Brief review of

1
Chapter 5

• Brief review of
• System, surroundings, universe
• Energy, work, heat
• Endothermic/exothermic
• Enthalpies of formation
• Heat capacity
• Calorimetry
• Foods and fuels

2
System and Surroundings

• The system includes the molecules we want to
  study (here, the hydrogen and oxygen molecules).
• The surroundings are everything else (here, the
  cylinder and piston).

3
Heat, Work, Energy

• Heat (q) is a form of energy that moves when
  objects of different temperatures contact
• Work is energy expended in moving an object
  against an opposing force
• DE q w
• E is a state function q and w are not

4
Changes in Internal Energy

• When energy is exchanged between the system and
  the surroundings, it is exchanged as either heat
  (q) or work (w).
• That is, ?E q w.

5
?E, q, w, and Their Signs
6
First Law of Thermodynamics

• Energy is neither created nor destroyed.
• In other words, the total energy of the universe
  is a constant if the system loses energy, it
  must be gained by the surroundings, and vice
  versa.

Use Fig. 5.5
7
Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the
  surroundings, the process is endothermic.

H, enthalpy, is qp (heat flow at
constant pressure)
8
Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the
  surroundings, the process is endothermic.
• When heat is released by the system to the
  surroundings, the process is exothermic.

H, enthalpy, is qp (heat flow at
constant pressure)
9
Guidelines for Using DH Values
intensive property value doesnt depend on the
amount of material present
a property whose value depends on the amount of
material involved

• Enthalpy is an extensive property. The size of
  DH will depend on the amount of matter involved
  in the chemical reaction
• For
• 2H2(g) O2(g) ? 2H2O(g)
• If 2 mol H2(g) are combusted, DHrxn -483.6 kJ
• If 4 mol H2(g) are combusted, DHrxn -967.2 kJ

enthalpy of reaction
10
Guidelines for Using DH Values

• The enthalpy change for a reaction is equal in
  magnitude, but opposite in sign, to DH for the
  reverse reaction
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l) DH -890
  kJ
• CO2(g) 2H2O(l) ? CH4(g) 2O2(g) DH 890 kJ

11
Guidelines for Using DH Values

• The enthalpy change for a reaction depends on the
  state of the reactants and products
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l) DH -890
  kJ
• CH4(g) 2O2(g) ? CO2(g) 2H2O(g) DH -802
  kJ
• (because 2H2O(l) ? 2H2O(g) DH 88 kJ)

12
Hesss Law
Because DH is a state function, its value is
independent of its history (i.e. it is a state
function)

• Overall enthalpy change is independent of the
  number of steps or the particular nature of the
  path by which the reaction is carried out
• Useful for calculating enthalpy changes for
  reactions without actually carrying out the
  experiment

13
Hesss Law

• For example, lets say we want to determine DH
  for the reaction
• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)
• And we know the following information
• H2(g) F2(g) ? 2HF(g) DH -537 kJ
• C(s) 2F2(g) ? CF4(g) DH -680 kJ
• 2C(s) 2H2(g) ? C2H4(g) DH 52.3 kJ

Using the guidelines for DH values, we should be
able to rearrange the above equations, add them
together to create the top (blue) equation, and
get DH for this reaction.
14
Hesss Law

• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)
• H2(g) F2(g) ? 2HF(g) DH -537 kJ
• C(s) 2F2(g) ? CF4(g) DH -680 kJ
• 2C(s) 2H2(g) ? C2H4(g) DH 52.3 kJ

multiply by 2
flip
Each of the three lower equations involves at
least one of the molecules in the top
equation Rearrange the three red equations
(multiplying by coefficients when necessary
so that they can be added together to yield the
blue equation
15
Hesss Law

• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)
• 2H2(g) 2F2(g) ? 4HF(g) DH 2(-537 kJ)
• 2C(s) 4F2(g) ? 2CF4(g) DH 2(-680 kJ)
• C2H4(g) ? 2C(s) 2H2(g) DH -52.3 kJ

16
Hesss Law

• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)
• 2H2(g) 2F2(g) ? 4HF(g) DH 2(-537 kJ)
• 2C(s) 4F2(g) ? 2CF4(g) DH 2(-680 kJ)
• C2H4(g) ? 2C(s) 2H2(g) DH -52.3 kJ
• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)

Adding, get DH -2.49x103 kJ
17
Enthalpies of Formation

• An enthalpy of formation, ?Hf, is defined as the
  enthalpy change for the reaction in which one
  mole of compound is made from its constituent
  elements in their elemental forms.
• C(graphite) O2(g) ? CO2(g)
• 2C(graphite) 2H2(g) O2(g) ? HC2H3O2(l)
• Mg(s) ½O2(g) ? MgO(s)

these are the elemental (natural) forms of carbon
and oxygen
qP DHf
DHf defined per one mole of product
18
Standard Enthalpies of Formation

