Thermochemistry Chapter 5.

Name: Thermochemistry Chapter 5.
Uploaded: 2017-07-19T22:41:03+00:00
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Description: Thermochemistry Chapter 5.

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. 1 2 KE =  mv2

Units of Energy The SI unit of energy is the joule (J). kg m2
An older, non-SI unit is still in widespread use: The calorie (cal). 1 cal = 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 Energy By 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.

First Law of Thermodynamics
Relating DE to Heat and Work DE = 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).

Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston. w = −PV

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 + PV

Enthalpy Since E = q + w and w = −PV, we can substitute these into the enthalpy expression: H = E + PV 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

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

For a reaction Enthalpy is an extensive property (magnitude DH is directly proportional to amount):

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802 kJ
Enthalpies of Reaction For a reaction Enthalpy is an extensive property (magnitude DH is directly proportional to amount): CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802 kJ

For a reaction Enthalpy is an extensive property (magnitude DH is directly proportional to amount): CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802 kJ 2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(g) DH = kJ

For a reaction Enthalpy is an extensive property (magnitude DH is directly proportional to amount): When we reverse a reaction, we change the sign of DH:

For a reaction Enthalpy is an extensive property (magnitude DH is directly proportional to amount): When we reverse a reaction, we change the sign of DH: CO2(g) + 2H2O(g)  CH4(g) + 2O2(g) DH = +802 kJ

For a reaction Enthalpy is an extensive property (magnitude DH is directly proportional to amount): When we reverse a reaction, we change the sign of DH: CO2(g) + 2H2O(g)  CH4(g) + 2O2(g) DH = +802 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802 kJ

For a reaction Enthalpy is an extensive property (magnitude DH is directly proportional to amount): When we reverse a reaction, we change the sign of DH: Change in enthalpy depends on state:

For a reaction Enthalpy is an extensive property (magnitude DH is directly proportional to amount): When we reverse a reaction, we change the sign of DH: Change in enthalpy depends on state: CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) DH = -802 kJ

For a reaction Enthalpy is an extensive property (magnitude DH is directly proportional to amount): When we reverse a reaction, we change the sign of DH: Change in enthalpy depends on state: 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

Problem: Mg(s) + O2(g)  2 MgO(s) DH = kJ 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) DH = 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) DH = 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) DH = 1204 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O2 at constant pressure?

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: The quantity of material (extensive). The difference in temperature. 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 (1C) 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.

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

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

Calorimetry Constant-Pressure Calorimetry
- Atmospheric pressure is constant DH = 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 DTsystem = - cwatermwater DTwater

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.

qrxn = -Ccalorimeter(DT)
Calorimetry Bomb Calorimetry (Constant-Volume Calorimetry) Special calorimetry for combustion reactions Substance of interest is placed in a “bomb” and filled to a high pressure of oxygen The sealed bomb is ignited and the heat from the reaction is transferred to the water This calculation must take into account the heat capacity of the calorimeter (this is grouped together with the heat capacity of water). qrxn = -Ccalorimeter(DT)

Calorimetry Problem 5.50 NH4NO3(s)  NH4+(aq) + NO3-(aq)
DTwater = 18.4oC – 23.0oC = -4.6oC mwater = 60.0g cwater = 4.184J/goC msample = 3.88g qsample = -qwater qsample = -cwatermwater DTwater qsample = -(4.184J/goC)(60.0g+3.88g)(-4.6oC) qsample = 1229J - Now calculate DH in kJ/mol

Calorimetry Problem 5.50 NH4NO3(s)  NH4+(aq) + NO3-(aq)
DTwater = 18.4oC – 23.0oC = -4.6oC mwater = 60.0g cwater = 4.184J/goC msample = 3.88g qsample = 1229J moles NH4NO3 = 3.88g/80.032g/mol = mol DH = qsample/moles DH = 1229J/ mol DH = 25.4 kJ/mol

Calorimetry A 1.800g sample of octane, C8H18, was burned in a bomb calorimeter whose total heat capacity is 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? 2 C8H O2  16 CO H2O DTwater = 28.78oC – 21.36oC = 7.42oC Ccal = 11.66kJ/oC msample = 1.80g

