6.1 Basic principles

6.2 measurement of heat flow– Calorimetry (Laboratory Measurement of Heats of Reaction)

6.3 Energy and Changes of State

6.5 Enthalpy change for chemical reaction

6.6 Hess’s Law

6.7 Standard enthalpy of formation

6.4 The first law of thermodynamics

6.8 Product or reactant-favored reaction and thermochemistry

change in potential energyEQUALSkinetic energy

A system of fuel and exhaust. A fuel is higher in chemical potential chemical potential energyenergy than the exhaust. As the fuel burns, some of its potential energy is converted to the kinetic energy of the moving car.

State A – Fuel in the tank

State B - Fuel burned and exhaust produced

6.1 Some basic principles6.16.1 Some basic principlesSome basic principles

When a chemical reaction takes place, we consider the substances involved. Therefore, the reactants and products are the system.

A chemical system and its surroundingsA chemical system and its surroundings

Thermochemistry is a branch of thermodynamics that deals withthe heat involved with chemical and physical changes.Thermochemistry is a branch of thermodynamics that deals withthe heat involved with chemical and physical changes.

Thermodynamics is the study of heat and its transformations.Thermodynamics is the study of heat and its transformations.

Fundamental premiseFundamental premise

When energy is transferred from one object to another, it appears as work and/or as heat.

For our work we must define a system to study; everything else then becomes the surroundings.

The system is composed of particles with their own internal energies (E or U).Therefore the system has an internal energy. When a change occurs, theinternal energy changes.

State propertiesState property of system is described by giving its composition, temperature, and pressure.State property depends on the state of the system, not on the way the system reaches the state.

A system transferring energy as heat onlyA system transferring energy as heat only

When q is negative (-), the heat flows out of the system into surrounding.

When q is positive (+), the heat flows into the system from surrounding.

A system losing energy as work onlyA system losing energy as work only

Energy, E

Zn(s) + 2H+(aq) + 2Cl-(aq)

H2(g) + Zn2+(aq) + 2Cl-(aq)

DE<0

When w is negative (-), work done by system.

When w is positive (+), work done on system.

work done onsurroundings

q w+ = ∆E

+

+

-

-

-

-

+

+

+

-

depends on sizes of q and w

depends on sizes of q and w

• For q (heat): + means system gains heat, Endothermic;

- means system loses heat. Exothermic.

• For w (work): + means work done on system;

- means work done by system.

The Sign Conventions* for q, w and ∆EThe Sign Conventions* for q, w and ∆E

In any process, we are interested in the direction of heat flow and heat magnitude.

We express heat, q, in the unit of joules (SI unit) and kilojoules.

The joules is named for James Joule who carried out the precise thermodynamic measurement.

Traditionally, chemists use the calorie as an energy unit.

Calorie is the amount of heat needed to raise 1.00 g water 1 °C.

English physicist

1 cal = 4.184 J

1 kcal = 4.184 kJ

Magnitude of heatMagnitude of heat

6.2 Specific Heat Capacity and Heat Transfer6.2 Specific Heat Capacity and Heat Transfer

Finding the Quantity of Heat from Specific Heat Capacity

PROBLEM: A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 25 0C to 300 0C? The specific heat capacity (c) of Cu is 0.387 J/g*K.

SOLUTION:

PLAN: Given the mass, specific heat capacity and change in temperature, we can use q = c x mass x DT to find the answer. DT in 0C is the same as for K.

q =0.387 J

g*K125 g (300-25) 0Cx x = 1.33x104 J= 1.33x104 J

It is important to discuss the magnitude of heat flow in chemical reactions of phase changes.

The equation, q = C × ∆T, express the relationship between the magnitude of heat flow and temperature change. ∆T = Tfinal - Tinitial

The quantity C is known as heat capacity of the system, having a unit J/°C.

It is important to discuss the magnitude of heat flow in chemical reactions of phase changes.

The equation, q = C × ∆T, express the relationship between the magnitude of heat flow and temperature change. ∆T = Tfinal - Tinitial

The quantity C is known as heat capacity of the system, having a unit J/°C.

Coffee-cup calorimeterCoffee-cup calorimeter

• The heat given out by a reaction is absorbed by water.

• The mass of water can be determined.