• Standard enthalpies of formation, ?Hof, are
  enthalpies of formation measured when all
  reactants and products are present in their
  standard states at (25C (298 K) and 1.00 atm
  pressure (for any gases present).
• The symbol o indicates standard conditions

19
Enthalpies of formation

• Thus, the enthalpy change that accompanies the
  formation of one mole of a compound from its
  constituent elements, with all substances in
  their standard states, is DHof
• e.g.
• 2C(graphite) 3H2(g) ½O2(g) ? C2H5OH(l)
  DHof -277.7 kJ
• This reaction represents DHofC2H5OH(l)
• Notice that all the reactants are pure elements
  (this reaction shows the formation of 1 mol of
  ethanol from the elements that make up an ethanol
  molecule, C, H, and O. These elements are
  written as they appear in nature)

20
Elements in their standard states

• Some examples
• Carbon C(s) (graphite)
• Oxygen O2(g)
• Hydrogen H2(g)
• Sodium Na(s)
• Chlorine Cl2(g)
• Bromine Br2(l)
• Iodine I2(s)
• Phosphorus P4(s)
• Iron Fe(s)
• Trés important Enthalpy of formation of any
  element in its standard state is zero

e.g. DHof of H2(g) would correspond to the
reaction H2(g) ? H2(g) DHof 0kJ/mol
21
Enthalpies of formation

• Which of the following represent standard
  enthalpy of formation reactions?
• 2K(l) Cl2(g) ? 2KCl(s)
• C6H12O6(s) ? 6C(diamond) 6H2(g) 3O2(g)
• 2Na(s) ½O2(g) ? Na2O(s)

(remember DHf enthalpy change that accompanies
the formation of one mole of a compound from its
constituent elements)
22
Enthalpy Changes for Chemical Reactions

• We calculated enthalpy changes for chemical
  reactions using Hesss Law in the first semester.
• Using enthalpies of formation, enthalpy changes
  for chemical reactions may also be calculated

23
Example

• For the question we solved earlier using Hesss
  Law
• C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)
• If we knew the formation enthalpies for the
  reactants and products involved, we could get
  DHrxn as follows
• DHof C2H4(g) 52.30 kJ/mol
• DHof F2(g) 0 kJ/mol
• DHof CF4(g) -679.9 kJ/mol
• DHof HF(g) -268.61 kJ/mol

Technically only valid under standard conditions
24
Heat Capacity and Specific Heat

• The amount of energy required to raise the
  temperature of a substance by 1 K (1?C) is its
  heat capacity

Calorimetry Heat evolved/consumed by eaction
causes heat change in surroundings (measurable)
observe a temperature change
Note 100oC ? 101oC DT 1oC 373.15K ? 374.15K
DT 1K DT same for celsius and Kelvin
25
Specific Heat and Molar Heat Capacity

• We define specific heat capacity (or simply
  specific heat) as the amount of energy required
  to raise the temperature of 1 g of a substance by
  1 K (or 1oC)
• The molar heat capacity is the heat required to
  raise the temperature of one mole of substance by
  1 K

26
Problem

• The specific heat of iron metal is 0.450 J/g.K.
  How many J of heat are required to raise the
  temperature of a 1.05-kg block of iron from
  25.0oC to 88.5oC?

again, just to show
27
Problem

• What is the heat capacity of 185 g of liquid
  water? (s 4.184 J/g.K)
• What is the molar heat capacity of water?

Have the specific heat of water (4.184 J/g.K).
Want to know how much heat is needed to raise the
temperature of 1 mol of water (18.0152g) by 1K
28
Calorimetry

• Since we cannot know the exact enthalpy of the
  reactants and products, we measure ?H through
  calorimetry, the measurement of heat flow.

29
Constant Pressure Calorimetry

• By carrying out a reaction in aqueous solution
  in a simple calorimeter such as this one, one can
  indirectly measure the heat change for the system
  by measuring the heat change for the water in the
  calorimeter.

this is an example of an open system
30
Constant Pressure Calorimetry

• Because the specific heat for water is well
  known (4.184 J/g.K), we can measure ?H for the
  reaction with this equation
• Heat that flows into solution causes a
  temperature increase qrxn-qsoln
• qrxn -msDT

m in this Equation is the mass of the solution
rxn system soln surroundings
For an exothermic reaction, qrxn will be
negative For an endothermic reaction, qrxn will
be positive
31
Problem

• In a coffee cup calorimetry experiment, two
  aqueous solutions are combined (50.0 mL of 0.100
  M AgNO3 and 50.0 mL of 0.100 M HCl). The
  following reaction occurs
• AgNO3(aq) HCl(aq) ? AgCl(s) HNO3(aq)
• The temperature is observed to increase from
  22.20oC to 23.11oC in the experiment. Calculate
  DH for this reaction in kJ/mol AgNO3 (assuming a
  total solution mass of 100.0g and a specific heat
  of 4.184 J/g.K)

