ThermodynamicsChapter 15

Part IMeasuring Energy Changes

Thermodynamic Terms System – Surroundings Thermodynamic State of System – noting

temperature, pressure, composition, and STATE of the system! (aq) (s) (l) (g)

State Function -

State Function - Change in energy of the system

The First Law of Thermodynamics The First Law of thermodynamics – The

amount of energy in the universe is constant Law of Conservation of Energy – Energy can

neither be created nor destroyed in ordinary chemical and physical changes

Our challenge is to predict which reactions will proceed spontaneously or need a push and which reactions will release energy and which will absorb energy

Enthalpy Changes Heat – form of energy that flows between

samples of matter due to a temperature difference

Enthalpy Change – the quantity of heat (enthalpy) transferred into or out of a system as it undergoes a chemical or physical change at constant pressure

Enthalpy Changes (H) Enthalpy of a system depends on the KE and PE of

the system (internal energy) Affected by pressure and temperature and volume Heat cannot be measured directly, only enthalpy

changes: H = HFinal - HInitial

H = Hsubstances produced – H Substances Consumed

Exothermic H < 0 Endothermic H > 0

Calorimetry The system is the reactants and products

(liquids and solids work the best) The surroundings is the calorimeter and the

water Assume pressure is constant, no gaseous

products.

Calorimetry Energy is neither created

nor destroyed…so it must go somewhere when released.

Heat released from chemical reaction or cooling object is absorbed by water in the cup.

Q = m c T m c T = m c T

Bomb Calorimetry Calorimetry – observing temperature

change when a system absorbs or releases energy.

Bigger devise with a bomb chamber inside a water reservoir, usually measuring combustion reactions.

http://www.youtube.com/watch?v=ohyA9amFfsc

Calibrating the Calorimeter It is silly to think that no heat is lost to the

environment, and to the calorimeter It is necessary to calculate the specific heat

capacity of the calorimeter by delivering a known quantity of heat and measuring T

(H Rxn) = (H calorimeter) + (H solution)

Calibrating the Calorimeter (EX1) Determining Heat Capacity of Calorimeter

6.850KJ of heat were added to a calorimeter with 100.g H2O. The temperature rises from 25.00oC to 37.54oC

Find Cp in J/oC

Calorimetry after Calibration (EX 2) Determining associated H of reaction

25.00mL of 0.500M NaOH and 25.00mL of 0.600M CH3COOH, both substances start at 23.00oC and the reaction ends at a maximum temperature of 25.947oC. How much heat is lost from the reaction? What is the H/mole reaction?

Part IIThermochemical Equations

Writing Equations and using H to calculate energy

Thermochemical Equations A balanced chemical equation with a value

for H Interpret coefficients as # of moles, not # of

molecules, so fractional coefficients are ok When stoichiometric amounts react, it is

called one mole of reaction (mol rxn)

Thermochemical EquationsC2H5OH (l)+ 3O2 (g) 2CO2 (g) + 3H2O (l) + 1367KJ

H = -1367KJ/mole rxn. The physical states are important to note

because physical changes involve heat changes.

Thermochemical Equations (EX 3) H of formation for H2SO4(l) is

-814.0KJ/mole rxn. Write a thermochemical equation for which H rxn is -814.0kJ/mole

Thermochemical Equations (EX 4) Find Hf/mole for the following reaction at

25oC and 1atm:

P4 (g)+ 6Cl2 (g) 4PCl3 (g)

Thermochemical Equations (EX 5) Calculate the change in enthalpy (H) for

the reaction of 23.0g of CaO at 25oC and 1atm.

CaO (s) + H2O (l) Ca(OH)2 (s) H =-65.3kJ/mole

Part IIIUsing Thermochemical Equations for Predicting Unknown Reactions

Hess’s Law The enthalpy change of a reaction is the

same whether it occurs by one step or any series of steps (state function)

Allows for calculation of H for difficult to measure reactions

Hess’s Law – (EX6)From the following enthalpies of reaction,

CaCO3 (s) CaO (s) + CO2 (g) H = 178.1

CaO(s) + H2O (l) Ca(OH)2 (s) H = -65.3

Ca(OH)2 (s) Ca2+ (aq) + 2OH1- (aq) H = -16.2

Calculate Hrxn for:

Ca2+ (aq) + 2OH1- (aq) + CO2 (g) CaCO3 (s) + H2O (l) H = ?

