JF Basic Chemistry Tutorial Thermodynamics

Recommended reading

• M.S. Silberberg, Chemistry, The Molecular Nature of Matter and Change, 3rd Ed., Chapter 6

• P. Atkins & L. Jones, Molecules, Matter and Change, 3rd Ed., Chapter 2

• Multiple choice tests: http://www.mhhe.com/silberberg6

Shane Plunkettplunkes@tcd.ie

1. Gas Laws and the behaviour of gases2. Enthalpy

- Exothermic and endothermic reactions- Hess’s Law- Calorimetry- Bond enthalpy

2. Entropy- Second and Third Laws of

Thermodynamics3. Spontaneity and Gibbs Free Energy

Laws of Thermodynamics - relationshipsCharles’ Law

Boyle’s Law

Avogadro’s Law

Ideal Gas Law

At constant temperature, the volume occupied by a fixed amount of gas is inversely proportional to the applied (external) pressure

V 1/P T and n fixed

At constant pressure, the volume occupied by a fixed amount of gas is directly proportional to the temperature

V T P and n fixed

At a fixed temperature and pressure, equal volumes of any gas contain equal numbers of particles (or moles)

V n T and P fixed

Combination of these laws…PV = nRT R is the universal gas constant

Gas Law ProblemA steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 21oC.

PV = nRT

Don’t forget to change the units!!V = 438 L

n = number of molesGiven mass = 0.885 kgn = mass in g/molar mass of O2

n = 885g/32gmol-1 = 27.7 mol of O2

R = 0.0821 atmLmol-1K-1

T = 21oC = 294K

P = nRT/VP = (27.7mol)(0.0821 atmLmol-1K-1)(294K)

(438L)

P = 1.53 atm

Dalton’s Law of Partial Pressure

Behaviour of gases

•Gases mix homogeneously in any proportions•Each gas in a mixture behaves as if it were the only gas present (as long as there is no chemical reaction)

Dalton’s Law: In a mixture of unreacting gases, the total pressure is the sum of the partial pressures of the individual gases

Ptotal = P1 + P2 + P3 + ….

Each component in a mixture contributes a fraction of the total number of moles in the mixture….the mole fraction (X) of that component

Pa = Xa x PtotalWhere a is the one component you are interested in

Xa = na

ntotal

QuestionA mixture of noble gases consisting of 5.5 g of He, 15.0 g of Ne and 35.0 g of Kr is place in an tank at STP. Calculate the partial pressure of each gas.

The gases are noble gases, therefore they are stable (inert) and will not react with each other. Need to work out mole fraction (X) of each gas.

Pa = Xa x Ptotal

Xa = na

ntotal

Molar mass of He = 4.003 gmol-1

Mass of He = 5.5 gnHe = 1.37 mol

Molar mass of Ne = 20.18 gmol-1

Mass of Ne = 15.0 gnNe = 0.74 mol

Molar mass of Kr = 83.8 gmol-1

Mass of Kr = 35.0 gnKr = 0.42 mol

ntotal = nHe + nNe + nKr

= 1.37 mol + 0.74 mol + 0.42 mol = 2.53 molPtotal = 1 atm

XHe = nHe / ntotal

= 1.37 / 2.53 = 0.54 PHe = XHe x Ptotal

=0.54 x 1 atm = 0.54 atm

XNe = nNe / ntotal

= 0.74 / 2.53 = 0.29PNe = XHe x Ptotal

= 0.29 x 1 atm = 0.29 atm

XKr = nKr / ntotal = 0.42 / 2.53 = 0.17PKr = XKr x Ptotal

= 0.17 x 1 atm = 0.17 atm

Kinetic-Molecular Theory

Behaviour of gases

An observation of the behaviour of gases at the molecular level….looking at their motion and their speed; also helps us to make sense of the gas laws.The theory is based on 3 assumptions.

1. Particle volume. Any gas consists of a large amount of individual particles. The volume of an individual particle is extremely small compared with the volume of the container – therefore the theory says that the gas particles have mass but no volume.

2. Particle motion. Gas particles are in contant, random, straight line motion, except when they collide with the container walls or each other.

3. Particle collisions. Collisions are elastic. This means that the colliding molecules exchange energy (but do not lose energy through friction). Therefore their total kinetic energy (Ek)is contant. In between collisions, the molecules do not influence each other at all.

