# Chapter 5 sem 2 by yk6k8j2

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• pg 1
```									                    Chapter 5
• Brief review of
– System, surroundings, universe
– Energy, work, heat
– Endothermic/exothermic

•   Enthalpies of formation
•   Heat capacity
•   Calorimetry
•   Foods and fuels
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).
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
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, DE = q + w.
DE, q, w, and Their Signs
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
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)
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)
Guidelines for Using DH Values
intensive property: value doesn’t depend         a property whose value depends
on the amount of material present               on the amount of material involved

1.    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)      “enthalpy of reaction”

If 2 mol H2(g) are combusted, DHrxn = -483.6 kJ
If 4 mol H2(g) are combusted, DHrxn = -967.2 kJ
Guidelines for Using DH Values
2. 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
Guidelines for Using DH Values

3. 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)
Hess’s 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
Hess’s Law
• For example, let’s 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.
Hess’s Law
C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)
multiply by 2

H2(g) + F2(g)  2HF(g)                 DH = -537 kJ
C(s) + 2F2(g)  CF4(g)                 DH = -680 kJ
flip
2C(s) + 2H2(g)  C2H4(g)               DH = +52.3 kJ
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
Hess’s 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
Hess’s 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
Enthalpies of Formation
An enthalpy of formation, DHf, 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.
these are the elemental (natural) forms of carbon and oxygen

C(graphite) + O2(g)  CO2(g)
2C(graphite) + 2H2(g) + O2(g)  HC2H3O2(l)                       qP = DHf

Mg(s) + ½O2(g)  MgO(s)
DHf defined per one mole of product
Standard Enthalpies of Formation
Standard enthalpies of formation, DHof, are enthalpies of
formation measured when all reactants and products are
present in their standard states at (25°C (298 K) and 1.00 atm
pressure (for any gases present).
The symbol “o” indicates standard conditions
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 DHof[C2H5OH(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)
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
Enthalpies of formation
• Which of the following represent standard
enthalpy of formation reactions?
a)2K(l) + Cl2(g)  2KCl(s)
b)C6H12O6(s)  6C(diamond) + 6H2(g) + 3O2(g)
c)2Na(s) + ½O2(g)  Na2O(s)

(remember: DHf: enthalpy change that accompanies the formation of one mole
of a compound from its constituent elements)
Enthalpy Changes for Chemical
Reactions
We calculated enthalpy changes for chemical
reactions using Hess’s Law in the first semester.
Using enthalpies of formation, enthalpy changes
for chemical reactions may also be calculated

DH rxn    DH f products   DH f reactants
Example
• For the question we solved earlier using Hess’s
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
DH rxn    DH f products   DH f reactants
DH rxn   2DH f CF4 ( g )  4DH f  HF( g )   DH f C2 H 4( g )  6DH f  F2( g ) 
DH rxn   2(679.9kJ / mol )  4(268.61kJ / mol )   (52.30kJ / mol )  6(0kJ / mol ) 
DH rxn   2486.54kJ / mol  2.49 x103 kJ / mol
*Technically only valid under standard conditions
Heat Capacity and Specific Heat
• The amount of energy required                  Calorimetry:
Heat evolved/consumed by
to raise the temperature of a           eaction causes heat change
substance by 1 K (1C) is its         in surroundings (measurable) –
observe a temperature change
heat capacity
Note:  100oC  101oC; DT = 1oC
373.15K  374.15K; DT = 1K
DT same for celsius and Kelvin
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)
(quantity _ of _ heat _ transferred )
specific _ heat 
( grams _ of _ subs tan ce)(temperature _ change)
q
s
(m)(DT )

• The molar heat capacity is the heat
required to raise the temperature of one
mole of substance by 1 K
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?

q                               again, just to show…
s           q  msDT
(m)(DT )                       DT  Tf  Ti
DT  88.5o C  25.0o C  63.5o C
       J 
q  1050g  0.450 . 63.5K 

      gK  
DT  361.65K  298.15K  63.5K

q  30003J  3.00 x10 4 J
Problem
• What is the heat capacity of 185 g of liquid
water? (s = 4.184 J/g.K)
       J 
 4.184 . 185g   774
J
      gK
                       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

       J  18.0152g _ H 2O             J
 4.184 . 