• The heat capacity of water is 4.18 J/g •°C.

• The temperature change can be measured by the thermometer.

• Heat flow can be calculated for the reaction.

• Equation, q = mass × c × ∆T, express the relationship of heat flow and temperature change.

• The heat flow for the reaction is equal in magnitude, but opposite in sign to that measured by calorimeter

• The heat given out by a reaction is absorbed by water.

• The mass of water can be determined.

• The heat capacity of water is 4.18 J/g •°C.

• The temperature change can be measured by the thermometer.

• Heat flow can be calculated for the reaction.

• Equation, q = mass × c × ∆T, express the relationship of heat flow and temperature change.

• The heat flow for the reaction is equal in magnitude, but opposite in sign to that measured by calorimeter

Determining the Heat of a ReactionDetermining the Heat of a Reaction

PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.00 0C and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C. After stirring, the final temperature is 27.21 0C.

Calculate qsoln (in J).

(Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specfic heat capacity as water: d = 1.00 g/mL and c = 4.18 J/g*K)

PLAN: 1. We need to determine the limiting reactant from the net ionic equation.

2. The moles of NaOH and HCl as well as the total volume can be calculated.

3. From the volume we use density to find the mass of the water formed.

4. At this point, qsoln can be calculated using the eqaution, q = mass × c × ∆T.

• The heat divided by the M of water will give us the heat per mole of water formed.

Total volume after mixing = 0.0750 L

0.0750 L x 103 mL/L × 1.00 g/mL = 75.0 g of water

Q = mass x specific heat x DT

= 75.0 g × 4.18 J/g* 0C × (27.21-25.00) °C = 693 J

SOLUTION:

For NaOH 0.500 M × 0.0500 L = 0.0250 mol OH-

For HCl 0.500 M × 0.0250 L = 0.0125 mol H+

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)H+(aq) + OH-(aq) H2O(l)

HCl is the limiting reactant.

0.0125 mol of H2O will form during the rxn.

Determining the Heat of a ReactionDetermining the Heat of a Reaction

Take-home message: The schematic diagram of A bomb calorimeterTake-home message: The schematic diagram of A bomb calorimeter

• The heat given out by a reaction is absorbed by water.

• The mass of water can be determined.

• The heat capacity of water is 4.18 J/g • °C.

• The temperature change can be measured by the thermometer.

• Heat flow can be calculated for the reaction.

• Equation, q = mass × c ×∆T, express the relationship of heat flow and temperature change.

• The heat flow for the reaction is equal in magnitude, but opposite in sign to that measured by calorimeter

6.6 Calorimetery6.6 Calorimetery

Calculating the Heat of CombustionCalculating the Heat of Combustion

PROBLEM: A manufacturer claims that its new dietetic dessert has “fewer than 10 KiloCalories per serving.”

To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it inO2 (the heat capacity of the calorimeter = 8.15 kJ/K).

The temperature increases 4.937 0C. Is the manufacturer’s claim correct?

SOLUTION:

PLAN: - q sample = qcalorimeter

qcalorimeter= heat capacity × ∆T

= 8.151 kJ/K × 4.937 K

= 40.24 kJ

40.24 kJ ×kcal

4.18 kJ= 9.63 Kilocalories

The manufacturer’s claim is true.The manufacturer’s claim is true.

A cooling curve for the conversion of gaseous water to ice.

Five stages – vapor cools, vapor condenses (constant temperature),vapor condenses (constant temperature), liquid water cools, liquid water freezes (constant tempterature),liquid water freezes (constant tempterature), solid water cools

6.3 Energy and changes of state6.3 Energy and changes of state

Within a phase, a change in heat is accompanied by a change in temperature which is associated with a change in average Ek as the most probable speed of the molecules changes.

