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ENTHALPY CHANGES A guide for A level students KNOCKHARDY PUBLISHING.

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1 ENTHALPY CHANGES A guide for A level students KNOCKHARDY PUBLISHING

2 ENTHALPY 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 Navigation is achieved by... either clicking on the grey arrows at the foot of each page orusing the left and right arrow keys on the keyboard

3 ENTHALPY CHANGES CONTENTS Thermodynamics Enthalpy changes Standard enthalpy values Standard enthalpy of formation Standard enthalpy of combustion Enthalpy of neutralisation Bond dissociation enthalpy Hesss 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

4 Before you start it would be helpful to… be able to balance equations be confident with simple arithmetical operations ENTHALPY CHANGES

5 THERMODYNAMICS First LawEnergy can be neither created nor destroyed but It can be converted from one form to another Energy changes all chemical reactions are accompanied by some form of energy change changes can be very obvious (e.g. coal burning) but can go unnoticed ExothermicEnergy is given out Endothermic Energy is absorbed ExamplesExothermiccombustion of fuels respiration (oxidation of carbohydrates) Endothermicphotosynthesis thermal decomposition of calcium carbonate

6 THERMODYNAMICS - ENTHALPY 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

7 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 ENTHALPY THERMODYNAMICS - ENTHALPY CHANGES

8 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 Enthalpy of reactants < products = - ive = + ive EXOTHERMIC Heat given out ENDOTHERMIC Heat absorbed REACTION CO-ORDINATE ENTHALPY REACTION CO-ORDINATE ENTHALPY THERMODYNAMICS - ENTHALPY CHANGES

9 Why a standard? 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 CHANGES

10 STANDARD 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) + O 2 (g) > CO 2 (g) H 2 (g) + ½O 2 (g) > H 2 O(l) 2C(graphite) + ½O 2 (g) + 3H 2 (g) > C 2 H 5 OH(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.

11 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) + O 2 (g) > CO 2 (g) H 2 (g) + ½O 2 (g) > H 2 O(l) C 2 H 5 OH(l) + 3O 2 (g) > 2CO 2 (g) + 3H 2 O(l) NotesAlways 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 COMBUSTION

12 Definition The enthalpy change when ONE MOLE of water is formed from its ions in dilute solution. Values Exothermic Equation H + (aq) + OH¯(aq) > H 2 O(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 NEUTRALISATION

13 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 Cl 2 (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 ENTHALPY

14 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 Cl 2 (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-F562 N-N163 C-C 346 H-Cl431 N=N409 C=C 611 H-Br366 N N944 C C 837 H-I299 P-P172 C-O 360 H-N388 F-F158 C=O 743 H-O463 Cl-Cl242 C-H 413 H-S338 Br-Br193 C-N 305 H-Si318 I-I151 C-F 484 P-H322 S-S264 C-Cl 338 O-O146 Si-Si176 BOND 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

15 The enthalpy change is independent of the path taken HowThe enthalpy change going from A to B can be found by adding the values of the enthalpy changes for the reactions A to X, X to Y and Y to B. r = + 2 + 3 HESSS LAW

16 The enthalpy change is independent of the path taken HowThe enthalpy change going from A to B can be found by adding the values of the enthalpy changes for the 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 of the enthalpy change. e.g. 2 = - + r - 3 The values of and 3 have been subtracted because the route involves going in the opposite direction to their definition. HESSS LAW

17 The enthalpy change is independent of the path taken Useapplying Hesss Law enables one to calculate enthalpy changes from other 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 HESSS LAW

18 TheoryImagine that, during a reaction, all the bonds of reacting species are broken and the individual atoms join up again but in the form of products. The overall energy change will depend on the difference between the energy required 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 enthalpies Step 1Energy is put in to break bonds to form separate, gaseous atoms Step 2The gaseous atoms then combine to form bonds and energy is released its value will be equal and opposite to that of breaking the bonds Applying Hesss Law r = Step 1 + Step 2

19 Enthalpy of reaction from bond enthalpies = bond enthalpies – bond enthalpies of reactants of products = bond enthalpies – bond enthalpies of reactants of products Alternative view Step 1Energy is put in to break bonds to form separate, gaseous atoms. Step 2Gaseous atoms then combine to form bonds and energy is released; its value will be equal and opposite to that of breaking the bonds r = Step 1 - Step 2 Because, in Step 2 the route involves going in the OPPOSITE DIRECTION to the defined change of bond enthalpy, its value is subtracted. SUM OFTHE BOND ENTHALPIES OF THE REACTANTS REACTANTS PRODUCTS ATOMS SUM OFTHE BOND ENTHALPIES OF THE PRODUCTS

20 Calculate the enthalpy change for the hydrogenation of ethene Enthalpy of reaction from bond enthalpies

21 Calculate the enthalpy change for the hydrogenation of ethene Enthalpy of reaction from bond enthalpies 2 1 x C=C bond @ 611= 611 kJ 4 x C-H bonds @ 413= 1652 kJ 1 x H-H bond @ 436= 436 kJ Total energy to break bonds of reactants= 2699 kJ 3 1 x C-C bond @ 346= 346 kJ 6 x C-H bonds @ 413= 2478 kJ Total energy to break bonds of products= 2824 kJ Applying Hesss Law 1 = 2 – 3 = (2699 – 2824) = – 125 kJ

