Thermodynamics

Temperature Vs. Heat

• Temperature– K, °C– Measure of average

KE of motion of particles

• Heat– kJ, kcal (Cal)– 1kcal=4.184 kJ– Measure of total energy transferred from an

object of high E to low E

Note: A change in T is accompanied by a transfer of energy

Specific Heat

• c or cp

• specific for each substance

= amount of energy required to raise the temperature of 1 g of a substance by 1°C

Thinking about specific heat

List the following substances in terms of specific heat, from lowest to highest:

water, gold, ethanol, granite

Thinking about specific heat

List the following substances in terms of specific heat, from lowest to highest:

Answer:

gold, granite, ethanol, water

0.129 0.803 2.44 4.184 J/g·oC

Relationship among temperature, heat and mass

Change in heat = specific heat x mass x change in T

q = cp x m x Δt

Units: heat absorbed or released, Jcp in J/g·°Cm in gΔT in °C or K

Example Problems

1. How much heat energy is released to your body when a cup of hot tea containing 200. g of water is cooled from 65.0oC to body temperature, 37.0oC? (cpH2O = 4.18 J/g·oC)

A: 23.4 kJ

Example Problems

2. A 4.50-g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0oC?

(cp of gold = 0.129 J/g·oC)

Hint: begin by solving for DT

A: 500. oC

Example Problems

3. A 155 g sample of an unknown substance was heated from 25.0oC to 40.0oC. In the process, the substance absorbed 5696 J of energy. What is the specific heat of the substance?

A: 2.45 J/g·oC

Using a Calorimeter (WS)

1. Put dried food sample in “bomb”

2. Burn it

3. Use ∆T of water to calculate ∆Hcomb of the food. 

4. The amount of energy in food is measured in Cal, rather than kJ.

5. 1 Calorie = 1 kcal = 4.184 kJ

Calorimetry

Step 1

Calculate q from calorimeter data.

q = cpH2O x mH2O x DtH2O

Step 2

Relate q to the quantity of substance in the calorimeter.

Calorimetry Problem

What is the heat of combustion, DHc , of sucrose, C12H22O11, if 1.500 g of sugar raises the temperature of water (3.00 kg) in a bomb calorimeter from 18.00oC to 19.97oC? (cpH2O = 4.18 J/g·oC)

A:

16.5 kJ/g sucrose, 5650 kJ/mol sucrose

Calorimetry Practice Problem

You burned 1.30 g of peanuts below an aluminum can which contained 200.0 mL of ice water (DH2O = 1.00 g/mL). The temperature of the water increased from 6.7oC to 28.5oC. Assume all of the heat energy produced by the peanuts went into the water in the can.

(cpH2O = 4.18 J/goC)

What was the energy content of peanuts, in kJ/g (i.e. DHcomb of peanuts?

A: 14.0 kJ/g

Calorimetry Problem, p. 881, #6(Honors)

• A 75.0 g sample of a metal is placed in boiling water until its temperature is 100.0oC.

• A calorimeter contains 100.0 g of water at a temperature of 24.4oC. • The metal sample is removed from the boiling water and

immediately placed in water in the calorimeter. • The final temperature of the metal and water in the calorimeter is

34.9oC. • Assuming that the calorimeter provides perfect insulation, what is

the specific heat of the metal?

• A: 0.900 J/g·oC

Heats of …

• ΔH = change in enthalpy (i.e. change in the amount of heat)

• The heat change which occurs during processes can be described as ΔHx,

x=process name

• ΔHcomb= … combustion

• ΔHfus= … fusion

• ΔHvap= … vaporization

• ΔHrxn= … reaction

• ΔHf°= … formation

• ΔHsol= … solution

Units: All in kJ/mol except heats of reaction and solution, which are just in kJ

Thermodynamics Word Problems Using Heats of _____ Data

Two of the following three factors will be given:

• amount of substance of interest

(usually in grams), • heat of _____ for the substance (kJ/g or kJ/mol)• total kJ

Thermodynamics Word Problems Using DHx Data

e.g. Calculate the heat required to melt 25.7 g of solid methanol at its melting point.

