Thermochemistry Chapter 5. Energy The ability to do work or transfer heat. –Work: Energy used to...

Thermochemistry

Chapter 5

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.

Potential Energy

Energy an object possesses by virtue of its position or chemical composition.

Kinetic Energy

Energy an object possesses by virtue of its motion.

12

KE = mv2

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

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).

Work• Energy 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.

Heat

• Energy can also be transferred as heat.

• Heat flows from warmer objects to cooler objects.

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

Internal Energy

The 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

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

Changes in Internal Energy

• If E > 0, Efinal > Einitial

– Therefore, the system absorbed energy from the surroundings.

– This energy change is called endergonic.


• If E < 0, Efinal < Einitial

– Therefore, the system released energy to the surroundings.

– This energy change is called exergonic.


• 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.

Exchange of Heat between System and Surroundings

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

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.

State Functions

Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

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.

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.

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.

Work

When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

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

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

Enthalpy• When the system changes at constant

pressure, the change in enthalpy, H, is

H = (E + PV)

• This can be written

H = E + PV

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.

Endothermicity and Exothermicity

• A process is endothermic, then, when H is positive.

Endothermicity and Exothermicity

• A process is endothermic when H is positive.

• A process is exothermic when H is negative.

Enthalpies of Reaction

The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants

Enthalpies of Reaction

This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

For a reaction

1. Enthalpy is an extensive property (magnitude H is

directly proportional to amount):

Enthalpies of ReactionEnthalpies of Reaction

For a reaction



CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

Enthalpies of Reaction Enthalpies of Reaction

For a reaction



CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(g) H = -1604 kJ


For a reaction



2. When we reverse a reaction, we change the sign of

H:


For a reaction




H:

CO2(g) + 2H2O(g) CH4(g) + 2O2(g) H = +802 kJ


For a reaction




H:

CO2(g) + 2H2O(g) CH4(g) + 2O2(g) H = +802 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ


For a reaction




H:

3. Change in enthalpy depends on state:


For a reaction




H:


CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) H = -802 kJ


For a reaction




H:


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


)reactants()products( HHH reaction


b) Calculate the amount of heat transferred when

2.4g of Mg reacts at constant pressure.


Problem: 5.33

2 Mg(s) + O2(g) 2 MgO(s) H = -1204 kJ




Problem: 5.33

2 Mg(s) + O2(g) 2 MgO(s) H = -1204 kJ

MgmolmolggMgmoles 10.0/3.24

4.2




Problem: 5.33

2 Mg(s) + O2(g) 2 MgO(s) H = -1204 kJ

producedheat theand

used Mgof moles thebetween ratioa is there

10.0/3.244.2 Mgmolmolg

gMgmoles



Enthalpies of ReactionEnthalpies of Reaction Problem: 5.33

2 Mg(s) + O2(g) 2 MgO(s) H = -1204 kJ

mol

x

MgmolmolggMgmoles

10.0Mg2

kJ1204-

10.0/3.244.2



Enthalpies of ReactionEnthalpies of Reaction Problem: 5.33

2 Mg(s) + O2(g) 2 MgO(s) H = -1204 kJ

kJx

mol

x

MgmolmolggMgmoles

60

10.0Mg2

kJ1204-

10.0/3.244.2

Problem: 5.33

2 Mg(s) + O2(g) 2 MgO(s) H = -1204 kJ

c) How many grams of MgO are produced during

an enthalpy change of 96.0 kJ?


Problem: 5.33

2 Mg(s) + O2(g) 2 MgO(s) H = -1204 kJ




kJ

x

kJ

MgO

0.961204

2

Problem: 5.33

2 Mg(s) + O2(g) 2 MgO(s) H = -1204 kJ



MgOmolxkJ

x

kJ

MgO

16.00.961204

2


Problem: 5.33

2 Mg(s) + O2(g) 2 MgO(s) H = -1204 kJ



)/3.40(16.0

16.00.961204

2

molgmolMgOg

MgOmolxkJ

x

kJ

MgO


Problem: 5.33

2 Mg(s) + O2(g) 2 MgO(s) H = -1204 kJ




g

molgmolMgOg

MgOmolxkJ

x

kJ

MgO

42.6

)/3.40(16.0

16.00.961204

2

Problem: 5.33

2 Mg(s) + O2(g) 2 MgO(s) H = -1204 kJ

d) How many kilojoules of heat are absorbed

when 7.50g of MgO is decomposed into Mg and

O2 at constant pressure?


