TOPIC 5ENERGETICS/THERMOCHEMISTRY

5.1MEASURING ENERGY CHANGES

INTERNATIONAL-MINDEDNESS

The SI unit of temperature is the Kelvin (K), but the Celsius scale (◦C), which has

the same incremental scaling, is commonly used in most countries. The

exception if the USA which continues to use the Fahrenheit scale (◦F) for all

non-scientific communication.

UNDERSTANDING/KEY IDEA 5.1.A

Heat is a form of energy.

ENERGY• All chemical reactions are

accompanied by energy changes.• Energy is a measure of the ability to

do work, that is to move an object against an opposing force.

• Examples: heat, light, sound, electricity and chemical energy which is the energy released or absorbed during chemical reactions.

HEAT• Heat is a form of energy which is

transferred as a result of a difference in temperature and produces an increase in disorder of the behavior of particles.

• Heat increases the average kinetic energy of the molecules in a disordered fashion.

Needed definitions• Enthalpy

• Heat content of a substance (at constant pressure)

• Enthalpy is also the internal energy stored in the reactants.

• The absolute value for the enthalpy of reactants and products cannot be known, but the what can be measured is the difference between the two.

• System – area of interest (Example: beaker and its contents)

• Surroundings – everything else in the universe

• An “open” system can exchange matter and energy with the surroundings.

• A “closed” system can only exchange energy, not matter, with the surroundings.

• The joule (J) is the SI unit for energy and work.

Difference between terms?

• Enthalpy: total potential energy in a system @ constant pressure• Energy, not always heat• State of the system• ΔH = change in enthalpy

• q = amount of heat transferred to a system • Always heat

• At constant pressure, q = ΔH

Temperature

• Def – measure of the average kinetic energy of the particles

• Temperature increase depends upon:• Mass of the object• Amount of heat added• Nature of the substance

• Different substances need different amounts of heat to increase the temp of a unit mass by 1K or 1ºC.

EXOTHERMIC REACTIONS

A reaction which results in a transfer of energy from the system to the surroundings.

• Heat is given off or produced.• Products have less energy or heat

content than the reactants.• ΔH is negative.• The bonds in the products are stronger

than the bonds in the reactants. • Note that the lower they are on the graph, the

stronger the bond and the lower the energy.

Exothermic reactionEnthalpy diagram

reactants

enthalpy H ΔH = negative

products

extent of reaction

ENDOTHERMIC REACTIONS

• A reaction which results in a transfer of energy from the surroundings to the system.• Heat is absorbed.• Reactants have less energy than the

products.• ΔH is positive.• The bonds in the reactants are stronger

than those in the products.• Note that the lower they are on the graph, the

stronger the bond and the lower the energy.

Endothermic reactionEnthalpy diagram

products

enthalpy H ΔH = positive

reactants

extent of reaction

All combustion reactions are exothermic processes.

All neutralization reactions are exothermic processes.

• There is a natural direction for change.

• The direction of change is in the direction of lower stored energy.• Large - Δ H (exothermic) = likely to be

spontaneous

• +Δ H (endothermic) = not likely to be spontaneous

Spontaneity of Reactions

Enthalpy Diagram for Hydrogen Peroxide (H2O2)

H2(g) + O2(g)

ΔH1

H2O2(l)

ΔH3

ΔH2Enthalpy

H2O(l) + 1/2 O2(g)

In this diagram, hydrogen peroxide in the middle is stable compared to H2 and O2, but unstable compared to the decomposition of water and oxygen on the bottom line.

UNDERSTANDING/KEY IDEA 5.1.E

The enthalpy change (ΔH) for chemical reactions is indicated in kJ mol-1.

Definitions

• Enthalpy change of combustion (ΔHc ◦) is

the enthalpy change for the complete combustion of one mole of a substance in its standard state in excess oxygen under standard conditions. (Always exothermic)

• Enthalpy change of formation (ΔHf ◦) is the

enthalpy change that occurs when one mole of a compound is formed from its elements in their standard states.

UNDERSTANDING/KEY IDEA 5.1.F

ΔH values are usually expressed under standard conditions, given by ΔH◦, including the standard states.

GUIDANCE

Standard state refers to the normal, most pure stable state of a substance measured at 100 kPa (1 atm). Temperature is not part of the definition of standard state, but 298K is commonly given as the temperature of interest.

