Transcript Chapter 06 - University of Windsor

Chemistry-140
Lecture 13
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.
1
2
EK = mv2 (m = mass, v = velocity)
 Potential energy (EP): energy stored by an object
due to its relative position.
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
Energy Transfer & Specific Heat
 Heat capacity of an object, C, is the amount of heat energy
required to raise its temperature by 1°C,
heat transferred = q = C x DT
 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,


quantity of heat supplied

C= 
 mass of object  temperature change  
Chemistry-140
Lecture 13
Energy Transfer & Specific Heat
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)
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 = 2.6 x 1013 cm3
Mass(H2O) = (2.6 x 1013 cm3)(1.0 g/cm3) = 2.6 x 1013 g
Since:


quantity of heat supplied

C= 
 mass of object  temperature change  
q = m x C x DT = (2.6 x 1013 g)(4.184 J/g-K)(1.0 K)
= 1.1 x 1011 kJ
Chemistry-140
Lecture 13
Energy and Changes of State
 Heat transfer: heat is
500 kJ
transferred to a
substance with no
structural change
1100oC
Iron, 1.0 kg, 0oC
 Phase changes or phase
transitions: a
500 kJ
substance's structure is
altered as in melting,
freezing, vaporizing,
and condensing.
Ice, 2.0 kg, 0oC
0oC
0oC
Chemistry-140
Lecture 13
Energy and Changes of State
 Heat of fusion: the enthalpy change associated with melting
a substance, DHfus (kJ/mol).
 Heat of vapourization; the enthalpy associated with
vapourizing a substance, DHvap (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 DT,
(specific heat, C, is different for each physical state)
 For the phase transitions: q = (# of moles) x DH.
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; DHfus = 6.01 kJ/mol and DHvap
= 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
Energy and Changes of State
Temperature oC
Heat liberated
5: Steam
4: Boiling
Heat absorbed
3: Water
2: Melting
1: Ice
Heat added (kJ)
Chemistry-140
Lecture 13
Energy and Changes of State
Segment 1: Heating the ice from -25 oC to 0 oC.
DH1 = (1.00 mol)(18.0 g/mol)(2.09 J/g-K)(25 K) = 940 J
Segment 2: Convert ice to water at 0 oC.
DH2 = (1.00 mol)(6.01 kJ/mol) = 6.01 kJ
Segment 3: Heating the water from 0 oC to 100 oC.
DH3 = (1.00 mol)(18.0 g/mol)(4.18 J/g-K)(100 K) = 7520 J
Chemistry-140
Lecture 13
Energy and Changes of State
Segment 4: Convert water to steam at 100oC.
DH4 = (1.00 mol)(40.67 kJ/mol) = 40.67 kJ
Segment 5: Heating the steam from 100oC to 125oC.
DH5 = (1.00 mol)(18.0 g/mol)(1.84 J/g-K)(25 K) = 830 J
Total Enthalpy Change:
DH = DH1 + DH2 + DH3 + DH4+ DH5
= (0.94 kJ) + (6.01 kJ) + (7.52 kJ) + (40.67 kJ) + (0.83 kJ)
= 55.97 kJ
Chemistry-140
Lecture 13
The First Law of Thermodynamics
 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, DE, can be measured.
DE = Efinal - Einitial
(+ve = gain by system & -ve = loss to surroundings)
DE =
q + w
work
internal
heat
energy
added done
(heat added is positive & heat withdrawn is negative)
(work done is positive & work done by is negative)
Chemistry-140
Lecture 13
The First Law of Thermodynamics
Energy transferred
from surroundings
to system
Energy transferred
from 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:
DE = q + w = (-550 J) + (-240 J) = -790 J
Chemistry-140
Lecture 13
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
Chemistry-140
Lecture 13
Duration: 75 minutes
Contents: SIX “Problems”!!
Covers Material From Chapters 4 - 6
A 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 14
The First Law of Thermodynamics
 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, DE, can be measured.
DE = Efinal - Einitial
(+ve = gain by system & -ve = loss to surroundings)
DE =
q + w
work
internal
heat
energy
added done
(heat added is positive & heat withdrawn is negative)
(work done is positive & work done by is negative)
Chemistry-140
Lecture 14
The First Law of Thermodynamics
Energy transferred
from surroundings
to system
Energy transferred
from 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:
DH = qp
DH = Hfinal - Hinitial
 Endothermic reaction: heat is added (positive value).
 Exothermic reaction: heat is withdrawn (negative value).
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.
DH = H(products) - H(reactants)
2 H2(g) + O2(g)
2 H2(g) + O2(g)
2 H2O(l) + energy
2 H2O(l)
DH = -483.6 kJ
Chemistry-140
Lecture 14
Enthalpies of Reaction
 Enthalpy is an extensive property. The magnitude of DH is
proportional to the amount of reactant consumed.
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
DH = -802 kJ
2 CH4(g) + 4 O2(g)
2 CO2(g) + 4 H2O(g)
DH = -1604 kJ
Chemistry-140
Lecture 14
Enthalpies of Reaction
 The enthalpy change is equal in magnitude but opposite in
sign to DH for the reverse reaction.
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
DH = -802 kJ
CO2(g) + 2 H2O(g)
CH4(g) + 2 O2(g)
DH = +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)
DH = -802 kJ
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(l)
DH = -890 kJ
2 H2O(g)
2 H2O(l)
DH = -88 kJ
Chemistry-140
Lecture 14
Enthalpies of Reaction
Question:
Ammonium nitrate can decompose by the reaction:
NH4NO3(s)
N2O(g) + 2 H2O(g)
DH = -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.
 1 mol NH 4 NO3
heat = (2.50 g) 
 80.06 g NH 4 NO3
= -1.16 kJ


