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Thermochemistry
Chapter 5Thermochemistry
John D. Bookstaver
St. Charles Community College
St. Peters, MO
2006, Prentice Hall, Inc.
Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Thermochemistry
December 16HOMEWORK
• Do questions 1,2,9,11,17,19,20
• First law of thermodynamics 23,25,29
• Enthalpy 31 to 45 odd
• Lab Notebook due by next Monday
Thermochemistry
Energy
• The ability to do work or transfer heat.Work: Energy used to cause an object that
has mass to move.Heat: Energy used to cause the
temperature of an object to rise.
Thermochemistry
Potential Energy
Energy an object possesses by virtue of its position or chemical composition.
Thermochemistry
Electrostatic PEArises from the interaction between charged
particles.
Eel = k Q1 Q2 k = 8.99x109 Jm/C2
d Q charge of particles
d distance
If charges are same sign they repel and Eel is positive. If they have opposite sign they attract and Eel is negative.
The lower the energy of system the more stable it is. Thus the more strongly opposite charged particles attract each other.
Thermochemistry
Kinetic Energy
Energy an object possesses by virtue of its motion.
12
KE = mv2
Thermochemistry
Units of Energy
• The SI unit of energy is the joule (J).
• An older, non-SI unit is still in widespread use: The calorie (cal).
1 cal = 4.184 J
1 J = 1 kg m2
s2
Thermochemistry
Joule
• A mass of 2 kg moving at a speed of 1 m/s possesses a kinetic energy of 1 J.
calorie: the amount of heat needed to increase the temperature of 1 g of water from 14.5 0C to 15.5 0C.
• 1 Cal = 1000 cal = 1 kcal
Thermochemistry
System and Surroundings• Consider the reaction inside
the piston
• 2H2(g) +O2->2H2O(g)+heat
• The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). Reactants an products.
• The surroundings are everything else (here, the cylinder and piston).
Thermochemistry
Work
• Energy is used to move an object over some distance.
• w = F d,
where w is work, F is the force, and d is the distance over which the force is exerted.
Thermochemistry
Heat
• Energy can also be transferred as heat.
• Heat flows from warmer objects to cooler objects.
• HEAT THE ENERGY TRANSFERRED BETWEEN A SYSTEM AND ITS SURROUNDINGS AS A RESULT OF THEIR
DIFFERENCE IN TEMPERATURE
Thermochemistry
December 17
• First law of thermodynamics
• Internal energy
• Exchange of heat – endo and exo processes
• State functions
• Enthalpy – Heat of a reaction
• Calorimetry
Thermochemistry
Transferal of Energy
a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.
Thermochemistry
Transferal of Energy
a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.
b) As the ball falls, its potential energy is converted to kinetic energy.
Thermochemistry
Transferal of Energy
a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.
b) As the ball falls, its potential energy is converted to kinetic energy.
c) When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat.
Thermochemistry
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
Thermochemistry
Internal EnergyThe internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.
Use Fig. 5.5
Thermochemistry
Internal EnergyBy definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:
E = Efinal − Einitial
Use Fig. 5.5
Thermochemistry
Changes in Internal Energy
• If E > 0, Efinal > Einitial
Therefore, the system absorbed energy from the surroundings.
This energy change is called endergonic.
Thermochemistry
Changes in Internal Energy
• If E < 0, Efinal < Einitial
Therefore, the system released energy to the surroundings.
This energy change is called exergonic.
Thermochemistry
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.
Thermochemistry
E, q, w, and Their Signs
Thermochemistry
Exchange of Heat between System and Surroundings
• When heat is absorbed by the system from the surroundings, the process is endothermic.
Thermochemistry
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.
Thermochemistry
State Functions
Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.
Thermochemistry
State Functions• However, we do know that the internal energy
of a system is independent of the path by which the system achieved that state. In the system below, the water could have reached
room temperature from either direction.