• Standard enthalpy change of reaction ΔHº• This is the heat or enthalpy change of a

reaction when carried out at standard conditions.• Temperature = 298K or 25ºC• Pressure = 101.3 kPa (1atm)• Solution concentration = 1mol dm-3 (1M)• All substances are in their standard states

(how they are found in nature)

Calculating ΔH

• This chapter mainly deals with calculating ΔH in a variety of ways.

• The first way to calculate ΔH is by using thermochemical equations where you are given the ΔH for the equation and asked to calculate ΔH for varying initial conditions.

THERMOCHEMICALEQUATIONS

Combustion of MethaneCH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -890 kJ mol-1

Photosynthesis6CO2(g) + 6H2O(l) C6H12O6(aq) + 6O2(g) ΔH = +2802.5 kJ mol-1

Thermite Reaction2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(s) ΔH = -841 kJ mol-1

Working a thermochemical equation problem

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -890 kJ mol-1

What is the heat released when 32.0g of oxygen is burned?

1. Change grams to moles32.0g x mol/32.00g = 1.00 mol O2

2. Multiply by the heat and divide by the given’s coefficient.

1.00 mol O2 x -890 kJ/ 2 mol O2 = -445 kJ

You must give the “state” symbols such as (s), (g), (l), (aq) in thermochemical equations because energy changes depend upon the state of the reactants and the products.

Another way to calculate ΔH

• The second method of calculating ΔH is using the specific heat equation and calorimetry.

• You will solve for heat “q” and divide by the number of moles to determine ΔH.

Specific Heat Capacity

• The following relationship allows the heat change in a material to be calculated from the temperature change.

q = mcΔT heat = mass x specific heat x ΔT

Specific heat capacity (c) is the heat needed to increase the temperature of a unit mass (usually 1g) by 1K or 1ºC.

Specific heat of water c = 4.18 J K-1g-1

• The actual amount of heat absorbed or produced in a chemical reaction depends upon several factors:• Nature of the reactants and products• The amount or concentration of the

reactants. (The greater the amount that reacts, the greater the heat change.)

• The states of the reactants and products. Changing states involves an enthalpy change so this will affect the total heat change.

• The temperature of the reaction.

GUIDANCE

Students can assume the specific heat and density of aqueous solutions are equal to those of water, but be aware of this limitation.

APPLICATION/SKILLSBe able to evaluate calorimetry experiments for the enthalpy of a reaction.There are 5 types of calorimetry experiments you should be familiar with.

1. Specific heat of metals 2. Combustion reactions 3. Reactions in solution (extrapolation)4. Neutralization reactions5. Reactions with constant volumes

GUIDANCE

Heat losses to the environment and the heat capacity of the calorimeter in experiments should be considered.

How to calculate heat changes from temp changes.

• When heat released by an exothermic reaction is absorbed by water, the temperature of the water increases.

• The heat produced by the reaction can be calculated if it is assumed that all the heat is absorbed by the water.

heat change of rxn = -heat change of water

= - mH2O x cH2O x ΔTH2O

As the water has gained the heat produced by the reaction, the heat change of

the reaction is negative when the temperature of the water increases.

Example Problem – Specific Heat of a Metal

• How much heat is released when 10.0g of copper with a specific heat of 0.385 J/gºC is cooled from 85.0ºC to 25ºC?

• Use q = mcΔT

• m=10.0g, c=0.385 J/gºC, ΔT=60.0ºC• q = (10.0)(.385)(60.0) = 231 J or

.231kJ

Combustion of Alcohol Enthalpy Change Demo

(Figure 5.7 pg 173)

Example Problem – Combustion #1

• Calculate the enthalpy of combustion of ethanol from the following data. Assume all heat from the reaction is absorbed by the water. Compare your value with the IB Data booklet value and suggest reasons for any differences.

• Mass of water in Cu Calorimeter = 200.00g• Temperature increase in water = 13.00ºC• Mass of ethanol burned = 0.45g heat change of rxn = - mH2O x cH2O x ΔTH2O

= -(200.00g)(4.18Jg-1ºC-1)(13.00ºC)

= -10868 J = -10870 J = -10.87 kJ

The heat calculated above is the heat gained by the water which is also the heat lost by the combustion of the ethanol.