The average kinetic evergy of the molecules is proportional to the absolute temperature. At a given temperature, the molecules of all gases have the same average kinetic energy.

Therefore, if two different gases are at the same temperature, their molecules have the same average kinetic energy. If the temperature of a gas is doubled, the average kinetic energy of its molecules is doubled.

Ek T

Behaviour of gases

Although the molecules in a sample of gas have an average kinetic energy and therefore an average speed, the individual molecules move at various speeds, i.e. they exhibit a distribution of speeds – some move fast, others relatively slowly. Collisions can change individual molecular speeds but the distribution of speeds remains the same.

Behaviour of gases

Ek T

Behaviour of gases

At the same temperature, lighter gases move on average faster than heavier gases.

Mean Free Path

Behaviour of gases

The mean free path (λ) of any molecule is the average distance travelled by a molecule between two successive collisions.

Every time we carry out a reaction in the lab, we get a change in energy taking place. We study thermodynamics in order to keep track of these energy changes.

Our first task is to identify the system in the universe we are interestedin observing Open system

-can exchange mass or energy with surroundings

Closed system-can exchange only energy with surroundings

Isolated system- No transfer of either energy or mass

System

×

Surroundings

ΔEsys + ΔEsurr = 0

= ΔEuniv

Law of conservation of energy states that energy cannot be made or destroyed, but can be coverted from one form to another

Conclusion: The total energy of the universe remains constant

The First Law of Thermodynamics

Enthalpy: method for measuring energy changes during chemical reactions

At constant pressure, the change in enthalpy of a system is the amountof energy released or absorbed

Standard reaction enthalpy, ΔH°rxn: change in enthalpy when reactants in their standard states are changed to products in their standard states

Standard enthalpy of formation, ΔH°form: change in enthalpy when one mole of product is formed from reactants in their standard states

Standard enthalpy of combustion, ΔH°comb: change in enthalpy when one mole of reactant is burned completely in oxygen

Exothermic reactions• release heat to their surroundings• negative enthalpy change, ΔH < 0

Endothermic reactions• take in heat from their surroundings• positive enthalpy change, ΔH > 0

Heat given out

reactants

products

reactants

productsHeat taken in

Hess’s Law• The overall reaction enthalpy is equal to the sum of the individual

enthalpies for the reactions which make it up

ExampleGiven that for the combustion of glucose

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ΔH = -2816 kJ

and for the combustion of ethanol,

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ΔH = -1372 kJ

Calculate ΔH (in kJ) for the fermentation of glucose:

C6H12O6(s) 2C2H5OH(l) + 2CO2(g) ΔH = ?

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ΔH = -2816 kJ

4CO2(g) + 6H2O(l) 2C2H5OH(l) + 6O2(g) ΔH = +2744 kJ

C6H12O6(s) 2C2H5OH(l) + 2CO2(g) ΔH = -72 kJ

Exothermic reaction

Example Given that ΔH° for formation (ΔH°f) of Pb3O4 is -175.6 kJ mol-1 and ΔH°

for the reaction 3PbO2 Pb3O4 + O2 is 22.8 kJ mol-1. What is the ΔH°f

for PbO2 (in kJ mol-1)?Step 1: Write equation for reactions given

Enthalpy of formation:

3Pb(s) + 2O2(g) Pb3O4(s) ΔH°f = -175.6 kJ mol-1

3PbO2(s) Pb3O4(s) + O2(g) ΔH° = 22.8 kJ mol-1

Step 2: Write down what you are looking forEnthalpy of formation of PbO2:Pb(s) + O2(g) PbO2 ΔH°f = ?

Step 3: Combine the equations we know to get the answer we require3Pb(s) + 2O2(g) Pb3O4(s) ΔH°f = -175.6 kJ mol-1

Pb3O4(s) + O2(g) 3PbO2(s) ΔH° = -22.8 kJ mol-1

3Pb(s) + 3O2(g) 3PbO2(s) ΔH°f = -198.4 kJ mol-1

Pb(s) + O2(g) PbO2(s) ΔH°f = -66.13 kJ mol-1

Question 1. Given the following information:

2C(s) + 3H2(g) C2H6(g) ΔH = -84.68 kJ mol-1

C(s) + O2(g) CO2(g) ΔH = -393.51 kJ mol-1

H2(g) + ½O2(g) H2O(l) ΔH = -285.83 kJ mol-1

Calculate the standard enthalpy of combustion of ethane:

