s) + H2(g) C2H2(g)
-(1) H = +1299.6 kJ
2 x (2)H = -787.0 kJ
(3)H = -285.9 kJ
Total H = +226.7 kJH = +226.7 kJ
Chemistry-140 Lecture 16
Enthalpies of Formation
We can calculate enthalpy associated with many changes.
Vapourization (Hvap for converting liquids to gas)
Fusion (Hfus for melting solids)
Combustion (Hcom for combusting in oxygen)
Enthalpy of the reaction that forms a substance from its
constituent elements is called
the enthalpy of formation, Hf, or the heat of formation.2 C(s) + H2(g) C2H2(g)
Hf = +226.7 kJ
Chemistry-140 Lecture 16
Enthalpies of Formation
Standard state: the state that is most stable for the substance at the temperature of interest, usually 298 K and 1 atm.
C(s) diamond graphite fullerene
Standard enthalpy of formation, H°f, is the enthalpy of the reaction that forms the substance from its constituent
elements in their standard states.
2 C(s) + H2(g) C2H2(g)
Hof = +226.7 kJ
Chemistry-140 Lecture 16
Enthalpies of Formation
Enthalpies of reaction can be calculated by applying Hess's law to the enthalpies of formation of the participants:
H°rxn = [nH°f(products)] - [mH°f(reactants)]
Chemistry-140 Lecture 16
Enthalpies of Formation
The standard enthalpy of formation for ethanol, C2H5OH, is:
2 C(graphite) + 3 H2(g) + O2(g) C2H5OH(l)
H°f = -277.7 kJ
By definition, H°f = 0 for the standard state of an element.
Thus, H°f = 0 for C(graphite)
H°f = 0 for H2(g)
H°f = 0 for O2(g)
1
2
Chemistry-140 Lecture 16
Enthalpies of FormationQuestion:
Calculate H for the combustion of propane, C3H8(g) using tables of standard enthalpies of formation.
TABLE 6.2 Selected StandardEnthalpies of Formation
Substance Hof (kJ/mol)
C3H8(g) -103.8
CO2(g) -393.5
H2O(l) -285.8
Chemistry-140 Lecture 16
Enthalpies of Formation Answer:
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
H°rxn = [nH°f(products)] - [mH°f(reactants)]
H°rxn = [3H°f(CO2) + 4H°f(H2O)] - [H°f(C3H8) + 5H°f(O2)]
= [3(-393.5 kJ/mol) + 4(-285.8 kJ/mol)] -
[(-103.8 kJ/mol) + 5(0.0 kJ/mol)]
= (-2324 kJ) - (-103.8) =
-2220 kJ/mol -2220 kJ/mol
Chemistry-140 Lecture 16
Enthalpies of Formation
Question:
The standard enthalpy change for the reaction
CaCO3(s) CaO(s) + CO2(g) is 178.1
kJ.
Calculate the standard enthalpy of formation of CaCO3(s) from the standard enthalpies of formation of CaO(s)
and CO2(g); -635.1 kJ and 393.5 kJ respectively.
Chemistry-140 Lecture 16
Enthalpies of Formation Answer:
H°rxn = [nH°f(products)] - [mH°f(reactants)]
H°rxn = [H°f(CaO) + H°f(CO2)] - [H°f(CaCO3)]
178.1 kJ = [(-635.1 kJ/mol) + (-393.5 kJ/mol)] -
[H°f(CaCO3)]
[H°f(CaCO3)] = -635.1 kJ - 393.5 kJ - 178.1 kJ
[H°f(CaCO3)] =
-1206.7 kJ/mol-1206.7 kJ/mol
Chemistry-140 Lecture 16
Textbook Questions From Chapter #6Specific Heat: 15, 18, 24
Changes of State: 30
Enthalpy: 34, 38
Hess’s Law: 40, 42, 44, 48, 54
Calorimetry: 62
General & Conceptual 70, 78
Textbook Questions From Chapter #6Specific Heat: 15, 18, 24
Changes of State: 30
Enthalpy: 34, 38
Hess’s Law: 40, 42, 44, 48, 54
Calorimetry: 62
General & Conceptual 70, 78
Chemistry-140 Lecture 14