Chapter 7. Thermodynamic Processes and Thermochemistry

Observations of the behavior of matter

Generalization: laws

Goal: To predict what types of chemical and physical processes are possible, under what conditions.To calculate quantitatively the properties of the equilibrium state

Macroscopic properties

Types of questions

• If hydrogen and nitrogen are mixed, is it possible for them to react? If so, in what percentage yield will they produce ammonia?

• How will a particular change in temperature or pressure affect the extent of the reaction?

• How can the conditions for the reaction be optimized to maximize its yield?

Thermodynamics

• Thermodynamics asserts that substances have specific measurable macroscopic properties, but it cannot explain why a substance has particular numerical values for these properties.

• Thermodynamics can determine whether a process can occur, but it cannot say how rapidly the process will occur. Ex) diamond to graphite

Equilibrium constant

Properties of pure substances in bulk

Properties of molecules

7.1 Systems, States, and Processes

• Thermodynamics uses operational models to represent real-world systems.

• A key step in applying the methods of thermodynamics to such diverse processes is to formulate the model.

• This requires precise definitions of thermodynamic terms.

Caution: may have different meaning than everyday usage

System

• Closed system: No flow in and out• Open system: the boundaries permit

such flow

A system is that part of the universe of immediate interest in a particular experiment or study

system surroundings

Thermodynamic universe

Macroscopic properties of systems

• Thermodynamics is concerned with macroscopic properties of systems and the ways in which they change.

• Extensive properties: sumex) energy, mass, volume

• Intensive properties: sameex) temperature, pressure

State• A thermodynamic state is a macroscopic

condition of a system whose properties are uniquely determined by its surroundings(such as a laboratory apparatus)

• Ex) A system comprising 2 mol of He gas can be held at 1.5 atm by a piston-cylinder apparatus and at 298 K by heat bath.

=> The properties P and T are constrained to the values 1.5 atm and 298 K.

Equilibrium

• After the system has been prepared by establishing a set of constraints in the surroundings, after all disturbances due to preparation cease, and none of its properties change with time, the system is said to have reached equilibrium.

• The same equilibrium state can be reached from different directions.

• The thermodynamic state of a system is fixed when any two of its independent properties are given. Ex) P and T for 1 mol of a pure gas fixes not merely the volume V, butall other properties of the material such as internal energy E.

Process

• A thermodynamic process leads to a changein the thermodynamic state of the system.

• Physical process• Chemical process

Reversible processes

Each point in the path can be explained by the equation of state

Each point is in equilibrium

IdealizationInfinite number of stepInfinite time

Irreversible process

Intermediate state is not a thermodynamic state

State functions

• Uniquely determined by the thermodynamic state of the system.

• Ex) Volume, temperature, pressure, the internal energy

Δ: change• Ex) ΔV = Vfinal – Vinitial (or Vf – Vi)• The change in any state function is

independent of path.

• ExampleA cylinder confines 2.00 L of gas under a pressure of 1.00 atm. The external pressure is also 1.00 atm. The gas is heated slowly, with the piston sliding freely to maintain the pressure of the gas close to 1.00 atm. Suppose the heating continues until a final volume of 3.50 L is reached. Calculate the work done on the gas and express it in joules.

7.2 The First Law of Thermodynamics:Energy, Work, and Heat

• Work: the product of the external force on a body times the distance through which the force acts

• Kinetic energy, Potential energy due to gravitational field, Pressure-volume work

• Internal energy: the total energy content of a system due to potential energy between molecules,due to the kinetic energy of molecular motions,due to chemical energy stored in chemical bonds

Potential energy between molecules appears as the lattice energy of solids and the attractive and repulsive interactions between molecules in gases and liquids.

Kinetic energy appears in the translation and the internal motions of individual molecules.

Heat, or thermal energy

• The amount of energy transferred between two objects initially at different temperatures is called heat, or thermal energy. Ex) A hot stone thrown in a cup of water.

• Heat is not a substance.• Instead, heat (like work) is a way in which energy is

exchanged between a system and its surroundings.

How can amounts of heat be measured?

Ice calorimeter

Specific heat capacity

• The specific heat capacity of a material is the amount of heat required to raise the temperature of a 1-gram mass by 1 degree.

q = McsΔT

-MgΔh

• ExampleSuppose a 10.00 kg mass drops through a height difference of 3.00 m, and the resulting work is used to turn a paddle in 200.0 g of water, initially at 15.00 oC. The final water temperature is found to be 15.35 oC. Assuming that the work done is used entirely to raise the water temperature, calculate the conversion factor between joules and calories.

The First Law of Thermodynamics

• Both heat and work are forms in which energy is transferred into and out of a system; they can be thought of as energy in transit.

• If the energy change is caused by mechanical contact of the system with its surroundings, work is done.

• If it is caused by thermal contact (leading to equalization of temperatures), heat is transferred.

• Heat and work are not state functions, but processes.ΔE = q + w

Although q and w depend individually on the path followed between a given pair of states, their sum does not.

qsys = -qsurr

wsys = -wsurr

ΔEsys = -ΔEsurr

ΔEuniv = ΔEsys + ΔEsurr = 0

7.3 Heat Capacity, Enthalpy, and Calorimetry

• The heat capacity C is defined as the amount of energy that must be added to the system to raise its temperature by 1K.