C2H6(g) + 3½O2 2CO2(g) + 3H2O(l)

Answer: -1559.8 kJ mol-1

2. Given the following information:

S(s) + O2(g) SO2(g) ΔH = -296.1 kJ mol-1

C(s) + O2(g) CO2(g) ΔH = -393.5 kJ mol-1

CS2(l) + 3O2(g) CO2(g) + 2SO2(g) ΔH = -1072 kJ mol-1

Calculate the enthalpy of formation of carbon disulfide, CS2:

C(s) + 2S(s) CS2(l)

Answer: +86.3 kJ mol-1

By using a bomb calorimeter in the lab, we can determine the reactionenthalpy

The equation to calculate the heat change isq = m s ΔT

where q is the heat changem is the mass of the samples is the specific heat of the sampleΔT is the temperature change during reaction

The specific heat of a substance is the amount of heat required to raise the temperature of one gram of that substance by one degree Celsius

To electrical source

Thermometer

Mechanical stirrer

Material combusted in oxygen

Example

A sample of 350g of water is heated from 10.5°C to 15.0°C. The specificheat of water is 4.184 J g-1 °C-1. Calculate the heat change.

q = m s ΔT

m = 350g s = 4.184 J g-1 °C-1 ΔT = (15.0 – 10.5)°C = 4.5°C

q = (350 g)(4.184 J g-1 °C-1)(4.5 °C) = 6589.8 J

= 6.59 kJ

Question

A 560g sample of mercury is heated from 40°C to 78°C. The specific heat of mercury is 0.139 J g-1 °C-1. What is the heat change for the reaction? Answer: 2.96 kJ

A 782g sample of water is cooled from 25°C to 1°C. The specific heat of water is 4.184 J g-1 °C-1. What is the heat change for the reaction?

Answer: -78.5 kJ

Bond Enthalpy

• Measure of stability of molecule

• Enthalpy change required to break a given bond in 1 mole of gaseous molecules

• Bond formation: exothermic process, negative sign enthalpy

• Bond breakage: endothermic process, positive sign enthalpy

• Average values since energy of given bond varies from molecule to molecule, e.g. due to electronegative atoms

Example

Given the following data:

H2(g) + I2(s) 2HI(g) ΔH°= +54 kJ mol-1

H2(g) 2H(g) ΔH°= +436 kJ mol-1

I2(g) 2I(g) ΔH°= +214 kJ mol-1

What is the bond dissociation energy for HI?

Step 1: Draw out structures for each molecule and decide what bonds are broken

The equation we are interested in is:

H H

hydrogen

I I

iodine

H I

hydrogen iodide

H I

hydrogen iodide

ΔH°= +54 kJ mol-1The information we are given is:

H H

hydrogen2H ΔH°= +436 kJ mol-1

I I

iodine2I ΔH°= +214 kJ mol-1

Bonds broken in equation 1:

1.

2.

3.

H H ΔH°= +436 kJ mol-1

I I ΔH°= +214 kJ mol-1

Bonds broken, endothermic, ΔH°= +650 kJ mol-1

Step 2: Draw out structure and decide what bonds are formed

H H

hydrogen

I I

iodine

H I

hydrogen iodide

H I

hydrogen iodide

ΔH°rxn = +54 kJ mol-1

Bonds formed, exothermic, ΔH°= -2 (H I)

ΔH°rxn = ΔH°bonds broken + ΔH°bonds formed

+54 kJ mol-1 = 650 kJ mol-1 – 2 ΔH° (H I)

2 ΔH° (H I) = (650 – 54) kJ mol-1

2 ΔH° (H I) = 596 kJ mol-1

ΔH° (H I) = 298 kJ mol-1

Question

Given the following data:

HCl(g) + F2(g) HF(g) + ClF(g) ΔH°= -232 kJ mol-1

ClF(g) Cl(g) + F(g) ΔH°= 256 kJ mol-1

HCl(g) H(g) + Cl(g) ΔH°= 427 kJ mol-1

F2(g) 2F(g) ΔH°= 158 kJ mol-1

What is the bond enthalpy for HF?