                              75.375
      g K  1mol _ H 2O 
                          mol. K
Calorimetry
Since we cannot
know the exact
enthalpy of the
reactants and
products, we measure
DH through
calorimetry, the
measurement of heat
flow.
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
this is an example
of an “open system”   change for the system
by measuring the heat
change for the water in
the calorimeter.
Constant Pressure Calorimetry

Because the specific heat
for water is well known
(4.184 J/g.K), we can
measure DH for the reaction
with this equation:

Heat that flows into solution
causes a temperature
increase: qrxn=-qsol’n     “m” in this
Equation is
the mass of
the solution
qrxn = -msDT

For an exothermic reaction, qrxn will be negative                      rxn = system
For an endothermic reaction, qrxn will be positive                 sol’n = surroundings
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)
qrxn  msDT
Problem
• Use the equation for the constant pressure
calorimeter to solve for qrxn:           DT = 0.91oC
q rxn  msDT
       J 
q rxn


 100g  4.184 o  23.11o C  22.2 o C   
      g. C 

q rxn    380.744J  0.380 kJ

• …and then divide by the number of moles of
AgNO3 (what the question asks you to do)

 0.38 kJ
DH                      76 .1 kJ / mol _ AgNO3
0.0050 mol _ AgNO3

Remember C = n/V  n = CV
For AgNO3 solution: n = (0.100M)(0.05000L) = 0.00500 mol AgNO3
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”

q  C cal DT

Ccal is the heat capacity of the calorimeter
(often given or needs to be determined separately)
Bomb Calorimetry

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

quinone?
Per gram
O
Per mole
q rxn  C cal DT                  56.0kJ
 25.4 54
kJ
2.200g             g
kJ
q rxn    (7.854 o )(7.13o C )
C                         kJ  108.098 g                       kJ
  25.4 54                        2.751 x103
q rxn    56.0kJ                 
           g  1mol _ C 6 H 4 O2 
                                 mol
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,
q rxn  C cal DT
Ccal is always positive (it’s the
q rxn    (26 .38 kJ )          kJ
amount of heat energy needed        C cal                          4.58 7 o
to raise the calorimeter’s                     DT         5.75 o C               C
temperature by 1oC)
2-Part Bomb Calorimeter Problem

Use this        2.50g of benzoic acid is combusted in a
info to        bomb calorimeter, causing the temperature
determine
Ccal
to increase by 8.35oC. (benzoic acid is
known to release 26.38 kJ/g).
In a separate experiment, 2.00g of glucose
After finding
Ccal, use it    (C6H12O6) is combusted using the same
with this info   calorimeter, producing a temperature
to find qrxn    increase of 3.95oC. What is qrxn for the
in kJ/mol
combustion of glucose, in kJ/mol?

q  Ccal DT
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?

        kJ 
  26.38 2.50g   65.95 kJ

        g 
First Part of Solution (Ccal)
• Using the heat evolved from the
combustion of 2.50g benzoic acid, we can
determine Ccal:

qrxn
Ccal  
DT
(65.95 kJ )   Ccal is always going
Ccal                       to be positive
8.35o C
kJ
Ccal    7.898 o
C
Second Part
• Now that the calibration constant is known,
can determine the heat evolved from the
combustion of 2.00g glucose
q rxn  C cal DT                       Part II: “2.00g of glucose
is combusted using the
kJ
q rxn  (7.898   o
)(3.95o C )       same calorimeter,
C                   producing a temperature
increase of 3.95oC”
q rxn    31.19 kJ

To find the heat in kJ/mol glucose:

 31 .19 kJ  180 .0 g                  kJ
qrxn                            2.80 7 x10 3
2g        1mol                      mol
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 body’s
breakdown of carbohydrates, fats, and
proteins
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
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 body’s energy reserve.
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
Energy in Foods
Most of the fuel in the
food we eat comes
from carbohydrates
and fats.
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

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