Problem: 5.33

2 MgO(s) 2 Mg(s) + O2(g) H = 1204 kJ





Problem: 5.33

2 MgO(s) 2 Mg(s) + O2(g) H = 1204 kJ





mol

molggMgOmoles

186.0

/3.4050.7

Problem: 5.33

2 MgO(s) 2 Mg(s) + O2(g) H = 1204 kJ





mol

x

MgO

kJ

mol

molggMgOmoles

186.02

1204

186.0

/3.4050.7

Problem: 5.33

2 MgO(s) 2 Mg(s) + O2(g) H = 1204 kJ





kJx

mol

x

MgO

kJ

mol

molggMgOmoles

112

186.02

1204

186.0

/3.4050.7

Calorimetry

Calorimetry

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

Calorimetry• Use a calorimeter (measures heat flow).

• Two kinds:

– Constant pressure calorimeter (called a coffee cup calorimeter)

– Constant volume calorimeter is called a bomb calorimeter.

Specific Heat Capacity and Heat Transfer

• The quantity of heat transferred depends upon three things:

1.The quantity of material (extensive).

2.The difference in temperature.

3.The identity of the material gaining/losing heat.

Heat Capacity and Specific Heat

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

• Molar Heat Capacity – The amount of heat required to raise the temperature of one mole of a substance by 1 Kelvin.

Specific Heat (Specific Heat Capacity, c) – The amount of heat required to raise 1.0 gram of a substance by 1 Kelvin.

Heat Capacity and Specific Heat

if TTetemperaturinchangeT

massm

heatspecificc

Tm

qc

,

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.

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 c T

- Atmospheric pressure is constantH = qP

qsystem = -qsurroundings

- The surroundings are composed of the water in the calorimeter and the calorimeter.

qsystem = - (qwater + qcalorimeter)

- For most calculations, the qcalorimeter can be ignored.qsystem = - qwater

csystemmsystem Tsystem = - cwatermwater Twater

Calorimetry Calorimetry Constant-Pressure Calorimetry

Bomb Calorimetry

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

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.

- Special calorimetry for combustion reactionsSpecial calorimetry for combustion reactions

- Substance of interest is placed in a “bomb” and filled to a high Substance of interest is placed in a “bomb” and filled to a high pressure of oxygenpressure of oxygen

- The sealed bomb is ignited and the heat from the reaction is The sealed bomb is ignited and the heat from the reaction is transferred to the watertransferred to the water

- This calculation must take into account the heat capacity of the This calculation must take into account the heat capacity of the calorimeter (this is grouped together with the heat capacity of water).calorimeter (this is grouped together with the heat capacity of water).

qrxn = -Ccalorimeter(T)

Calorimetry Calorimetry

Bomb Calorimetry (Constant-Volume Calorimetry)

Problem 5.50NH4NO3(s) NH4

+(aq) + NO3-(aq)

Twater = 18.4oC – 23.0oC = -4.6oCmwater = 60.0gcwater = 4.184J/goCmsample = 3.88g

qsample = -qwater

qsample = -cwatermwater Twater

qsample = -(4.184J/goC)(60.0g+3.88g)(-4.6oC)qsample = 1229J

- Now calculate H in kJ/mol


Problem 5.50NH4NO3(s) NH4

+(aq) + NO3-(aq)

Twater = 18.4oC – 23.0oC = -4.6oCmwater = 60.0gcwater = 4.184J/goCmsample = 3.88g


qsample = 1229J

moles NH4NO3 = 3.88g/80.032g/mol = 0.04848mol

H = qsample/moles

H = 1229J/0.04848molH = 25.4 kJ/mol

A 1.800g sample of octane, C8H18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/oC. The temperature of the calorimeter plus contents increased from 21.36oC to 28.78oC. What is the heat of combustion per gram of octane? Per mole of octane?