DHfus of methanol = 3.22 kJ/mol (from Table 16-6 on p. 502)

1. Convert mass of methanol to moles.

2. Multiply moles of methanol by DHfus.

A: 2.58 kJ

Thermodynamics Word Problems Using DHx Data

• How much heat is needed to vaporize 343 g of acetic acid (CH3COOH)?

DHvap = 38.6 kJ/mol

(A = 220. kJ)

Calculating E change using ΔHx

Lab Data Determining ΔHx

1. Measure total # kJ (calorimetry) using q = cp x m x ΔT

2. Measure mass of substance

3. Divide # kJ by mass of substance kJ/g

4. Convert to kJ/mol using massmole conversion

Combustion of a Candle Lab

What is the DHcomb of paraffin wax?

1. Measure total # kJ using

qH2O = cpH2O x mH2O x ΔTH2O

(i.e. q of water in can)

2. Measure mass of substance:

mwax burned = mcandle+card before and after expt.

Combustion of a Candle Lab

What is the DHcomb of paraffin wax?

3. Divide # kJ by mass of substance:#1 ¸ #2 kJ/g (note units include both kJ and g)

4. Convert to kJ/mol using massmole conversion__ kJ (MM of paraffin in g) ____ kJ/mol paraffin

g 1 mol paraffin

paraffin = C26H54

Temperature Changes of Water (WS)

Using Heat of Fusion

Use Table 16-6 on p. 502 for DHvapo and DHfus

o for different substances.

Calculate the energy required to melt 8.5 g of ice at 0oC. The molar heat of fusion for ice is 6.02 kJ/mol.

A: 2.8 kJ

Using Heat of Vaporization

Calculate the total energy change (in kJ) required to a) heat 25 g of liquid water from 25oC to 100.oC,

then b) change it to steam at 100.oC.

cpliquid water = 4.18 J/goC, DHvapH2O = 40.6 kJ/mol.

A: a) 7.8 kJ b) 56 kJTotal = 64 kJ

Thermochemical Stoichiometry

• The amount of energy (in kJ) can be incorporated into mole ratios.

• C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + 2870 kJ

• ΔH = -2870 kJ/mol glucose• Mole Ratios Examples

1 mol C6H12O6 6 mol CO2

6 mol O2 6 mol O2

2870 kJ 6 mol H2O

1 mol C6H12O6 2870 kJ

Thermochemical StoichiometryProblems

C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + 2870 kJ

1. How much energy (in kJ) will be released when 675 g of glucose is burned?

A: 10,800 kJ

Thermochemical StoichiometryProblems

C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + 2870 kJ

2. If 398 kJ is released when a certainamount of glucose is burned, how many grams of oxygen are consumed?

A: 26.6 g

Thermochemical StoichiometryProblems

C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + 2870 kJ

3. If 5782 kJ is released when a certain amount of glucose is burned, how many liters of carbon dioxide are released, assuming the reaction takes place at STP?

A: 271 L

Enthalpy

• enthalpien= to warm (Greek)

• Enthalpy (H) = total E of a substance

• Enthalpy change (ΔH) = comparison between H of products and H of reactants in a reaction (ΔH=Hproducts - Hreactants)

Enthalpy

• ΔHrxn is + if endothermic (energy of products > energy of reactants)

• ΔHrxn is - if exothermic (energy of reactants > energy of products)

2 H2 + O2 2 H2O + 483.6 kJDH = - 483.6 kJ

Hess’s Law

• The total enthalpy change for a chemical or physical change is the same whether it takes place in one or several steps.

• “state function”

Hess’s Law

• Version 1: Add equations together

• Version 2: Use heats of formation (DHfo)

of products and reactants

Hess’s Law Version 1Example:What is the change in enthalpy for converting graphite to diamond?Given: C(graphite) + O2(g) CO2(g) DH = -393.5 kJ/mol CC(diamond) + O2(g) CO2(g) DH = -395.4 kJ/mol C

Step 1: Write rxn: C(graphite) C(diamond)

Step 2: Add up rxns to get total rxn, canceling similar terms.C(graphite) + O2(g) CO2(g) DH = -393.5 kJ/mol CCO2(g) C(diamond) + O2(g) DH = +395.4 kJ/mol C

______________________________________________ C(graphite) C(diamond) DH = +1.9 kJ/mol C

Note: Focus is on the DH. Adding equations is the way you get there.