Quantitative Aspects of Phase ChangesQuantitative Aspects of Phase Changes

During a phase change, a change in heat occurs at a constant temperature, which is associated with a change in Ep, as the average distance between molecules changes.

q = (amount)(molar heat capacity)(∆T)

q = (amount)(enthalpy of phase change)

The Meaning of Enthalpy

The Meaning of Enthalpy

w = - P∆Vw = - P∆V

∆ H = DE + P ∆ V∆ H = DE + P ∆ V

qp = ∆ E + P ∆ V = ∆ Hqp = ∆ E + P ∆ V = ∆ H

∆ H ≈ ∆ E in

1. Reactions that do not involve gases.

2. Reactions in which the number of moles of gas does not change.

3. Reactions in which the number of moles of gas does change but q is >>> P ∆ V.

H = E + PVH = E + PVWhere H is enthalpy

6.4 The first law of thermodynamics6.4 The first law of thermodynamics

Enthalpy diagrams for exothermic and endothermic processes.

Enth

alpy

, H

Enth

alpy

, H

CH4 + 2O2

CO2 + 2H2O

Hinitial

HinitialHfinal

Hfinal

H2O(l)

H2O(g)

heat out heat in∆H < 0 ∆H > 0

A Exothermic processq < 0, Hproduct < H reactant

B Endothermic process

q > 0, Hproduct > H reactant

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g)

If a chemical reaction occurs under a constant pressure, the difference in enthalpy between product and reactant equals the heat flow for the reaction.

Qreaction at a constant pressure = ∆E + P∆V = ∆HQreaction at a constant pressure = ∆E + P∆V = ∆H

Drawing Enthalpy Diagrams and Determining the Sign of ∆HDrawing Enthalpy Diagrams and Determining the Sign of ∆H

PROBLEM: In each of the following cases, determine the sign of DH, state whether the reaction is exothermic or endothermic, and draw and enthalpy diagram.

SOLUTION:

PLAN: Determine whether heat is a reactant or a product. As a reactant, the products are at a higher energy and the reaction is endothermic. The opposite is true for an exothermic reaction

(a) H2(g) + 1/2O2(g) H2O(l) + 285.8kJ

(b) 40.7kJ + H2O(l) H2O(g)

(a) The reaction is exothermic.(a) The reaction is exothermic.

H2(g) + 1/2O2(g) (reactants)

H2O(l) (products)EXOTHERMICEXOTHERMIC

(products)H2O(g)

(reactants)H2O(l)∆H = -285.8kJ ∆H = +40.7kJENDOTHERMICENDOTHERMIC

(b) The reaction is endothermic.(b) The reaction is endothermic.

Some Important Types of Enthalpy Change

Heat of combustion (∆Hcomb)Heat of combustion (∆Hcomb)

Heat of formation (∆Hf)Heat of formation (∆Hf)

Heat of fusion (∆Hfus)Heat of fusion (∆Hfus)

Heat of vaporization (∆Hvap)Heat of vaporization (∆Hvap)

C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g)

K(s) + 1/2Br2(l) KBr(s)

NaCl(s) NaCl(l)

C6H6(l) C6H6(g)

6.5 Enthalpy change for chemical reactionsThermal chemical reaction – shows the enthalpy relationship between reactants and products

6.5 Enthalpy change for chemical reactionsThermal chemical reaction – shows the enthalpy relationship between reactants and products

Rules of thermochemistryv The magnitude of ∆H is directly proportional to the amount of reactants or products.v ∆H for a reaction is equal in magnitude but opposite in sign for the reverse reaction.v The value of for a reaction is the same whether it occurs in one step or multi-steps.v Hess law ∆H = ∆H1 + ∆H2 + ∙∙∙∙

Using the Heat of Reaction (∆Hrxn) to Find AmountsUsing the Heat of Reaction (∆Hrxn) to Find Amounts

SOLUTION:PLAN:

PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by

If aluminum is produced this way, how many grams of aluminum canform when 1.000x103 kJ of heat is transferred?

Al2O3(s) 2Al(s) + 3/2O2(g) ∆Hrxn = 1676 kJ

Heat (kJ)Heat (kJ)

mol of Almol of Al

g of Alg of Al

1676 kJ = 2 mol Al

X M

1.000x103 kJ x 2 mol Al

1676 kJ

26.98 g Al

1 mol Al

= 32.20 g Al

6.7 Hess’s Law --- To Calculate an Unknown ∆H6.7 Hess’s Law --- To Calculate an Unknown ∆H

SOLUTION:

PLAN:

PROBLEM: Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation:

CO(g) + NO(g) CO2(g) + 1/2N2(g) ∆ H = ?

