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Definitions An exothermic reaction is one where heat is given out. E.g. burning of a fuel ΔH = negative An endothermic reaction is one where heat is taken in. ΔH = positive

Definitions Heat of reaction The heat of reaction is the heat change which takes place when substances react in molar quantities according to a balanced equation.

Definitions Heat of Combustion. The heat of combustion of a substance is the heat change when one mole of a substance is completely burned in excess oxygen

Bomb Calorimeter A closed metal calorimeter, called a bomb calorimeter, is used to measure heats of combustion, to compare efficiencies of fuels and to measure the energy content of foodstuffs. The bomb is a thick-walled steel vessel into which oxygen at high pressure can be pumped. This ensures that the substance burns in an excess of oxygen and the normal oxide is formed. The sample is held in a platinum crucible and ignited electrically. The platinum will not burn and will not react with the sample even under the conditions of high temperature and pressure which exist inside the bomb during reaction. The bomb is immersed in a vessel of water in a container designed to eliminate Energy transfer to or from the surroundings. The bomb and surrounding water may then be considered to be the universe, the combustion reaction taking place in the bomb is the system and the water and insulated container with lid act as the surroundings.

Bomb calorimeter

Definitions Calorific value of food. This is the heat released when 100g of food is burned. Kilogram Calorific value of a fuel. This is the heat released when 1 kg of a fuel is burned. Heat of Neutralisation. This is the heat change when one mole of H+ ions from an acid react with one mole of OH- ions from a base to form one mole of water.

Bond Energy Bond Energy. The bond energy is the energy required to break one mole of covalent bonds and to completely separate the atoms. In a reaction, energy is absorbed to break bonds and energy is released when new bonds are formed.

Enthalpy of reaction from bond enthalpies Calculate the enthalpy change for the hydrogenation of ethene DH2 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 DH3 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 Hess’s Law DH1 = DH2 – DH3 = (2699 – 2824) = – 125 kJ

Heat of NEUTRALISATION This is the heat change when one mole of H+ ions from an acid react with one mole of OH- ions from a base to form one mole of water.

HEAT OF NEUTRALISATION Definition The HEAT 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

MEASURING Heat of Neutralisation 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 Heat of Neutralisation. [The specific heat capacity (c) of water is 4.18 kJ K -1 kg -1] NaOH + HCl ——> NaCl + H2O Temperature rise (DT) = 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 DT = 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 (DH) = heat energy / moles of water = 2.717 / 0.05 = 54.34 kJ mol -1

Definitions Heat Of Formation The Heat of Formation of a compound is the heat change that takes place when one mole of a substance is formed from its elements in their standard states.

STANDARD HEAT OF FORMATION Definition The hEAT change when ONE MOLE of a compound is formed in its standard state from its elements in their standard states. Symbol DH°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.

Hess’s Law The heat change for a reaction is equal to the sum of heat changes for intermediate reactions.

“The enthalpy change is independent of the path taken” HESS’S LAW “The enthalpy change is independent of the path taken” How The 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. DHr = DH1 + DH2 + DH3 If you go in the opposite direction of an arrow, you subtract the value of the enthalpy change. e.g. DH2 = - DH1 + DHr - DH3 The values of DH1 and DH3 have been subtracted because the route involves going in the opposite direction to their definition.

Enthalpy of reaction from enthalpies of combustion Sample calculation Calculate the standard heat of formation of methane; the standard heatsof combustion of carbon, hydrogen and methane are -394, -286 and -890 kJ mol-1 . C(graphite) + 2H2(g) ———> CH4(g) DH =  DHc of reactants –  DHc of products By applying Hess’s Law ... The Standard Enthalpy of Reaction DH°r will be... REACTANTS PRODUCTS [ (1 x DHc of C) + (2 x DHc of H2) ] minus [ 1 x DHc of CH4O ] DH°r = 1 x (-394) + 2 x (-286) - 1 x (-890) ANSWER = - 74 kJ mol-1

THERMODYNAMICS First Law Energy 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 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

STANDARD HEAT OF FORMATION Definition The hEAT change when ONE MOLE of a compound is formed in its standard state from its elements in their standard states. Symbol DH°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.

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 DH , “delta H ” Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants

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 DH , “delta H ” Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants Enthalpy of reactants > products DH = - ive EXOTHERMIC Heat given out REACTION CO-ORDINATE ENTHALPY

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 DH , “delta H ” Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants Enthalpy of reactants > products Enthalpy of reactants < products DH = - ive DH = + ive EXOTHERMIC Heat given out ENDOTHERMIC Heat absorbed REACTION CO-ORDINATE ENTHALPY ENTHALPY REACTION CO-ORDINATE

STANDARD ENTHALPY CHANGES 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 DH . Enthalpy Change Standard Enthalpy Change (at 298K)

STANDARD HEAT OF FORMATION Definition The hEAT change when ONE MOLE of a compound is formed in its standard state from its elements in their standard states. Symbol DH°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.