Hess’s Law (EX7) Horxn = Ho

f products - Hof reactants

Calculate Horxn using Hof data for:

SF6 (g) + 3H2O(l) 6HF(g) + SO3(g)

Hess’s Law (EX8) Can be used to calculate the heat of

formation if Horxn is known Calculate the heat of formation of Fe3O4 if

the thermite reaction has a Horxn of -3350kJ/mole

8Al (s) + 3Fe3O4(s) 4Al2O3 (s) + 9Fe(s)

Hess’s Law – Bond Energies Bond Energy – the energy to break a

chemical bond or separate atoms in the gaseous phase

HCl (g) + 432kJ H (g) + Cl (g)

Hess’s Law – Bond Energies Greater Bond Energy = Greater Bond

Stength (always + values) Bond energy is an approximation of bond

enthalpy Can use bond energies to estimate

thermodynamic values not readily available, but ONLY FOR GAS PHASE!

Hess’s Law – Bond Energies (EX9)

Horxn = BE reactants - BE products Use known bond energy values to calculate

the heat of reaction for: CH4 + Cl2 CH3Cl + HCl See page 610 for posted values

Changes in internal energy (E) Internal energy is the sum of all the energy

in the system Kinetic energy of the particles + potential

energy of the bonds = total energy Cannot be measured directly, only changes

(just like enthalpy) E = E products – E reactants

Changes in internal energy (E)E = q + w

W = -PV

PV = nRT

W = - nRT

if n = 0 , work is not done If n = +, work is –,

Work done by the system If n = -, work is +

Work done by the surroundings

Changes in internal energy (E) CO (g) + H2O (g) H2 (g) + CO2 (g) 2 Moles 2 moles n = 0 Work = 0

Changes in internal energy (E) Zn (s) + 2HCl (aq) H2 (g) + ZnCl2 (aq) 0 Moles 1 moles n = + Work = -, done by system (system

expands)

Changes in internal energy (E) N2 (g) + 3H2 (g) 2NH3 (g) 4 Moles 2 moles n = - Work = +, done by surroundings (system is

compressed by surroundings)

Changes in internal energy (E) E = q + w

+ q = heat absorbed by system - q = heat released + w = work done on system (compression) - w = work done by system (expansion)

H & E H = E + PV H = q + w + PV Useful for physical changes involving

changes in volume w = -PV at constant pressure and temperature

H = q Works for reactions with no change in moles

H & E For other reactions where the number of

moles of gas changes, the correction factor cannot be ignored

H = E + PV PV = nRT H = E + nRT Better for chemical processes where there

is a change in moles of a gas

H & E In a bomb calorimeter, v is not allowed,

therefore no work can be done so H = E In the standard environment, V cannot be

ignored, work term cannot be ignored

H & E (EX10)

Calculate the E and work for the combustion of n-pentane at 25oC. H = -3523kJ/mole rxn.

Part IVSpontaneity

Spontaneity Spontaneous – occurs under specified

conditions without any continued outside influence

Implies that the products are thermodynamically more stable than the products

States nothing about the speed of the change

Spontaneity vs Enthalpy Not all exothermic changes are

spontaneous Not all spontaneous changes are

exothermic EX. The melting of water is spontaneous at

temperatures over 0oC and is endothermic

2nd Law of thermodynamics The universe moves toward a state of

greater disorder Spontaneity is favored when heat is released

during a change Spontaneity is favored when the change creates

an increase in disorder

Spontaneity (S) + S = Spontaneity is favored - S = Spontaneity is not favored (S)universe = (S)system + (S)surroundings

Entropy Entropy – a measure of the disorder of the

system (S) Third Law of Thermodynamics – the

entropy of a pure crystalline substance is zero as it is perfectly ordered.

S (g) > S (l) > S (s)

Entropy Calculations (EX11) Sorxn = So products - So reactants Calculate Sorxn at 25C: 2NO2 (g) N2O4 (g)

Spontaneity

If S is negative, H is likely to be negative if the reaction is spontaneous.

Part VFree Energy Change (G) and

Spontaneity

Free Energy G relates heat change and entropy change

and indicates the spontaneity of the process, not the speed.

G (Gibbs Free Energy) – a measure of energy available to be used for work on the surroundings Affected by Temp, Pressure, and concentration

of reactants (specific conditions)

Free Energy If G < 0, the reaction is spontaneous, the

system would be at a lower energy level after the reaction

If G = 0, the reaction is at equilibrium If G > 0, the reaction is not spontaneous,

the system will be at a higher energy level after the reaction.

Free Energy Go = ΣnGo

product - ΣnGoreactant

G = H - TS Go = Ho - TSo

Free Energy (EX12) Is the oxidation of Magnesium spontaneous

at 25oC?

Free Energy (EX13) Use thermodynamic data to estimate the

normal b.p. of water.

H2O (l) H2O (g)

SummaryH S G

Forward reaction spontaneous at all temperature

- + -

Forward reaction non spontaneous at all temperature

+ - +

Forward reaction spontaneous at low temperature

- - ?

Forward reaction spontaneous at high temperature

+ + ?