22 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 formation

23 Step 1Energy is released as reactants are formed from their elements. Step 2Energy is released as products are formed from their elements. r = - Step 1 + Step 2 orStep 2 - Step 1 In Step 1 the route involves going in the OPPOSITE DIRECTION to the defined enthalpy change, its value is subtracted. SUM OFTHE ENTHALPIES OF FORMATION OF THE REACTANTS REACTANTS PRODUCTS ELEMENTS SUM OFTHE ENTHALPIES OF FORMATION OF THE PRODUCTS Enthalpy of reaction from enthalpies of formation = f of products – f of reactants

24 Sample calculation Calculate 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 2H 2 O(l) + 4NO 2 (g) + O 2 (g) > 4HNO 3 (l) By applying Hesss Law... The Standard Enthalpy of Reaction ° r will be... PRODUCTS REACTANTS [ 4 x f of HNO 3 ] minus [ (2 x f of H 2 O) + (4 x f of NO 2 ) + (1 x f of O 2 ) ] ° r = 4 x (-173) - 2 x (-286) + 4 x (+33) + 0 ANSWER = - 252 kJ Enthalpy of reaction from enthalpies of formation = f of products – f of reactants

25 Enthalpy of reaction from enthalpies of combustion If you burned all the products you should get the same amounts of oxidation products such a CO 2 and H 2 O as if you burned the reactants. Enthalpy of combustion is an exothermic process

26 = c of reactants – c of products Enthalpy of reaction from enthalpies of combustion Step 1Energy is released as reactants undergo combustion. Step 2Energy is released as products undergo 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, its value is subtracted. SUM OFTHE ENTHALPIES OF COMBUSTION OF THE REACTANTS REACTANTS PRODUCTS OXIDATION PRODUCTS SUM OFTHE ENTHALPIES OF COMBUSTION OF THE PRODUCTS

27 Enthalpy of reaction from enthalpies of combustion Sample calculation Calculate 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) + 2H 2 (g) > CH 4 (g) By applying Hesss Law... The Standard Enthalpy of Reaction ° r will be... REACTANTS PRODUCTS [ (1 x c of C) + (2 x c of H 2 ) ] minus [ 1 x c of CH 4 O ] ° r = 1 x (-394) + 2 x (-286) - 1 x (-890) ANSWER = - 74 kJ mol -1 = c of reactants – c of products

28 Calorimetryinvolves the practical determination of enthalpy changes usually involves heating (or cooling) known amounts of water water is heated upreaction is EXOTHERMIC water cools downreaction is ENDOTHERMIC CalculationThe energy required to change the temperature of a substance can be calculated using...q = m x c x where q= heat energykJ m= masskg c= Specific Heat Capacity kJ K -1 kg -1 [ water is 4.18 ] = change in temperatureK MEASURING ENTHALPY CHANGES

29 Calorimetryinvolves the practical determination of enthalpy changes usually involves heating (or cooling) known amounts of water water is heated upreaction is EXOTHERMIC water cools downreaction is ENDOTHERMIC CalculationThe energy required to change the temperature of a substance can be calculated using...q = m x c x where q= heat energykJ m= masskg c= Specific Heat Capacity kJ K -1 kg -1 [ water is 4.18 ] = change in temperatureK Example On complete combustion, 0.18g of hexane raised the temperature of 100g 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 kJ Moles of hexane burned = mass / M r = 0.18 / 86 = 0.00209 Enthalpy change = heat energy / moles= 10.45 / 0.00209 ANS= 5000 kJ mol -1 MEASURING ENTHALPY CHANGES

30 Example 1 - graphical The temperature is taken every half minute before mixing the reactants. Reactants are mixed after three minutes. Further readings are taken every half minute as the reaction mixture cools. Extrapolate the lines as shown and calculate the value of. MEASURING ENTHALPY CHANGES

31 Example 1 - graphical The temperature is taken every half minute before mixing the reactants. Reactants are mixed after three minutes. Further readings are taken every half minute as the reaction mixture cools. Extrapolate the lines as shown and calculate the value of. Example calculation When 0.18g of hexane underwent complete combustion, it raised the temperature of 100g (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= 0.00209 Enthalpy change = heat energy / moles= 10.45 / 0.00209= 5000 kJ mol -1 MEASURING ENTHALPY CHANGES

32 Example 2 25cm 3 of 2.0M HCl was added to 25cm 3 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 + H 2 O MEASURING ENTHALPY CHANGES

33 Example 2 25cm 3 of 2.0M HCl was added to 25cm 3 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 + H 2 O Temperature rise ( ) = 306K – 293K= 13K Volume of resulting solution= 25 + 25 = 50cm 3 = 0.05 dm 3 Equivalent mass of water= 50g= 0.05 kg (density is 1g per cm 3 ) Heat absorbed by the water (q)= m x c x 0.05 x 4.18 x 13= 2.717 kJ MEASURING ENTHALPY CHANGES

34 Example 2 25cm 3 of 2.0M HCl was added to 25cm 3 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 + H 2 O Temperature rise ( ) = 306K – 293K= 13K Volume of resulting solution= 25 + 25 = 50cm 3 = 0.05 dm 3 Equivalent mass of water= 50g= 0.05 kg (density is 1g per cm 3 ) 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 MEASURING ENTHALPY CHANGES

35 REVISION 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 Hesss 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? YES NO

36 You need to go over the relevant topic(s) again Click on the button to return to the menu

37 WELL DONE! Try some past paper questions

38 © 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING ENTHALPY CHANGES THE END


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