- 37.0 kJ


 1 mol NH 4 NO3 
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,
heat transferred = q = C x DT
 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, DH.
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
DH 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 


 m x DT 
q = C x m x DT
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
DT = (27.5 - 21.0 oC) = 6.5 oC
and
C = 4.18 J/g-oC
Chemistry-140
Lecture 14
Calorimetry
Answer:
then
qp = (4.18 J/g-oC) (100 g) (6.5 oC)
= -2700 J = -2.7 kJ
Since moles in solution were (0.050 L)(1.0 M ) = 0.050 mol
 - 2.7 kJ 
DH = 
 = -54 kJ/mol
 0.050 L 
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.
DE = qevolved = -Ccalorimeter x DT
Chemistry-140
Lecture 14
Bomb Calorimetry
Chemistry-140
Lecture 14
A Student “Bomb” Calorimetry
Thermometer
Cardboard or
Styrofoam Lid
Nested
Styrofoam Cups
Exothermic Reaction
Occurs 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:
DE = qevolved = -Ccalorimeter x DT
qevolved = (-5.510 kJ/oC) x (3.51 oC)
qevolved = -19.3 kJ
Calculate this on a per mole basis:
qevolved
 - 19.3 kJ   32.0 g N 2 H 4 
 = -618 kJ/mol
=