Thermochemistry
State Functions• Therefore, internal energy is a state function.
• It depends only on the present state of the system, not on the path by which the system arrived at that state.
• And so, E depends only on Einitial and Efinal.
Thermochemistry
State Functions
• However, q and w are not state functions.
• Whether the battery is shorted out or is discharged by running the fan, its E is the same.But q and w are different
in the two cases.
Thermochemistry
Work
PV work is the work associated with a change of
volume ( V) that occurs against a resisting pressure.
Thermochemistry
WorkWe can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston.
w = −PV
Thermochemistry
Enthalpy
• If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system.
• Enthalpy is the internal energy plus the product of pressure and volume:
H = E + PV
Thermochemistry
Enthalpy
• When the system changes at constant pressure, the change in enthalpy, H, is
H = (E + PV)
• This can be written
H = E + PV
Thermochemistry
Enthalpy
• Since E = q + w and w = −PV, we can substitute these into the enthalpy expression:
H = E + PV
H = (q+w) − w
H = q
• So, at constant pressure the change in enthalpy is the heat gained or lost.
Thermochemistry
Endothermicity and Exothermicity
• A process is endothermic, then, when H is positive.
Thermochemistry
Endothermicity and Exothermicity
• A process is endothermic when H is positive.
• A process is exothermic when H is negative.
Thermochemistry
Enthalpies of Reaction
The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:
H = Hproducts − Hreactants
Thermochemistry
Enthalpies of Reaction
This quantity, H, is called the enthalpy of reaction, or the heat of reaction.
Thermochemistry
The Truth about Enthalpy
1. Enthalpy is an extensive property.
2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.
3. H for a reaction depends on the state of the products and the state of the reactants.
Thermochemistry
Practice problem– Change in enthalpy – Heat of reaction
3CuCl2 + 2Al 3Cu + 2AlCl3 ΔH = -793.5 kJ
1. What is ΔH for the following reaction?
6CuCl2 + 4Al 6Cu + 4AlCl3
2. What is ΔH for the following reaction?
9 Cu + 6 AlCl3 9 CuCl2 + 6 Al
3. Given the equation at the top of the page, how much energy will be given off by the reaction of 2.07 g of Al with excess CuCl2?
Thermochemistry
December 20
• Enthalpy practice
• Calorimetry
Thermochemistry
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.
Thermochemistry
Heat Capacity and Specific Heat• Calorimeter = apparatus that measures heat flow.• Heat capacity = the amount of energy required to raise
the temperature of an object (by one degree). (J °C-1)
• Molar heat capacity = heat capacity of 1 mol of a substance. (J mol-1 °C-1)
• Specific heat = specific heat capacity = heat capacity of 1 g of a substance. (J g-1 °C-1)
Tq substance of gramsheat specific
Thermochemistry
• Remember a change in T in K and in 0C are equal in magnitude
• T in K = T in 0C
Thermochemistry
Molar heat capacity
• The heat capacity of one mole of substance. The amount of heat required to increase the temperature of 1 mole of substance 1 K.
Thermochemistry
Constant Pressure Calorimetry• Atmospheric pressure is constant!
• HEAT LOST = HEAT GAINED
T
qH P
solution of grams
solution ofheat specificsolnrxn
TmCTmC Object 1 Object 2
Thermochemistry
Constant Pressure Calorimetry
Because the specific heat for water is well known (4.184 J/mol-K), we can measure H for the reaction with this equation:
q = m s T
Thermochemistry
Heat Capacity and Specific Heat
Specific heat, then, is
Specific heat =heat transferred
mass temperature change
s =q
m T
Thermochemistry
Sample problem 1
Find the specific heat of water if 209 J is required to increased the temperature of 50.0 g of water by 1.00K s=q / m T
Thermochemistry
Sample Problem 2
• How much heat is needed to warm 250g of water from 25 0C to 75 0C.
• Specific heat of water 4.18 J/gK
• Find the molar heat capacity of water.
