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CHAPTER 11 - ENTHALPY OF CHEMICAL REACTIONS 11.1 Introduction All substances possess a certain amount of kinetic energy; their constituent particles may be moving with vibrational, rotational or translational energy (or all three) depending on whether the substance is in the solid, liquid or gaseous state. We say that substances have „thermal energy‟ and the amount of this energy can be measured using the „temperature‟ scale. As we have noted earlier the two main temperature scales we use are the „Celsius‟ scale and the „Kelvin‟ scale, where 0 C = 273 K. In this chapter we shall be looking at the changes in energy which may occur during a chemical reaction and how we may quantify these changes by measuring the „change in enthalpy‟ of the reaction. 11.2 The concept of enthalpy: H notation The enthalpy (symbol H) or „heat content‟ of a chemical is a measure of its thermal energy. When a chemical reaction proceeds there is almost always a change in the total energy of the reactants compared to the total energy of the products. This change in overall energy can be easily measured as a change in temperature where, if the reactants have more energy than the products there will be a net energy loss and the temperature will increase. If however, the reactants have less energy than the products the extra energy needed will come from the surroundings and the overall temperature will decrease. Due to the ease with which it can be measured, therefore, chemists talk about the change in enthalpy from reactants to products, where: change in enthalpy = enthalpy of the products - enthalpy of the reactants H = Hproducts - H reactants 11.3. Heats of reaction A reaction in which energy (or „heat’) is given off is known as an exothermic reaction - the enthalpy of the reactants was greater than that of the products and the H value is negative. As has been noted previously, an exothermic reaction is accompanied by an increase in the temperature of the surroundings. Similarly, a reaction in which energy (heat) must be taken from the surroundings is known as an endothermic reaction and the H value is positive. These most important of concepts may be represented on a „Heat change diagram‟. Fig. 1 Exothermic (H = -ve) Endothermic (H = + ve) Hr Hp H H Hp Hr These simple diagrams do not attempt to show what is happening as the reactants change into products - merely the end result of the transformation. The amount of energy that must be supplied to actually initiate the reaction is known as the activation energy and can also be represented on heat change diagrams: Fig.2 Exothermic (H = -ve) Endothermic (H = + ve) Activation energy Hp Hr Activation Hr energy Hp The larger the activation energy (or „hump‟ on the diagram) the more energy that must be supplied to initiate the reaction. Some reactions, such as that of nitroglycerine reacting with oxygen from the air in an explosion, have very small activation energies. Others, such as the reaction of nitrogen and oxygen gases to produce the noxious gas nitrogen monoxide, only occur to any appreciable extent at high temperatures - have high activation energies. In industry it is often desirable to make a reaction proceed more easily, and hence at lower cost, by using a suitable catalyst. The function of a catalyst is to provide an alternate reaction pathway with a smaller activation energy. By doing this the rate of the reaction is increased. Fig. 3 Reaction pathway without catalyst Reaction pathway with catalyst Activation Activation energy reduced energy Hr Hr Hp Hp Note that the magnitude and sign of the H value is not affected by the catalyst. By reducing the activation energy the reaction is able to proceed at a faster rate - the position of equilibrium and the equilibrium constant remain unchanged. 11.4 Specific Heat Capacity Could the well-known phrase “A watched pot never Table 11.1 Specific heats of boils!” be alluding to the fact that water requires an selected substances extraordinarily large input of energy before its Substance Specific Heat content temperature increases significantly? When supplied (J. C-1.g-1) with an equivalent amount of heat energy, water (due Water 4.184 to its strong hydrogen bonding between molecules) Ammonia 2.124 Ethanol 1.413 will have a lesser increase in temperature than almost Alumina 0.778 any other common substance, as is demonstrated in Iron 0.448 Table 11.1. Once more we see that water possesses Copper 0.386 physical and chemical properties that set it apart from Gold 0.128 most other substances. The „specific heat capacity‟ is defined as the amount of heat energy required to raise the temperature of 1 gram of a substance by 1 C or 