Given the following information, calculate the unknown ∆ H:

Equation A: CO(g) + 1/2O2(g) CO2(g) ∆ HA = -283.0 kJ

Equation B: N2(g) + O2(g) 2NO(g) ∆ HB = 180.6 kJ

Equations A and B have to be manipulated by reversal and/or multiplication by factors in order to sum to the first, or target, equation.

Multiply Equation B by 1/2 and reverse it.

∆ HB = -90.3 kJ

CO(g) + 1/2O2(g) CO2(g) ∆ HA = -283.0 kJNO(g) 1/2N2(g) + 1/2O2(g)

∆ Hrxn = -373.3 kJCO(g) + NO(g) CO2(g) + 1/2N2(g)

6.8 Standard enthalpies of Formation6.8 Standard enthalpies of Formation

PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include DH 0f.

SOLUTION:

PLAN:

(a) Silver chloride, AgCl, a solid at standard conditions.

(b) Calcium carbonate, CaCO3, a solid at standard conditions.

Use the table of heats of formation for values.

(c) Hydrogen cyanide, HCN, a gas at standard conditions.

∆H 0f = -127.0 kJ(a) Ag(s) + 1/2Cl2(g) AgCl(s)

∆H 0f = -1206.9 kJ(b) Ca(s) + C(graphite) + 3/2O2(g) CaCO3(s)

∆H 0f = 135 kJ(c) 1/2H2(g) + C(graphite) + 1/2N2(g) HCN(g)

The general process for determining ∆H 0rxn from ∆H0f values.The general process for determining ∆H 0rxn from ∆H0f values.

Ent

halp

y, H

Elements

Reactants

Products

∆H0rxn = Σ m∆H0

f(products) - Σ n∆H0f(reactants)

deco

mpo

sitio

n-∆H0

f ∆H0f

form

atio

n∆H0

rxn

Hinitial

Hfinal

Calculating the Heat of Reaction from Heats of FormationCalculating the Heat of Reaction from Heats of Formation

SOLUTION:

PLAN:

PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:

Calculate DH0rxn from DH 0f values.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Look up the DH0f values and use Hess’s Law to find DHrxn.

DHrxn = S mDH0f (products) - S nDH0

f (reactants)

DHrxn = [4(DH0f NO(g) + 6(DH0

f H2O(g)] - [4(DH0f NH3(g) + 5(DH0

f O2(g)]

= (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) -

[(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]

∆Hrxn = -906 kJ∆Hrxn = -906 kJ Lectrue 3

Specific Heat Capacities of Some Elements, Compounds, and MaterialsSpecific Heat Capacities of Some Elements, Compounds, and Materials

Specific Heat Capacity (J/g*K)

SubstanceSpecific Heat Capacity (J/g*K)

Substance

Compounds

water, H2O(l)ethyl alcohol, C2H5OH(l)ethylene glycol, (CH2OH)2(l)carbon tetrachloride, CCl4(l)

4.184

2.46

2.42

0.864

Elementsaluminum, Algraphite,Ciron, Fecopper, Cugold, Au

0.9000.7110.4500.3870.129

woodcementglassgranitesteel

Materials1.760.880.840.790.45

Selected Standard Heats of Formation at 250C(298K)Selected Standard Heats of Formation at 250C(298K)

Formula ∆H0f(kJ/mol)

calciumCa(s)CaO(s)CaCO3(s)

carbonC(graphite)C(diamond)CO(g)CO2(g)CH4(g)CH3OH(l)HCN(g)CSs(l)

chlorineCl(g)

0-635.1

-1206.9

01.9

-110.5-393.5-74.9

-238.6135

87.9

121.0

hydrogen

nitrogen

oxygen

Formula ∆H0f(kJ/mol)

H(g)H2(g)

N2(g)NH3(g)NO(g)

O2(g)O3(g)H2O(g)

H2O(l)

Cl2(g)

HCl(g)

0

0

0

-92.30

218

-45.990.3

143-241.8

-285.8

107.8

Formula ∆H0f(kJ/mol)

silverAg(s)AgCl(s)

sodium

Na(s)Na(g)NaCl(s)

sulfurS8(rhombic)S8(monoclinic)SO2(g)

SO3(g)

0

0

0

-127.0

-411.1

2-296.8

-396.0