Hess’s Law Version 1 Practice

Calculate the heat of formation of pentane (C5H12).Given:

C(s) + O2(g) CO2(g) DH = -393.51 kJ/mol

H2(g) + ½ O2(g) H2O(l) DH = -285.83 kJ/mol

C5H12(l) + 8 O2(g) 5 CO2(g) + 6 H2O(l)

DH = -3536.1 kJ/mol

Final equation:

5 C(s) + 6 H2(g) C5H12(l) DHfo = ?

(A: -146.4 kJ)

Hess’s Law (Version 2)

The enthalpy change of a rxn is equal to the heats of formation of the products minus the heats of formation of the reactants (and taking stoichiometric relationships into account).

DH = SDHfo

(products) – SDHfo (reactants)

Hess’s Law (Version 2)

Example: Calculate ΔH for this reactionCH4(g) + 2O29g) CO2(g) + 2H2O(l)

Given:

DHfo CO2(g)= -393.5 kJ/mol

DHfo H2O(l) = -285.8 kJ/mol

DHfo CH4(g) = -74.86 kJ/mol

DHfo O2(g) = 0 (Note: DHf

o of free elements = 0)

Hess’s Law (Version 2)

Step 1: Calculate the total DHfo for the products and the

reactants.

DHproducts = (-393.5 kJ)(1 mol CO2) + (-285.8 kJ)(2 mol H2O) 1 mol 1 mol

= -965.1 kJ

DHreactants = (-74.86 kJ)(1 mol CH4) 1 mol = -74.86 kJ

Hess’s Law (Version 2)

Step 2: Plug these values into the following equation.

DHrxn = DHproducts – DHreactants

= -965.1 kJ – (-74.86 kJ)

= -890.2 kJ

Hess’s Law (Version 2) Practice

Use standard enthalpies of formation from Appendix C, Table C-13 on p. 921 to calculate DHo

rxn for the following reaction:

4 NH3(g) + 7 O2(g) 4 NO2(g) + 6 H2O(l)

A: -1397.9 kJ

Hess’s Law (Version 2) Practice

Use standard enthalpies of formation from Appendix C, Table C-13 to calculate DHo

rxn for the thermite reaction:

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

A: - 851.5 kJ

Entropy: Sen- (Greek, = in) trope (Greek, = a turning)

Entropy: degree of disorder or chaos

In nature, things tend toward disorder

It is usually much easier to go from low S to high S

All chemical/physical changes involve changes in entropy.

This natural tendency to disorder can be overcome by enthalpy.

Change in Entropy: ∆S

Entropy change (∆S) = comparison in entropy of the products and reactants in a chemical equation

∆S = Sproducts – Sreactants

+∆S increased entropy: S products > S reactants

-∆S decreased entropy: S products < S reactants

Units of S = J/mol•K Always use K in S calculations

Factors that Increase Entropy

• state change from solid to liquid to gas• increased temperature (more collisions, more

disorder)• more particles of product than reactant

(decomposition, dissolving)

Calculating ∆S from So ΔS = S°Products - S°Reactants

Use the information in the table below to calculate ∆S for the following reaction:

2 SO2(g) + O2(g) 2 SO3(g)

[2 mol SO3(257 J/mol K)] –

[2 mol SO2 (248 J/mol K) + 1 mol O2(205 J/mol K)] =

514 J/K – 701 J/K = -187 J/K Would this be favored?

∆Hfo (kJ/mol) So (J/K.mol)

SO2(g) -297 248

SO3(g) -396 257

O2(g) 0 205

SpontaneitySpontaneous = happens

Spontaneity depends on: ∆S and ∆H of reaction

Example: Combustion of octane is spontaneous

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) + kJ

Exothermic Reaction (-∆H)

Entropy increased (+∆S) due to

– More particle energy (l) (g)

– More product particles

Spontaneous Reactions

Spontaneity depends on

1. Entropy of reaction (DS)

2. Enthalpy of reaction (DH)

3. Temperature (T in K)

Gibbs Free Energy calculation

Example

• Burn octane:

2 C8H18(l) + 25 O2(g)

16 CO2(g) + 18 H2O(g) + kJ

• increased entropy due to (l) (g) favored and more product particles

• exothermic - favored

Gibbs Free Energy

ΔG = ΔH - (TΔS)

ΔG = - kJ; yes! spontaneous

ΔG = + kJ; no! nonspontaneous

Gibbs Free Energy: ∆G = ∆H - T∆SReaction is spontaneous if the sign of ∆G is negative.