Answer: +561 kJ mol-1

Entropy

• Measure of disorder of a system

• May be increased by increasing number of ways of arranging components. Explained by Boltzmann equation:

S = k lnW

where S = entropy of system

k = Boltzmann constant

W = number of possible arrangements

• Has relationship with spontaneous change. Second Law of Thermodynamics: spontaneous processes (those which occur naturally without any external influence) are accompanied by an increase in entropy of the universe

• Absolute entropies may be determined from Third Law of Thermodynamics: At zero degrees Kelvin, the entropy of a perfect crystal is zero

• Because this starting point exists, can measure standard molar entropies: entropy change for 1 mol of a pure substance at 1 atm pressure (usually 25°C)

Predicting Entropy Changes

• An increase in temperature leads to greater kinetic energy of moving particles, more motion and hence greater S°

• Going from solid to liquid to gas (i.e. to less ordered systems) leads to an increase in S°

• For spontaneous change, ΔS must be greater than zero. For negative ΔS values, the process is spontaneous in the reverse direction

Example

Predict whether the entropy change for the following reaction will be positive or negative:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

6 gas molecules 3 gas molecules + 4 liquid molecules

A decrease of the more disordered gas system indicates the entropy change for the reaction should be negative

Given the following information, calculate ΔS° for the reaction

= [(3 mol)(213.7 J/molK) + (4 mol)(69.9 J/molK)] –

[(1 mol)(269.9 J/molK) + (5 mol)(205.0 J/molK)]

S° (J / mol K)

CO2 213.7

H2O 69.9

C3H8 269.9

O2 205.0

ΔS°rxn = ΣS°products - ΣS°reactants

= [(3 mol CO2)(S° of CO2) + (4 mol H2O)(S° of H2O)] – [(1 mol C3H8)(S° of C3H8) + (5 mol O2)(S° of O2)]

= -374.2 J/K

= 920.7 J/K – 1294.9 J/K

= 641.1 J/K + 279.6 J/K – (269.9 J/K + 1025 J/K)

Question

Predict the sign of ΔS° and calculate its value from the following:

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

S° (J / mol K)

NO 210.65

O2 205.0

NO2 239.9

Answer: should be negative

ΔS° = -146.5 J/K

Spontaneity and Gibbs Free Energy• Gibbs Free energy is a measure of the spontaneity of a process• ΔG is the free energy change for a reaction under standard state

conditions• At constant temperature and pressure:

ΔG = ΔH – TΔS– an increase in ΔS leads to a decrease in ΔG– if ΔG < 0, the forward reaction is spontaneous– if ΔG > 0, the forward reaction is nonspontaneous– if ΔG = 0, the process is in equilibrium

– if ΔH and ΔS are positive, then ΔG will be negative only at high T, i.e. forward reaction spontaneous

– if ΔH is positive and ΔS is negative, ΔG will always be positive,

i.e. forward reaction nonspontaneous

– if ΔH is negative and ΔS is positive, ΔG will always be negative,

i.e. forward reaction spontaneous

– if ΔH and ΔS are negative, ΔG is negative only at low T, i.e. forward reaction spontaneous

Example

At 27°C, a reaction has ΔH = +10 kJ mol-1 and ΔS = +30 J K-1 mol-1.

What is the value of ΔG?

ΔG = ΔH – TΔS T = 300 K

ΔG = (+10 kJ mol-1) – (300 K)(+30 J K-1 mol-1)

ΔG = +10 kJ mol-1 – 9000 J mol-1

ΔG = +10 kJ mol-1 – 9 kJ mol-1

ΔG = + 1.0 kJ mol-1

Question

For the reaction

4KClO3(s) 3KClO4(s) + KCl(s)

Calculate ΔG for the process at 298 K if ΔH = -144.3 kJ and ΔS = -36.8 J K-1

Answer: -133.3 kJ

Kinetics

Equilibrium

Refers to reaction rates – the speed of the reaction, i.e. the change in concentration of reactants (or products) with respect to time

Refers to the extent of the reation – when no further change occurs, i.e. the (final) concentration of the product given unlimited time

At equilibrium, the rate of the forward reaction = the rate of the backwards reaction

An equilibrium reaction:

Reactants Products

Rateforward

Ratebackward

∆G0 = -RTlnK

∆G0 = Standard Gibb’s free energy changeR = gas contantT = temperatureK = equilibrium constant

Le Chatelier’s PrincipleWhen a chemical system at equilibrium is disturbed, the system shifts to counteract the change