32
Problem

• Use the equation for the constant pressure
  calorimeter to solve for qrxn
• and then divide by the number of moles of AgNO3
  (what the question asks you to do)

DT 0.91oC
Remember C n/V ? n CV For AgNO3 solution n
(0.100M)(0.05000L) 0.00500 mol AgNO3
33
Bomb Calorimetry
Constant Volume Calorimetry

• Reactions can be carried out in a sealed bomb,
  such as this one, and measure the heat absorbed
  by the water.

This is an example of a closed system
Ccal is the heat capacity of the
calorimeter (often given or needs to be
determined separately)
34
Bomb Calorimetry

• Because the volume in the bomb calorimeter is
  constant, what is measured is really the change
  in internal energy, ?E, not ?H.
• For most reactions, the difference is very small.

35
A bomb calorimeter problem
In this problem, you are given Ccal

• A 2.200-g sample of quinone was burned in a bomb
  calorimeter, whose heat capacity is 7.854 kJ/oC.
  The temperature of the calorimeter changes from
  23.44oC to 30.57oC. What is the heat of
  combustion per gram (and mole) of quinone?

Per gram
Per mole
36
Calibration of the Bomb

• Bomb calorimeter equations involve the Ccal term,
  which is determined through an experiment in
  which a quantity of a substance that releases a
  known amount of heat per gram is combusted in the
  calorimeter.
• Example, combustion of 1.000g benzoic acid is
  known to release 26.38 kJ of heat. If combustion
  of 1.000g benzoic acid causes the temperature in
  a bomb calorimeter to rise by 5.75oC,

Ccal is always positive (its the amount of heat
energy needed to raise the calorimeters temperatu
re by 1oC)
37
2-Part Bomb Calorimeter Problem

• 2.50g of benzoic acid is combusted in a bomb
  calorimeter, causing the temperature to increase
  by 8.35oC. (benzoic acid is known to release
  26.38 kJ/g).
• In a separate experiment, 2.00g of glucose
  (C6H12O6) is combusted using the same
  calorimeter, producing a temperature increase of
  3.95oC. What is qrxn for the combustion of
  glucose, in kJ/mol?

Use this info to determine Ccal
After finding Ccal, use it with this info to find
qrxn in kJ/mol
38
Solution

• Must solve the problem in two basic parts
• Ccal determination
• qrxn for glucose combustion

Remember that combustion of 1.000g benzoic acid
releases 26.38 kJ of heat how much heat does
combustion of 2.50g benzoic acid create?
39
First Part of Solution (Ccal)

• Using the heat evolved from the combustion of
  2.50g benzoic acid, we can determine Ccal

Ccal is always going to be positive
40
Second Part

• Now that the calibration constant is known, can
  determine the heat evolved from the combustion of
  2.00g glucose

Part II 2.00g of glucose is combusted using
the same calorimeter, producing a
temperature increase of 3.95oC
To find the heat in kJ/mol glucose
41
Food and Fuels

• Our body requires energy for
• maintenance of body temperature
• muscle contraction
• tissue construction/repair
• We are able to get energy from the bodys
  breakdown of carbohydrates, fats, and proteins

42
Carbohydrates

• Carbohydrates (e.g. starch) are transformed into
  sugars in the intestines (e.g. glucose, C6H12O6),
  which is absorbed into the blood and transported
  to the cells
• C6H12O6 6O2 ? 6CO2 6H2O
• The breakdown of carbohydrates is fast, so this
  energy gets used quickly (carbohydrates tend not
  to get stored by the body)
• Average fuel value of carbohydrates is 17kJ/g

43
Fats

• Fats also break down into CO2 and H2O. For
  example, the combustion of C57H110O6
• 2C57H110O6 163O2 ? 114CO2 110H2O
• Fats are water-insoluble, and so they are easily
  stored.
• Also, they have a higher average fuel value than
  carbohydrates (38kJ/g), and so are well-suited as
  a bodys energy reserve.

44
Proteins

• The combustion of proteins in a bomb calorimeter
  yields N-containing products (released as NO2)
• In the body, nitrogen is released mainly as urea
  (NH2)2CO
• Proteins are mainly used as building materials
  for organ walls, skin, hair, muscle, etc.
• The average fuel value for proteins is 17kJ/g

45
Energy in Foods

• Most of the fuel in the food we eat comes from
  carbohydrates and fats.

46
Fuels

• The vast majority of the energy consumed in this
  country comes from fossil fuels.

The greater the percentage of C and H in a fuel,
the higher its fuel value