Thermochemistry
Sample problem 3
25.0 mL of 0.100 M NaOH is mixed with 25.0 mL of 0.100 M HNO3, and the temperature of the mixture increases from 23.1°C to 49.8°C. Calculate ΔH for this reaction in kJ mol-1. Assume that the specific heat of each solution is equal to the specific heat of water, 4.184 J g-1 °C-1.
Thermochemistry
Sample problem 4
An unknown metal is to be analyzed for specific heat. A 12.84-g piece of the metal is placed in boiling
water at 101.5°C until it attained that temperature. The metal is then removed from the boiling water
and placed into a styrofoam cup containing 54.92 g of water at 23.2°C. The final temperature of the
water was 26.1°C. What is the specific heat of the metal sample, in J g-1 °C-1?
Thermochemistry
Bomb Calorimetry
A more sophisticated model is the bomb calorimeter, it has a chamber where a chemical reaction takes place and a device to start the reaction.
Thermochemistry
HEAT CAPACITY OF THE CALORIMETER
• Heat is released when combustion occurs. The heat is absorbed by the calorimeter and the T of the water rises.
• THE HEAT CAPACITY OF THE CALORIMETER HAS TO BE CALCULATED. It is done by measuring T in a reaction that releases a known amount of heat. The heat capacity of the calorimeter Ccal= amount of heat absorbed / T
Thermochemistry
TCq calrxn
• Reaction carried out under constant volume.
• Use a bomb calorimeter.• Usually study combustion.
Bomb Calorimetry (Constant Volume Calorimetry)
Thermochemistry
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.
Thermochemistry
Sample problem 5
A 0.5269 g sample of octane is placed in a bomb calorimeter with a heat capacity of 11.3 kJ °C-1. The original temperature of the water in the calorimeter is 21.93°C. After combustion, the temperature increases to 24.21°C. Determine the enthalpy of combustion per mole and per gram of octane.
Thermochemistry
Sample problem 6
• After 50 mL of 1.0 M HCl and 50mL of 1.0M NaOH are mixed in a coffe cup calorimeter, the temperature of the solution increases form 21.0 0C to 27.5 0C. Calculate the enthalpy change for the rx in kJ/mol of HCl assuming that the calorimeter loses only a negligible quantity of heat, total volume of solution is 100mL, that its density is 1.0 g/mL and the specific heat is 4.18J/gk
Thermochemistry
December 21
• MORE HESS LAW
• ENTHALPIES OF FORMATION
Thermochemistry
Hess’s Law
H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested.
• However, we can estimate H using H values that are published and the properties of enthalpy.
Thermochemistry
Hess’s Law
Hess’s law states that “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”
Thermochemistry
Hess’s Law
Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.
Thermochemistry
Example
The bombardier beetle uses an explosive discharge as a defensive measure. The chemical reaction involved is the oxidation of hydroquinone by hydrogen peroxide to produce quinone and water:
C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l)
Thermochemistry
C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g) ΔH = +177.4 kJ
H2 (g) + O2 (g) H2O2 (aq) ΔH = -191.2 kJ
H2 (g) + ½ O2 (g) H2O (g) ΔH = -241.8 kJ
H2O (g) H2O (l) ΔH = -43.8 kJ
C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l)
ΔH = ???
Thermochemistry
Enthalpies of Formation
An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.
Thermochemistry
Hof .
• Standard conditions (standard state): 1 atm and 25 oC (298 K).
• Standard enthalpy, Ho, is the enthalpy measured when everything is in its standard state.
• Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states.
• If there is more than one state for a substance under standard conditions, the more stable one is used.
• Standard enthalpy of formation of the most stable form of an element is zero.
Thermochemistry
Enthalpies of FormationEnthalpies of Formation
Thermochemistry
CONVENTIONAL DEFINITIONS OF STANDARD STATES FOR
ELEMENTS
• Is the form in which the element exist under conditions of 1 atm and 25 C.
