ENTHALPY CHANGES A guide for A level students KNOCKHARDY PUBLISHING.

ENTHALPY ENTHALPY CHANGESCHANGES

A guide for A level studentsA guide for A level students

KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING

ENTHALPY CHANGESENTHALPY CHANGES

INTRODUCTION

This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards.

Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available.

Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm

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CONTENTS• Thermodynamics

• Enthalpy changes

• Standard enthalpy values

• Standard enthalpy of formation

• Standard enthalpy of combustion

• Enthalpy of neutralisation

• Bond dissociation enthalpy

• Hess’s law

• Calculating enthalpy changes using bond enthalpies

• Calculating enthalpy changes using enthalpy of formation values

• Calculating enthalpy changes using enthalpy of combustion values

• Practical measurement

• Check list

Before you start it would be helpful to…

• be able to balance equations

• be confident with simple arithmetical operations


THERMODYNAMICSTHERMODYNAMICS

First Law Energy can be neither created nor destroyed but It can be converted from one form to another

Energychanges all chemical reactions are accompanied by some form of energy change

changes can be very obvious (e.g. coal burning) but can go unnoticed

Exothermic Energy is given out

Endothermic Energy is absorbed

Examples Exothermic combustion of fuels

respiration (oxidation of carbohydrates)

Endothermic photosynthesis

thermal decomposition of calcium carbonate

THERMODYNAMICS - THERMODYNAMICS - ENTHALPY CHANGESENTHALPY CHANGES

Enthalpy a measure of the heat content of a substance at constant pressure

you cannot measure the actual enthalpy of a substance

you can measure an enthalpy CHANGE

written as the symbol , “delta H ”

Enthalpy change () = Enthalpy of products - Enthalpy of reactants






Enthalpy of reactants > products

= - ive

EXOTHERMIC Heat given out

REACTION CO-ORDINATE

EN

TH

AL

PY







Enthalpy of reactants > products Enthalpy of reactants < products

= - ive = + ive

EXOTHERMIC Heat given out ENDOTHERMIC Heat absorbed


EN

TH

AL

PY


EN

TH

AL

PY


Why astandard? enthalpy values vary according to the conditions

a substance under these conditions is said to be in its standard state ...

Pressure:- 100 kPa (1 atmosphere)

A stated temperature usually 298K (25°C)

• as a guide, just think of how a substance would be under normal lab conditions

• assign the correct subscript [(g), (l) or (s) ] to indicate which state it is in

• any solutions are of concentration 1 mol dm-3

• to tell if standard conditions are used we modify the symbol for .

Enthalpy Change Standard Enthalpy Change(at 298K)

STANDARD ENTHALPY CHANGESSTANDARD ENTHALPY CHANGES

STANDARD ENTHALPY OF FORMATIONSTANDARD ENTHALPY OF FORMATION

Definition The enthalpy change when ONE MOLE of a compound is formed in its standard state from its elements in their standard states.

Symbol °f

Values Usually, but not exclusively, exothermic

Example(s) C(graphite) + O2(g) ———> CO2(g)

H2(g) + ½O2(g) ———> H2O(l)

2C(graphite) + ½O2(g) + 3H2(g) ———> C2H5OH(l)

Notes Only ONE MOLE of product on the RHS of the equation

Elements In their standard states have zero enthalpy of formation.

Carbon is usually taken as the graphite allotrope.

Definition The enthalpy change when ONE MOLE of a substance undergoes complete combustion under standard conditions. All reactants and products are in

their standard states.

Symbol °c

Values Always exothermic

Example(s) C(graphite) + O2(g) ———> CO2(g)

H2(g) + ½O2(g) ———> H2O(l)

C2H5OH(l) + 3O2(g) ———> 2CO2(g) + 3H2O(l)

Notes Always only ONE MOLE of what you are burning on the LHS of the equation

To aid balancing the equation, remember...you get one carbon dioxide molecule for every carbon atom in the original and one water molecule for every two hydrogen atoms

When you have done this, go back and balance the oxygen.

STANDARD ENTHALPY OF COMBUSTIONSTANDARD ENTHALPY OF COMBUSTION

Definition The enthalpy change when ONE MOLE of water is formed from its ions in dilute solution.