Since ΔHc is per mole of substance combusted, you have to find moles of ethanol burned and divide the heat it lost by the moles.

To find moles, divide mass by molar mass:

0.45g / 46.08 g mol-1 = .0098mol

ΔHc = -10.87kJ / .0098mol = -1109 kJ/mol

Using sig figs, the answer is limited by the mass = -1100 kJ/mol

• The IB Data booklet value is -1367 kJ/mol.

• What are some reasons for the difference in values?• Not all the heat produced by the

combustion is transferred to the water.• Some is needed to heat the Cu

calorimeter and some has passed to the surroundings.

• The combustion of ethanol is unlikely to be complete due to the limited oxygen available.

• Some alcohol could have evaporated.

Example Problem – Combustion #2

• When 0.824 g of ethanol was burned, it produced a temperature rise of

15.7 K in 275g of water. • a. Draw the calorimetry set up.• b. Write the ΔHc reaction.• c. Calculate the ΔH for the reaction.• d. The Data Book value is -1367 kJ/mol.

Include three reasons for the discrepancy.

• It is important to record qualitative as well as quantitative data.

• When asked to evaluate experiments and suggest improvements, avoid giving trivial answers such as incorrect measurements.

• Incomplete combustion can be reduced by burning the fuel in oxygen and heat loss can be reduced by insulating the apparatus.

Example Problem – Reaction in Solution

• The enthalpy changes of reaction in solution can be calculated by carrying out the reaction in an insulated system such as a polystyrene cup or Styrofoam calorimeter.

• The heat released or absorbed by the reaction can be measured by the change in temp of the water.

• The largest source of error in this type of experiment is heat loss to the environment.

• To compensate for the heat lost by the water in exothermic reactions to the surroundings as the reaction proceeds, a plot of temperature vs time can be drawn.

• We want the highest temperature that would have been reached had the reaction been instantaneous.

• If we graph the temp vs time, we can extrapolate back to determine the highest temperature.

T2

extrapolation at same rate of cooling

T1

∆T ∆T for rxn = T2-T

0

Temp (K or ⁰C)

T0

point where reactants are mixed

time (s)

Compensating for Heat Loss

T0 = initial temp of reactants T1 = highest temp actually reached T2 = temp that would have been reached if no heat were lost to the surroundings

50.0 cm3 of 0.200 mol dm-3 copper (II) sulfate was placed in a polystyrene cup. After 2 min, 1.20 g of powdered zinc was added. The temperature was taken every 30 seconds and the following graph was obtained. Calculate the enthalpy change for the reaction taking place.

T2

extrapolation at same rate of cooling

T1

∆T ∆T for rxn = T2-T

0

Temp (⁰C)

T0

point where zinc was added

time (min)

Temp Change for the Reaction of Copper II Sulfate and Zinc T0 = initial temp of reactants = 17⁰C T1 = highest temp actually reached = 26.5 ⁰C T2 = temp that would have been reached if no heat were lost to the surroundings = 27.4 ⁰C

• From the graph, we are able to extrapolate back and determine that T2 would have been 27.4⁰C.

• That gives us a ∆T of 27.4⁰C - 17.0⁰C = 10.4⁰C• Next we should recognize that we are mixing

two given amounts which gives us a limiting reactant problem.

CuSO4(aq) + Zn(s) Cu(s) + ZnSO4(aq)

1.20 g Zn x mol/65.4g = 0.0183mol Zn given

50.0cm3 x 1dm3/1000cm3 x 0.200mol/dm3 = 0.0100mol CuSO4 given

0.0183mol Zn x 1/1 = 0.0183mol CuSO4 needed 0.0100mol CuSO4 x 1/1 = 0.0100mol Zn needed We need 0.0183 mol CuSO4 and are only given 0.0100 mol so CuSO4 is

the limiting reactant.

• We know that the reaction will stop after the 0.0100 mol of CuSO4 is used up so the heat evolved is per 0.0100 mol of CuSO4.

• q = mc∆T = (50.0 cm3 x 1.00 g/cm3)(4.18J/g⁰C)(10.4⁰C)

= 2170 J = 2.17 kJ

• ∆H = -2.17kJ / 0.0100 mol = -217 kJ mol-1

Comments on problem• Why was there a wait time at the beginning?• The highest temp was reached 5 min after the

zinc had been added. What exactly was happening at this point?