• For O2 is gas.
• For Na is solid
• For Hg is liquid
• For C is graphite
• By definition the H of formation of the most stable form of an element is zero.
Thermochemistry
Examples – write formation reactions for each:
1. H2SO4
2. C2H5OH
3. NH4NO3
Thermochemistry
Examples
Use the tables in the back of your book to calculate ΔH for the following reactions:
1. 4 CuO (s) 2 Cu2O (s) + O2 (g)
2. C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
3. NH3 (g) + HCl (g) NH4Cl (s)
Thermochemistry
Using Enthalpies of Formation of Calculate Enthalpies of Reaction
• We use Hess’ Law to calculate enthalpies of a reaction from enthalpies of formation.
Thermochemistry
Using Enthalpies of Formation of Calculate Enthalpies of Reaction
• For a reaction
reactantsproductsrxn ff HmHnH
Thermochemistry
Calculation of H
We can use Hess’s law in this way:
H = nHf(products) - mHf(reactants)
where n and m are the stoichiometric coefficients.
Thermochemistry
Calculation of H
• Imagine this as occurringin 3 steps:
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
C3H8 (g) 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) 3 CO2 (g)
4 H2 (g) + 2 O2 (g) 4 H2O (l)
Thermochemistry
Calculation of H
• Imagine this as occurringin 3 steps:
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
C3H8 (g) 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) 3 CO2 (g)
4 H2 (g) + 2 O2 (g) 4 H2O (l)
Thermochemistry
Calculation of H
• Imagine this as occurringin 3 steps:
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
C3H8 (g) 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) 3 CO2 (g)
4 H2 (g) + 2 O2 (g) 4 H2O (l)
Thermochemistry
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
C3H8 (g) 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) 3 CO2 (g)
4 H2 (g) + 2 O2 (g) 4 H2O (l)
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
Calculation of H
• The sum of these equations is:
Thermochemistry
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
Calculation of H
H = [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)]
= [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]
= (-2323.7 kJ) - (-103.85 kJ)
= -2219.9 kJ
Thermochemistry
Energy in FoodsMost of the fuel in the food we eat comes from carbohydrates and fats.
Thermochemistry
Fuels
The vast majority of the energy consumed in this country comes from fossil fuels.
Thermochemistry
HW
• PRELAB FOR HESS LAW LAB DUE TOMORROW (Lab 6 – page 25)
• Remember test on monday
• Hw for Thursday up to 5.77 odd
Thermochemistry
Example
Calculate the total energy produced from the combustion of one gram of gasoline (assume it is pure octane, C8H18). Compare that to the energy produced from the combustion of one gram of ethanol (a renewable fuel – C2H5OH)
Thermochemistry
• Hess’s law: if a reaction is carried out in a number of steps, H for the overall reaction is the sum of H for each individual step.
• For example:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ
2H2O(g) 2H2O(l) H = -88 kJ
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ
Thermochemistry
Note that: H1 = H2 + H3
Thermochemistry
JANUARY 8
• Prelab
• Foods and fuels. Read from txbook
• Hess law problems
Thermochemistry
Foods• Fuel value = energy released when 1 g of substance is
burned.• 1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal.• Energy in our bodies comes from carbohydrates and fats
(mostly).• Intestines: carbohydrates converted into glucose:
C6H12O6 + 6O2 6CO2 + 6H2O, H = -2816 kJ
• Fats break down as follows:2C57H110O6 + 163O2 114CO2 + 110H2O, H = -75,520 kJ
• Fats: contain more energy; are not water soluble, so are good for energy storage.
Thermochemistry
Fuels• In 2000 the United States consumed 1.03 1017 kJ of
fuel.• Most from petroleum and natural gas.• Remainder from coal, nuclear, and hydroelectric.• Fossil fuels are not renewable.
Thermochemistry
Sources of Sources of energy energy
consumed in consumed in USAUSA