Values Exothermic

Equation H+(aq) + OH¯(aq) ———> H2O(l)

Notes A value of -57kJ mol-1 is obtained when strong acids react with strong alkalis.

See later slides for practical details of measurement

ENTHALPY OF NEUTRALISATIONENTHALPY OF NEUTRALISATION

Definition Energy required to break ONE MOLE of gaseous bonds to form gaseous atoms.

Values Endothermic - Energy must be put in to break any chemical bond

Examples Cl2(g) ———> 2Cl(g) O-H(g) ———> O(g) + H(g)

Notes • strength of bonds also depends on environment; MEAN values quoted • making bonds is exothermic as it is the opposite of breaking a bond • for diatomic gases, bond enthalpy = 2 x enthalpy of atomisation • smaller bond enthalpy = weaker bond = easier to break

BOND DISSOCIATION ENTHALPYBOND DISSOCIATION ENTHALPY

Definition Energy required to break ONE MOLE of gaseous bonds to form gaseous atoms.

Values Endothermic - Energy must be put in to break any chemical bond

Examples Cl2(g) ———> 2Cl(g) O-H(g) ———> O(g) + H(g)

Notes • strength of bonds also depends on environment; MEAN values quoted • making bonds is exothermic as it is the opposite of breaking a bond • for diatomic gases, bond enthalpy = 2 x enthalpy of atomisation • smaller bond enthalpy = weaker bond = easier to break

Mean Values H-H 436 H-F 562 N-N 163 C-C 346 H-Cl 431 N=N 409

C=C 611 H-Br 366 NN 944 CC 837 H-I 299 P-P 172

C-O 360 H-N 388 F-F 158 C=O 743 H-O 463 Cl-Cl 242 C-H 413 H-S 338 Br-Br 193 C-N 305 H-Si 318 I-I 151 C-F 484 P-H 322 S-S 264 C-Cl 338 O-O 146 Si-Si 176

BOND DISSOCIATION ENTHALPYBOND DISSOCIATION ENTHALPY

Average (mean) values are quoted because the actual value depends on the environment of the bond i.e. where it is in the molecule

UNITS = kJ mol-1

“The enthalpy change is independent of the path taken”

How The enthalpy change going fromA to B can be found by adding thevalues of the enthalpy changes forthe reactions A to X, X to Y and Y to B.

r = + 2 + 3

HESS’S LAWHESS’S LAW


How The enthalpy change going fromA to B can be found by adding thevalues of the enthalpy changes forthe reactions A to X, X to Y and Y to B.

r = + 2 + 3

If you go in the opposite direction of an arrow, you subtract the value ofthe enthalpy change.

e.g. 2 = - + r - 3

The values of and3 have been subtracted because the route

involves going in the opposite direction to their definition.



Use applying Hess’s Law enables one to calculate enthalpy changes fromother data, such as...

changes which cannot be measured directly e.g. Lattice Enthalpy

enthalpy change of reaction from bond enthalpy

enthalpy change of reaction from °c

enthalpy change of formation from °f


Theory Imagine that, during a reaction, all the bonds of reacting species are brokenand the individual atoms join up again but in the form of products. Theoverall energy change will depend on the difference between the energyrequired to break the bonds and that released as bonds are made.

energy released making bonds > energy used to break bonds ... EXOTHERMIC

energy used to break bonds > energy released making bonds ... ENDOTHERMIC

Enthalpy of reaction from bond enthalpiesEnthalpy of reaction from bond enthalpies

Step 1 Energy is put in to break bonds to form separate, gaseous atoms

Step 2 The gaseous atoms then combine to form bonds and energy is releasedits value will be equal and opposite to that of breaking the bonds

Applying Hess’s Law r = Step 1 + Step 2


= bond enthalpies – bond enthalpies of reactants of products

= bond enthalpies – bond enthalpies of reactants of products

Alternative view

Step 1 Energy is put in to break bonds to form separate, gaseous atoms.

Step 2 Gaseous atoms then combineto form bonds and energy is released; its value will be equaland opposite to that of breaking the bonds

r = Step 1 - Step 2

Because, in Step 2 the route involvesgoing in the OPPOSITE DIRECTIONto the defined change of bond enthalpy,it’s value is subtracted.