STANDARD HEAT OF COMBUSTION Definition The heat change when ONE MOLE of a substance undergoes complete combustion in excess Oxygen under standard conditions. All reactants and products are in their standard states. Symbol DH°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.

HEAT OF NEUTRALISATION Definition The HEAT 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

BOND 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

BOND 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 NN 944 CC 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 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” HESS’S LAW “The enthalpy change is independent of the path taken” How The 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. DHr = DH1 + DH2 + DH3

“The enthalpy change is independent of the path taken” HESS’S LAW “The enthalpy change is independent of the path taken” How The 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. DHr = DH1 + DH2 + DH3 If you go in the opposite direction of an arrow, you subtract the value of the enthalpy change. e.g. DH2 = - DH1 + DHr - DH3 The values of DH1 and DH3 have been subtracted because the route involves going in the opposite direction to their definition.

“The enthalpy change is independent of the path taken” HESS’S LAW “The enthalpy change is independent of the path taken” Use applying Hess’s 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 DH°c enthalpy change of formation from DH°f

Enthalpy of reaction from bond enthalpies Theory Imagine 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 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 released its value will be equal and opposite to that of breaking the bonds Applying Hess’s Law DHr = Step 1 + Step 2

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

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

Enthalpy of reaction from bond enthalpies Calculate the enthalpy change for the hydrogenation of ethene DH2 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 DH3 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 Hess’s Law DH1 = DH2 – DH3 = (2699 – 2824) = – 125 kJ

Enthalpy of reaction from enthalpies of formation 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 Step 1 Energy is released as reactants are formed from their elements. Step 2 Energy is released as products DHr = - 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 DH SUM OFTHE ENTHALPIES OF FORMATION OF THE PRODUCTS DH =  DHf of products –  DHf of reactants

Enthalpy of reaction from enthalpies of formation 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 2H2O(l) + 4NO2(g) + O2(g) ———> 4HNO3(l) DH =  DHf of products –  DHf of reactants By applying Hess’s Law ... The Standard Enthalpy of Reaction DH°r will be... PRODUCTS REACTANTS [ 4 x DHf of HNO3 ] minus [ (2 x DHf of H2O) + (4 x DHf of NO2) + (1 x DHf of O2) ] DH°r = 4 x (-173) - 2 x (-286) + 4 x (+33) + 0 ANSWER = - 252 kJ

Enthalpy 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

Enthalpy of reaction from enthalpies of combustion Step 1 Energy is released as reactants undergo combustion. Step 2 Energy is released as products DHr = 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 OXIDATION DH SUM OFTHE ENTHALPIES OF COMBUSTION OF THE PRODUCTS DH =  DHc of reactants –  DHc of products

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) + 2H2(g) ———> CH4(g) DH =  DHc of reactants –  DHc of products By applying Hess’s Law ... The Standard Enthalpy of Reaction DH°r will be... REACTANTS PRODUCTS [ (1 x DHc of C) + (2 x DHc of H2) ] minus [ 1 x DHc of CH4O ] DH°r = 1 x (-394) + 2 x (-286) - 1 x (-890) ANSWER = - 74 kJ mol-1

MEASURING ENTHALPY CHANGES Calorimetry involves the practical determination of enthalpy changes usually involves heating (or cooling) known amounts of water water is heated up reaction is EXOTHERMIC water cools down reaction is ENDOTHERMIC Calculation The energy required to change the temperature of a substance can be calculated using... q = m x c x DT where q = heat energy kJ m = mass kg c = Specific Heat Capacity kJ K -1 kg -1 [ water is 4.18 ] DT = change in temperature K

MEASURING ENTHALPY CHANGES Calorimetry involves the practical determination of enthalpy changes usually involves heating (or cooling) known amounts of water water is heated up reaction is EXOTHERMIC water cools down reaction is ENDOTHERMIC Calculation The energy required to change the temperature of a substance can be calculated using... q = m x c x DT where q = heat energy kJ m = mass kg c = Specific Heat Capacity kJ K -1 kg -1 [ water is 4.18 ] DT = change in temperature K Example On complete combustion, 0.18g of hexane raised the temperature of 100g water from 22°C to 47°C. Calculate its HEAT of combustion. Heat absorbed by the water (q) = 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 ANS = 5000 kJ mol -1

MEASURING ENTHALPY CHANGES 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 DT.

MEASURING ENTHALPY CHANGES 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 DT. 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 HEAT of combustion. Heat absorbed by the water (q) = m C DT = 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 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

MEASURING ENTHALPY CHANGES 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 Temperature rise (DT) = 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 DT = 0.05 x 4.18 x 13 = 2.717 kJ

MEASURING ENTHALPY CHANGES 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 Temperature rise (DT) = 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 DT = 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 (DH) = heat energy / moles of water = 2.717 / 0.05 = 54.34 kJ mol -1

What should you be able to do? 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 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? YES NO

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