• Copper II sulfate solution is blue. Zinc sulfate solution is colorless. Zinc is a silver – grey metal and copper is a reddish brown metal. State what you would think you would observe as the reaction proceeds.

• What are three assumptions you have made in arriving at your answer?

• The literature value is -218 kJ/mol. Comment on the validity of your assumptions you stated.

Example problem - Neutralization

• 50.0 cm3 of 1.00 mol/dm3 hydrochloric acid was added to 50.0 cm3 of 1.00 mol/dm3 sodium hydroxide solution in a polystyrene beaker. The initial temperature of both solutions was 16.7ºC. After stirring and accounting for the heat loss, the highest temperature reached was 23.5ºC. Calculate the enthalpy change for the reaction.

• Step 1 – write the equation HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

• Step 2 – calculate molar quantities Anytime you are given a volume and a molarity,

multiply them together (watch your units) to get moles. Convert cm3 to dm3 first.

50.0cm3 x 1dm3/1000cm3 = 0.0500 dm3

Then multiply the volume by the molarity. 0.0500dm3 x 1.00 mol/dm3 = 0.0500 mol

• Step 3 – calculate the heat evolved• Assumptions:

• The solution has the same density as water which is 1.00 g/cm3.

• The solution has the same specific heat as water which is 4.18 J/gK.

• Total volume of solution 50.0 cm3 + 50.0 cm3 = 100.0 cm3

• Using the density of water, calculate the mass of the solution

100.0 cm3 x 1.00 g/cm3 = 100.0 g• Temperature change 23.5ºC – 16.7ºC = 6.8ºC = 6.8 K• The heat evolved from the reaction = mcΔT. q = (100.g)(4.18 J/gK)(6.8 K) = 2840 J = -2.8kJ• The ΔH is per mole so ΔH = -2.8kJ/.0500mol = -56 kJ/mol

Example Problem – Constant Volume Neutralization

• The experiment involves the neutralization of sodium hydroxide with sulfuric acid to form water and sodium sulfate.

• Assumptions:• The NaOH and H2SO4 are 1M (1mol dm-3).• The total volume is kept constant at 120 cm3

with varying volumes of NaOH and H2SO4.• No heat is lost from the system.• All heat is transferred to the water.• Assume 120cm3 of solution contains 120cm3

of water.

• Your task is to evaluate the following when the temperature changes are measured when different volumes of NaOH and H2SO4 are mixed:• Determine the volumes of the solutions which

produce the largest increase in temperature.• Calculate the heat produced by the reaction when

the maximum temperature was produced.• Calculate the heat produced for one mole of NaOH.• Calculate the %error and suggest a reason for the

discrepancy between the experimental and literature values.

• The literature value at standard conditions is -57.5 kJ/mol-1.

353433323130292827 2625

0 20 40 60 80 100 120

volume of NaOH cm3

Temp ºC

□

• The highest increase in temp is produced at 80 cm3 of NaOH which means that there is 120-80 = 40 cm3 of H2SO4.• The maximum temp where the lines cross on

the graph was 33.5ºC. q = -mcΔT = -(120.0g)(4.18J/gºC)(33.5ºC-25.0ºC) = -4264J

• To calculate the heat produced per mole: 80.0cm3 x 1dm3/1000cm3 x 1mol/dm3 = .0800mol

ΔH = -4264J/.0800mol = -53300Jmol-1 = -53.3kJmol-1

• Percent Error %error = (-57.5--53.3)/-57.5 x 100% = 7%• There are uncertainties in temp, vol and

concentration measurements and it is assumed all heat is transferred to the water and no heat is lost from the system.

Final comments on calculating ∆H

• There are 5 types of experiments that you should be familiar with.• Specific heat of metals• Combustion calorimetry• Extrapolation for heat loss• Neutralization rxns• Finding answers from a graph when

2 solutions are mixed and volumes are kept constant.

• Make sure you always record qualitative as well as quantitative results.

• Be sure to state the assumptions you are making which are often that no heat is lost from the system, all heat is transferred to the water and that the mass of the solution is assumed to be water so you are using the specific heat of water in your calculations.