SUM OFTHE BOND ENTHALPIES OF THE REACTANTS

REACTANTS

PRODUCTS

ATOMS

SUM OFTHE BOND ENTHALPIES OF THE PRODUCTS

Calculate the enthalpy change for the hydrogenation of ethene


Calculate the enthalpy change for the hydrogenation of ethene


2 1 x C=C bond @ 611 = 611 kJ4 x C-H bonds @ 413 = 1652 kJ1 x H-H bond @ 436 = 436 kJ

Total energy to break bonds of reactants = 2699 kJ

3 1 x C-C bond @ 346 = 346 kJ6 x C-H bonds @ 413 = 2478 kJ

Total energy to break bonds of products = 2824 kJ

Applying Hess’s Law 1 = 2 – 3 = (2699 – 2824) = – 125 kJ

If you formed the products from their elements you should need the same amounts of every substance as if you formed the reactants from their elements.

Enthalpy of formation tends to be an exothermic process

Enthalpy of reaction from enthalpies of formationEnthalpy of reaction from enthalpies of formation

Step 1 Energy is released as reactantsare formed from their elements.

Step 2 Energy is released as productsare formed from their elements.

r = - Step 1 + Step 2

or Step 2 - Step 1

In Step 1 the route involves going in the OPPOSITE DIRECTION to the defined enthalpy change, it’s value is subtracted.

SUM OFTHE ENTHALPIES OF FORMATION OF

THE REACTANTS

REACTANTS

PRODUCTS

ELEMENTS

SUM OFTHE ENTHALPIES OF FORMATION OF THE PRODUCTS


= f of products – f of reactants = f of products – f of reactants

Sample calculationCalculate the standard enthalpy change for the following reaction, given that the standard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286, +33 and -173 kJ mol-1 respectively; the value for oxygen is ZERO as it is an element

2H2O(l) + 4NO2(g) + O2(g) ———> 4HNO3(l)

By applying Hess’s Law ... The Standard Enthalpy of Reaction °r will be...

PRODUCTS REACTANTS

[ 4 x f of HNO3 ] minus [ (2 x f of H2O) + (4 x f of NO2) + (1 x f of O2) ]

°r = 4 x (-173) - 2 x (-286) + 4 x (+33) + 0

ANSWER = - 252 kJ


= f of products – f of reactants = f of products – f of reactants

Enthalpy of reaction from enthalpies of combustionEnthalpy of reaction from enthalpies of combustion

If you burned all the products you should get the same amounts of oxidation products such a CO2 and H2O as if you burned the reactants.

Enthalpy of combustion is an exothermic process

= c of reactants – c of products = c of reactants – c of products


Step 1 Energy is released as reactantsundergo combustion.

Step 2 Energy is released as productsundergo combustion.

r = Step 1 - Step 2

Because, in Step 2 the route involves going in the OPPOSITE DIRECTION to the defined change of Enthalpy of Combustion, it’s value is subtracted.

SUM OFTHE ENTHALPIES OF COMBUSTION OF THE REACTANTS

REACTANTS

PRODUCTS

OXIDATIONPRODUCTS

SUM OFTHE ENTHALPIES OF

COMBUSTION OF THE PRODUCTS


Sample calculationCalculate the standard enthalpy of formation of methane; the standard enthalpies of combustion of carbon, hydrogen and methane are -394, -286 and -890 kJ mol-1 .

C(graphite) + 2H2(g) ———> CH4(g)

By applying Hess’s Law ... The Standard Enthalpy of Reaction °r will be...

REACTANTS PRODUCTS

[ (1 x c of C) + (2 x c of H2) ] minus [ 1 x c of CH4O ]

°r = 1 x (-394) + 2 x (-286) - 1 x (-890)

ANSWER = - 74 kJ mol-1

= c of reactants – c of products = c of reactants – c of products

Calorimetry involves the practical determination of enthalpy changesusually involves heating (or cooling) known amounts of waterwater is heated up reaction is EXOTHERMICwater cools down reaction is ENDOTHERMIC

Calculation The energy required to change the temperatureof a substance can be calculated using... q = m x c x

where q = heat energy kJ m = mass kg c = Specific Heat Capacity kJ K -1 kg -1 [ water is 4.18 ]