CalorimetryCalorimetry

2 C8H18 + 25O2 16 CO2 + 18 H2O

Twater = 28.78oC – 21.36oC = 7.42oC

Ccal = 11.66kJ/oC

msample = 1.80g

2 C8H18 + 25O2 16 CO2 + 18 H2O Twater = 28.78oC – 21.36oC = 7.42oC

Ccal = 11.66kJ/oCmsample = 1.80g

qrxn = -Ccal (Twater)





qrxn = -Ccal (Twater)qrxn = -11.66kJ/oC(7.42oC)

CalorimetryCalorimetry A 1.800g sample of octane, C8H18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/oC. The temperature of the calorimeter plus contents increased from 21.36oC to 28.78oC. What is the heat of combustion per gram of octane? Per mole of octane?



qrxn = -Ccal (Twater)qrxn = -11.66kJ/oC(7.42oC) = -86.52kJ



Hcombustion(in kJ/g) Hcombustion = -86.52kJ/1.80g =


Hcombustion(in kJ/g) Hcombustion = -86.52kJ/1.80g = -48.1 kJ/g



Hcombustion(in kJ/mol) Hcombustion = -86.52kJ/0.01575mol =




Hcombustion(in kJ/mol) Hcombustion = -86.52kJ/0.01575mol = -5492 kJ/mol


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.

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.”

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.

Hess’s Law– rxns in one step or multiple steps are additive

because they are state functions• eg.

• CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = - 802 kJ

• CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = - 890 kJ

2H2O(g) 2H2O(l) H = - 88 kJ

Practice – Calculate H for the conversion of graphite to

diamond:• Cgraphite Cdiamond

• Cgraphite + O2(g) CO2(g) H = -393.5 kJ

• Cdiamond + O2(g) CO2(g) H = -395.4 kJ

• Cgraphite + O2(g) CO2(g) H = -393.5 kJ

• CO2(g) Cdiamond + O2(g) H = 395.4 kJ

Cgraphite Cdiamond H = + 1.9 kJ

Enthalpies of Formation

- There are many type of H, depending on what you want to know

Hvapor – enthalpy of vaporization (liquid gas)

Hfusion – enthalpy of fusion (solid liquid)

Hcombustion – enthalpy of combustion

(energy from burning a substance)


- A fundamental H is the Standard Enthalpy of Formation ( )o

fH


- A fundamental H is the Standard Enthalpy of Formation ( )

Standard Enthalpy of Formation ( ) – The enthalpy change that accompanies the formation of one mole of a substance from the most stable forms of its component elements at 298 Kelvin and 1 atmosphere pressure.

ofH

ofH

Enthalpies of Formation- A fundamental H is the Standard Enthalpy of

Formation ( )

Standard Enthalpy of Formation ( ) – The enthalpy change that accompanies the formation of one mole of a substance from the most stable forms of its component elements at 298 Kelvin and 1 atmosphere pressure.

“The standard enthalpy of formation of the most stable form on any element is zero”

ofH

ofH

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)

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:

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.

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


Problem 5.72a) N2O4(g) + 4 H2(g) N2(g) + 4 H2O(g)

N2O4(g) 9.66 kJ/molH2(g) 0 kJ/molN2(g) 0 kJ/molH2O(g) -241.82 kJ/mol

H = [1mol(H(N2)) + 4mol(H(H2O))] – [1mol(H(N2O4)) + 4mol(H(H2))]H = [1mol(0kJ/mol) + 4mol(-241.82kJ/mol)] – [1mol(9.66kJ/mol) + 4mol(0kJ/mol)] = -976 kJ


Problem 5.72b) 2 KOH(s) + CO2(g) K2CO3(s) + H2O(g)

KOH(s) -424.7 kJ/mol CO2(g) -393.5 kJ/mol K2CO3(s) -1150.18 kJ/mol H2O(g) -241.82 kJ/mol

H = [1mol(H(K2CO3)) + 1mol(H(H2O))] – [2mol(H(KOH)) + 1mol(H(CO2))]H = [1mol(-1150.18kJ/mol) + 1mol(-241.82kJ/mol)] –

[2mol(-424.7kJ/mol) + 1mol(-393.5kJ/mol)] = -149.1kJ

Energy in Foods

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

Foods• 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.

Foods and FuelsFoods and Fuels

Fuels

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

Fuels• Fuel value = energy released when 1 g of substance is

burned.• Most from petroleum and natural gas.• Remainder from coal, nuclear, and hydroelectric.• Fossil fuels are not renewable.• In 2000 the United States consumed 1.03 1017 kJ of

fuel. • Hydrogen has great potential as a fuel with a fuel value

of 142 kJ/g.

Foods and FuelsFoods and Fuels

Thermochemistry Chapter 5. Energy The ability to do work or transfer heat. –Work: Energy used to...

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