Calorimetry A 1.800g sample of octane, C8H18, was burned in a bomb calorimeter whose total heat capacity is 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? 2 C8H O2  16 CO H2O DTwater = 28.78oC – 21.36oC = 7.42oC Ccal = 11.66kJ/oC msample = 1.80g qrxn = -Ccal (DTwater)

Calorimetry A 1.800g sample of octane, C8H18, was burned in a bomb calorimeter whose total heat capacity is 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? 2 C8H O2  16 CO H2O DTwater = 28.78oC – 21.36oC = 7.42oC Ccal = 11.66kJ/oC msample = 1.80g qrxn = -Ccal (DTwater) qrxn = kJ/oC(7.42oC)

Calorimetry A 1.800g sample of octane, C8H18, was burned in a bomb calorimeter whose total heat capacity is 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? 2 C8H O2  16 CO H2O DTwater = 28.78oC – 21.36oC = 7.42oC Ccal = 11.66kJ/oC msample = 1.80g qrxn = -Ccal (DTwater) qrxn = kJ/oC(7.42oC) = kJ

Calorimetry A 1.800g sample of octane, C8H18, was burned in a bomb calorimeter whose total heat capacity is 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? DHcombustion(in kJ/g) DHcombustion = kJ/1.80g =

Calorimetry A 1.800g sample of octane, C8H18, was burned in a bomb calorimeter whose total heat capacity is 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? DHcombustion(in kJ/g) DHcombustion = kJ/1.80g = kJ/g

Calorimetry A 1.800g sample of octane, C8H18, was burned in a bomb calorimeter whose total heat capacity is 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? DHcombustion(in kJ/g) DHcombustion = kJ/1.80g = kJ/g DHcombustion(in kJ/mol) DHcombustion = kJ/ mol =

Calorimetry A 1.800g sample of octane, C8H18, was burned in a bomb calorimeter whose total heat capacity is 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? DHcombustion(in kJ/g) DHcombustion = kJ/1.80g = kJ/g DHcombustion(in kJ/mol) DHcombustion = kJ/ mol = 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) D H = kJ CH4(g) + 2O2(g) ® CO2(g) + 2H2O(l) D H = kJ 2H2O(g) ® H2O(l) D H = kJ

Practice Calculate D H for the conversion of graphite to diamond:
Cgraphite ® Cdiamond Cgraphite + O2(g) ® CO2(g) D H = kJ Cdiamond + O2(g) ® CO2(g) D H = kJ CO2(g) ® Cdiamond + O2(g) D H = kJ Cgraphite ® Cdiamond D H = 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 ( )

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.

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”

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring in 3 steps: 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)

The sum of these equations is: 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 We can use Hess’s law in this way:
H = nHf(products) - mHf(reactants) where n and m are the stoichiometric coefficients.  

H = [3( kJ) + 4( kJ)] - [1( kJ) + 5(0 kJ)] = [( kJ) + ( kJ)] - [( kJ) + (0 kJ)] = ( kJ) - ( kJ) = kJ

Problem 5.72 a) N2O4(g) + 4 H2(g)  N2(g) + 4 H2O(g) N2O4(g) 9.66 kJ/mol H2(g) 0 kJ/mol N2(g) 0 kJ/mol H2O(g) kJ/mol DH = [1mol(DH(N2)) + 4mol(DH(H2O))] – [1mol(DH(N2O4)) + 4mol(DH(H2))] DH = [1mol(0kJ/mol) + 4mol( kJ/mol)] – [1mol(9.66kJ/mol) + 4mol(0kJ/mol)] = -976 kJ

Problem 5.72 b) 2 KOH(s) + CO2(g)  K2CO3(s) + H2O(g) KOH(s) kJ/mol CO2(g) kJ/mol K2CO3(s) kJ/mol H2O(g) kJ/mol DH = [1mol(DH(K2CO3)) + 1mol(DH(H2O))] – [2mol(DH(KOH)) + 1mol(DH(CO2))] DH = [1mol( kJ/mol) + 1mol( kJ/mol)] – [2mol(-424.7kJ/mol) + 1mol(-393.5kJ/mol)] = kJ

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

Foods and Fuels 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, DH = kJ Fats break down as follows: 2C57H110O O2  114CO H2O, DH = -75,520 kJ Fats contain more energy; are not water soluble, so are good for energy storage.

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

Foods and 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.

Thermochemistry Chapter 5.

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