= change in temperature K

MEASURING ENTHALPY CHANGESMEASURING ENTHALPY CHANGES

Calorimetry involves the practical determination of enthalpy changesusually involves heating (or cooling) known amounts of waterwater is heated up reaction is EXOTHERMICwater cools down reaction is ENDOTHERMIC

Calculation The energy required to change the temperatureof a substance can be calculated using... q = m x c x

where q = heat energy kJ m = mass kg c = Specific Heat Capacity kJ K -1 kg -1 [ water is 4.18 ]

= change in temperature K

Example On complete combustion, 0.18g of hexane raised the temperature of100g water from 22°C to 47°C. Calculate its enthalpy of combustion.

Heat absorbed by the water (q) = 0.1 x 4.18 x 25 = 10.45 kJMoles of hexane burned = mass / Mr = 0.18 / 86

= 0.00209

Enthalpy change = heat energy / moles = 10.45 / 0.00209

ANS = 5000 kJ mol -1


Example 1 - graphicalThe temperature is taken every halfminute before mixing the reactants.

Reactants are mixed after three minutes.

Further readings are taken every halfminute as the reaction mixture cools.

Extrapolate the lines as shown andcalculate the value of.


Example 1 - graphicalThe temperature is taken every halfminute before mixing the reactants.

Reactants are mixed after three minutes.

Further readings are taken every halfminute as the reaction mixture cools.

Extrapolate the lines as shown andcalculate the value of.

Example calculationWhen 0.18g of hexane underwent complete combustion, it raised the temperature of100g (0.1kg) water from 22°C to 47°C. Calculate its enthalpy of combustion.

Heat absorbed by the water (q) = m C = 0.1 x 4.18 x 25 = 10.45 kJ

Moles of hexane burned = mass / Mr = 0.18 / 86 = 0.00209Enthalpy change = heat energy / moles = 10.45 / 0.00209 = 5000 kJ mol -1


Example 2

25cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial temperature of both solutions was 20°C. The highest temperature reached by the solution was 33°C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat capacity (c) of water is 4.18 kJ K -1 kg -1]

NaOH + HCl ——> NaCl + H2O


Example 2



Temperature rise () = 306K – 293K = 13K

Volume of resulting solution= 25 + 25 = 50cm3 = 0.05 dm3

Equivalent mass of water = 50g = 0.05 kg(density is 1g per cm3)

Heat absorbed by the water (q) = m x c x 0.05 x 4.18 x 13 = 2.717 kJ


Example 2



Temperature rise () = 306K – 293K = 13K

Volume of resulting solution= 25 + 25 = 50cm3 = 0.05 dm3

Equivalent mass of water = 50g = 0.05 kg(density is 1g per cm3)

Heat absorbed by the water (q) = m x c x 0.05 x 4.18 x 13 = 2.717 kJ

Moles of HCl reacting = 2 x 25/1000 = 0.05 mol

Moles of NaOH reacting = 2 x 25/1000 = 0.05 mol

Moles of water produced = 0.05 mol

Enthalpy change per mol () = heat energy / moles of water = 2.717 / 0.05

= 54.34 kJ mol -1


REVISION CHECKREVISION CHECK

What should you be able to do?

Explain the difference between an endothermic and exothermic reaction

Understand the reasons for using standard enthalpy changes

Recall the definitions of enthalpy of formation and combustion

Write equations representing enthalpy of combustion and formation

Recall and apply Hess’s law

Recall the definition of bond dissociation enthalpy

Calculate standard enthalpy changes using bond enthalpy values

Calculate standard enthalpy changes using enthalpies of formation and combustion

Know simple calorimetry methods for measuring enthalpy changes

Calculate enthalpy changes from calorimetry measurements

CAN YOU DO ALL OF THESE? CAN YOU DO ALL OF THESE? YES YES NONO

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© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING

ENTHALPY ENTHALPY CHANGESCHANGES

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ENTHALPY CHANGES A guide for A level students KNOCKHARDY PUBLISHING.

Documents

enthalpy of products

actual enthalpy

enthalpy changes introduction

enthalpy of formation

delta h enthalpy change

standard conditions

standard states

form of energy change