• Heat capacity at constant volume: CV

• Heat capacity at constant pressure: CP

• Molar heat capacity: cV, cP

• Gas vs. condensed phase• Specific heat capacities: per gram of substance

q = CΔT

qP = n cP ΔTqV = n cV ΔT

Example 7.3• A piece of iron weighing 72.4 g is

heated to 100.0 oC. And plunged into 100.0 g of water that is initially at 10.0 oC in a Styrofoam cup calorimeter. Assume no heat is lost to Styrofoam cup or to the environment. Calculate the final temperature that is reached.

Heat Transfer at Constant Volume: Bomb CalorimetersΔE = qV

Heat Transfer at Constant Pressure: Enthalpy

7.4 Illustrations of the First law of Thermodynamics in Ideal Gas Processes

Heat Capacities of Ideal Gasesthe kinetic theory of gasesthe ideal gas equation of state

Constant Volume

Constant Pressure

A Deeper Look• Equipartition

• Rotation

For any ideal gas process,

Example 7.4

• Suppose that 1.00 kJ of heat is transferred to 2.00 mol of argon (at 298 K, 1 atm). What will the final temperature Tf be if the heat is transferred (a) at constant volume, or (b) at constant pressure? Calculate the energy change ΔE in each case.

Heat and Work for Ideal Gases

Thermochemistry

2 Al(s) + Fe2O3(s) 2 Fe(s) + Al2O3(s)

Heat in the course of chemical reaction

Enthalpies of Reaction

Endothermic Reaction

Ba(OH)2·8H2O(s) + 2 NH4NO3(s) Ba(NO3)2(aq) + 2 NH3(aq) + 10 H2O(l)

Example 7.5• Red phosphorous reacts with liquid bromine in an exothermic reaction.

2 P(s) + 3 Br2(l) 2 PBr3(g) ΔH = - 243 kJCalculate the enthalpy change when 2.64 g of phosphorus reacts with an excess of bromine in this way.

Hess’s Law

If two or more chemical equations are added to give another chemical equation, the corresponding enthalpies of reaction must be added.

The corresponding energy change ΔE

For reactions in which only liquids and solids are involved, orthose in which the number of moles of gas does not change, the enthalpy and energy changes are almost equal and their difference can be neglected.

Phase Changes

Example 7.6

• To vaporize 100.0 g of carbon tetrachloride at its normal boiling point, 349.9 K, and P = 1 atm, 19.5 kJ of heat is required. Calculate ΔHvap for CCl4 and compare it with ΔE for the same process.

Standard-State Enthalpies• Standard-State

For solids and liquids, the standard state is the thermodynamically stable state at a pressure of 1 atm and at a specified temperature.For gases, the standard state is the gaseous phase at a pressure of 1 atm, at a specified temperature and exhibiting ideal gas behavior.For dissolved species, the standard state is a 1 M solution at a pressure of 1 atm, at a specified temperature and exhibiting ideal solution behavior.

Once standard states have been defined, the zero of the enthalpy scale is defined by arbitrarily setting the enthalpies of selected reference substances to zero in their standard states.

The chemical elements in standard states at 298.15 K have zero enthalpy.

Standard entahlpy (ΔHo): The enthalpy change for a chemical reaction in which all reactants and products are in their standard states and at a specified temperature.

Standard enthalpy of formation (ΔHfo) of a compound is the

nethalpy change for the reaction that produces one mole of the compound from its elements in their stable states, all at 25oC and 100,000 Pa pressure.

For dissolved speciesBoth positive and negative ions form; it is impossible to produce one without the other.

A further arbitrary choice is therefore made; ΔHfo of H+(aq)

is set to zero.

Example 7.7

• Using Appendix D, calculate ΔHo for the following reaction at 25oC and 1 atm pressure.

2 NO(g) + O2(g) 2 NO2(g)

General pattern

Bond Enthalpy

Example 7.8• Estimate the standard enthalpy of formation

of dichlorodifluoromethane, CCl2F2(g). This compound is also known as Freon-12 has been used as a refrigerant because of its low reactivity and high volatility. It and other related chlorofluorocarbons (CFCs) are being phased out because of their role in depleting the ozone layer in the outer atmosphere.

7.6 Reversible Processes in Ideal GasesStrategy: Use a reversible process to calculate the change between the initial and final states connected by an irreversible process.

Isochoric process: at constant volumeIsobaric process: at constant pressureIsothermal: at constant temperatureAdiabatic: no heat (q = 0)

Isothermal Processes

Example 7.9

• Calculate the heat and the work associated with a process in which 5.00 mol of gas expands reversibly at constant temperature T = 298 K from a pressure of 10.0 atm to 1.00 atm.

Adiabatic Processes• No transfer of heat into or out of the system

Example 7.10• Suppose 5.00 mol of an ideal monatomic gas at tan initial temperature

of 298 K and pressure of 10.0 atm is expanded adiabatically and reversibly until the pressure has dropped to 1.00 atm. Calculate the final volume and temperature, the energy, and enthalpy changes, and the work done.

1.0 mol 의 단원자 (monoatomic) 이상 기체로 이루어진 시스템에 대해 다음과같은 state diagram 이 주어졌다고 하자. 그림 위의 모든 점은 평형 상태를만족시키고 따라서 과정 a, b, c, d 는 가역적 과정이다. 단원자 이상기체의molar heat capacity 는 Cv = (3/2)R 이다. 1 L atm = 100 J의 관계식을사용하라. R = 0.08206 L atm mol-1K-1, Cp-Cv = R.

A B

C

D

0.0 1.00.0

2.0

1.0

3.0

2.0

3.0

V (L)

P (atm)

a

b

c

d

A B

C

D

0.0 1.00.0

2.0

1.0

3.0

2.0

3.0

V (L)

P (atm)

a

b

c

d