Is this reaction spontaneous @ 25oC?

2 SO2(g) + O2(g) 2 SO3(g)

To take both the change in enthalpy AND entropy into account, calculate ∆G:

∆G = ∆H - T∆S

= -198 kJ – 298 K (-0.187 kJ/K) watch the units…

= -198 kJ – (-55.7 kJ) = -142 kJ

So, is this reaction spontaneous at 298 K? YES.

T 298 K Favored?

∆H - 198 kJ Y

∆S -187 J/K N

Gibbs Free Energy: ∆G = ∆H - T∆SIs this reaction spontaneous @ 2000.oK?

2 SO2(g) + O2(g) 2 SO3(g)

∆G = ∆H - T∆S

= -198.2 kJ – 2000. K (-0.187 kJ/K)

= -198.2 kJ – (-374 kJ) = 176 kJ

So, is this reaction spontaneous at 2000. K? NO.

The reaction is spontaneous at low temps (enthalpy wins) but not spontaneous at very high temps (entropy wins) (decomposition occurs….)

T 2000 K Favored?

∆H - 198.2 kJ Y

∆S -187 J/K N

Gibbs Free Energy• Will the thermite reaction occur

spontaneously at room temperature (298 K)?

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

DH = -851.5 kJ

DS = -0.038 kJ/K

DG = ?

Will the thermite reaction occur spontaneously at room temp (298 K)?

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

∆H = -851.5 kJ

∆S = -0.038 kJ/K

∆G = ?

-851.5 kJ – [298 K (-0.038 kJ/K)] = - 840 kJ

Yes, ∆G is negative

Gibbs Free Energy: ∆G = ∆H - T∆S

Free Energy Practice Problems(extra)

1. For the reaction

NH4Cl(s) NH3(g) + HCl

at 25oC, DH = 176 kJ/mol,

DS = 0.285 kJ/mol.K

Calculate DG and decide if the reaction will be spontaneous in the forward direction at 25oC.

Free Energy Practice Problemsextra

2. When graphite (C) reacts with hydrogen at 27oC,

DH = -74.8 kJ/mol and

DS = -0.0809 kJ/mol.K.

Will this reaction occur spontaneously?

Free Energy Practice Problemsextra

3. Calculate the standard entropy change, DS, that occurs when 1 mol H2O(g) is condensed to 1 mol H2O(l) at 25oC.

So H2O(g) = 188.7 J/mol.K

So H2O(l) = 69.94 J/mol.K

Free Energy Practice Problemsextra

4. For the reaction

H2O(g) + C(s) CO(g) + H2O(g),

DH = 131.3 kJ/mol, DS = 0.134 kJ/mol.K

a) Is the reaction spontaneous at 1350 K?

b) At what temperature is the reaction spontaneous?

Relating Entropy and Enthalpy to Spontaneity: DG = DH – (TDS)

DH DS Spontaneity(Is DG negative?)

Example

(-)exothermic

(+) disorder

yes 2 K(s) + 2 H2O(l) 2 KOH(aq) + H2(g)

(-) valueexothermic

(-) order

Yes at low T H2O(g) H2O(l)

(+) valueendothermic

(+) disorder

Yes at high T (NH4)2CO3(s)

2 NH3(g) + CO2(g) + H2O(g)

(+) valueendothermic

(-) order

no 16 CO2(g) + 18 H2O(l) 2 C8H18(l)+25 O2(g)

Reaction Pathways

Why don’t exothermic reactions occur spontaneously (e.g. with no spark)?

Energy is needed to overcome the mutual repelling of the atoms involved in the reaction.

Reaction Pathways

Activation energy (Ea) = initial output of energy

Note: Different reactions have different activation energies.

Reaction Profiles (WS)

(a) Activation energy(b) Energy released by activated complex

(c) ∆Hrxn = Hproducts – Hreactants

Effect of Adding a Catalyst on Reaction Rates