 1.00 g N 2 H 4   1 mol N 2 H 4 
Chemistry-140
Lecture 14
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
Chemistry-140
Lecture 14
Duration: 75 minutes
Contents: SIX “Problems”!!
Covers Material From Chapters 4 - 6
A Simple Periodic Table & ALL
Required Constants will be Supplied
Chemistry-140
Lecture 16
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, DH 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)
DH = -802 kJ
(2)
2 H2O(g)
2 H2O(l)
DH = -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)
DH = (-802 - 88) kJ = -890 kJ
Net Equation:
CH4(g) +
2 O2(g)
CO2(g) + 2 H2O(l)
DH = -890 kJ
Chemistry-140
Lecture 16
Hess’s Law
Question:
Calculate DH for the reaction:
2 C(s)
+
H2(g)
C2H2(g)
Chemistry-140
Lecture 16
Hess’s Law
Question:
Given the following:
(1)
C2H2(g) +
5
O2(g)
2
2 CO2(g) + H2O(l)
DH = -1299.6 kJ
(2)
C(s) + O2(g)
CO2(g)
DH = -393.5 kJ
(3)
1
H2(g) +
O2(g)
2
H2O(l)
DH = -285.9 kJ
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 DH:
-(1)
2 CO2(g) + H2O(l)
C2H2(g) +
5
O2(g)
2
DH = +1299.6 kJ
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)
DH = -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)
1
H2(g) +
O (g)
2 2
H2O(l)
DH = -285.9 kJ
Chemistry-140
Lecture 16
Hess’s Law
Step 4: Add the three equations from steps 1,2 and 3:
-(1)
2 x (2)
(3)
2 CO2(g) + H2O(l)
2 C(s) + 2 O2(g)
H2(g) +
1
O (g)
2 2
C2H2(g) +
5
O2(g)
2
2 CO2(g)
H2O(l)
Total
2 CO2(g) + H2O(l) + 2 C(s) + 2 O2(g) + H2(g) +
1
O2(g)
2
5
C2H2(g) +
O2(g) + 2 CO2(g) + H2O(l)
2
Chemistry-140
Lecture 16
Hess’s Law
Net
2 C(s) + H2(g)
C2H2(g)
-(1)
2 x (2)
(3)
DH = +1299.6 kJ
DH = -787.0 kJ
DH = -285.9 kJ
Total
DH = +226.7 kJ
Chemistry-140
Lecture 16
Enthalpies of Formation
We can calculate enthalpy associated with many changes.
Vapourization
(DHvap for converting liquids to gas)
Fusion
(DHfus for melting solids)
Combustion
(DHcom for combusting in oxygen)
 Enthalpy of the reaction that forms a substance from its
constituent elements is called
the enthalpy of formation, DHf, or the heat of formation.
2 C(s)
+
H2(g)
C2H2(g)
DHf = +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, DH°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)
DHof = +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:
DH°rxn =
S [nDH°f(products)] - S [mDH°f(reactants)]
Chemistry-140
Lecture 16
Enthalpies of Formation
The standard enthalpy of formation for ethanol, C2H5OH, is:
1
2 C(graphite) + 3 H2(g) + O2(g)
2
C2H5OH(l)
DH°f = -277.7 kJ
 By definition, DH°f = 0 for the standard state of an element.
Thus,
DH°f = 0 for C(graphite)
DH°f = 0 for H2(g)
DH°f = 0 for O2(g)
Chemistry-140
Lecture 16
Enthalpies of Formation
Question:
Calculate DH for the combustion of propane, C3H8(g)
using tables of standard enthalpies of formation.
TABLE 6.2 Selected Standard
Enthalpies of Formation
o
Substance
H
D f (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)
DH°rxn =
3 CO2(g) + 4 H2O(l)
S [nDH°f(products)] - S [mDH°f(reactants)]
DH°rxn = [3DH°f(CO2) + 4DH°f(H2O)] - [DH°f(C3H8) + 5DH°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
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:
DH°rxn =
S [nDH°f(products)] - S [mDH°f(reactants)]
DH°rxn = [DH°f(CaO) + DH°f(CO2)] - [DH°f(CaCO3)]
178.1 kJ = [(-635.1 kJ/mol) + (-393.5 kJ/mol)] - [DH°f(CaCO3)]
[DH°f(CaCO3)] = -635.1 kJ - 393.5 kJ - 178.1 kJ
[DH°f(CaCO3)] = -1206.7 kJ/mol
Chemistry-140
Lecture 14
Textbook Questions From Chapter #6
Specific 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