1K. Example 11.1 A calorimeter is filled with 100 ml of water and its temperature is recorded as 18.28 C. A current of 2.41A at a potential difference of 5.92V is then passed through the water for exactly 2 minutes. What temperature will the water reach, presuming the calorimeter is well insulated? Solution First, we must calculate the amount of energy supplied to the water. E = V.I.t E = 5.92 x 2.41 x 120 E = 1712 J The specific heat capacity of water is 4.184 J. C-1.g-1 ie. to raise 1 g of water by 1 C requires the input of 4.184 J of energy to raise 100 g of water requires 100 x 4.184 J = 418.4 J 1712 Temperature increase = = 4.09 C 418.4 11.5 The calorimeter Calorimeters are devices used to measure the thermal energy released or absorbed as a chemical reaction proceeds. They can be used in a wide variety of applications, including reactions involving aqueous solutions, gases and even in the combustion of foods. 11.5.1 Components of calorimeters All calorimeters basically consist of a well insulated outer container enclosing an inner chamber where the reaction will take place. They have an electrically heated coil to allow for calibration, an accurate thermometer to measure any change in temperature and a stirring device to ensure even mixing of the reactants. If the reactants involve gases, the inner chamber will contain a sealed glass vessel (or „bomb‟) into which the gaseous reactants are placed. When the reaction is to take place, the glass bulb is broken to allow contact of all reactant species. Fig. 4 - Solution and bomb calorimeters 11.5.2 Calibration of calorimeters Before any calorimeter can be used to measure the enthalpy of a reaction it must be calibrated ie. we must know how much energy is required to raise the calorimeter and its contents by 1 C. In most cases this is achieved by passing an electric current at a known voltage through a heating coil within the calorimeter and measuring the increase in temperature that results. The formula for calculating the calorimeter constant is: E VIt Calorimeter constant = Units: J.C-1 T T Example 11.2 A student wishes to calibrate a calorimeter in order to perform an experiment involving the heat change of a particular reaction. To do this she fills the calorimeter with exactly 100.0 ml of deionised water and then passes a current of 1.88A through the heating coil for exactly 2 minutes. The potential difference across the coil is measured at 4.13V and the increase in temperature that occurs is 3.81 C. Solution E VIt 413 188 120 . . Calorimeter constant = 245 J.C-1 T T . 381 11.6 Thermochemical equations A thermochemical equation is one which gives the relevant H value for that particular equation exactly as it is written. For example, the thermochemical equation for the combustion of methane in air can be written as: CH4 (g) + 2O2 (g) ----> CO2 (g) + 2H2O (l) H = -890 kJ mol-1 This equation literally means that “when 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water that 890 kJ of energy is released”. If the equation were doubled, the H value would also double. If the equation were reversed, the H value would also change sign: 2CH4 (g) + 4O2 (g) ----> 2CO2 (g) + 4H2O (l) H = -1780 kJ mol-1 CO2 (g) + 2H2O (l) ----> CH4 (g) + 2O2 (g) H = +890 kJ mol-1 Thermochemical equations can be used to perform a range of stoichiometric calculations involving heats of reaction. Example 11.3 Calculate the amount of energy released when 500 ml of methane gas at S.L.C. reacts with excess air according to the equation CH4 (g) + 2O2 (g) ----> CO2 (g) + 2H2O (l) H = -890 kJ mol-1 Note: VM of any gas at S.L.C. = 24.5L Solution V 0.500 n(CH4) = 0.0204 mol VM 24.5 From the equation, 1 mol of yields 890 kJ 0.0204 mol yields „x‟ kJ By ratio, x = 890 x 0.0204 = 18.16 The energy released by the combustion reaction is 18.16 kJ. Example 11.4 What mass of propanol must be burnt in excess air to produce 10,000 kJ of energy, according to the reaction 2C3H7OH (l) + 9O2 (g) ----> 6CO2 (g) + 8H2O (l) H = -4034 kJ mol-1 Solution From the equation, 2 mol of propanol yields 4034 kJ of energy „x‟ mol yields 10,000 kJ By ratio, 2 x 10,000 = „x‟ x 4034 x = 4.96 mol M (C3H7OH) = 60 g.mol-1 mass of propanol = 297 g Example 11.5 When hydrogen sulfate is mixed with water to produce sulfuric acid a substantial amount of heat is evolved. If 100 ml of hydrogen sulfate (density = 1.834 g.cm-3) is added to 1 litre of water at 18.5 C, calculate the temperature that the water will reach after dissolution. The relevant equation is : H2SO4 (l) water H2SO4 (aq) H = -7.2 kJ mol-1 Solution m d(H2SO4) = m (H2SO4) = 1.834 x 100 = 183.4 g V m 1834 . n(H2SO4) = = 1.87 mol M . 981 From the equation, 1 mol of H2SO4 yields 7.2 kJ 1.87 mol of H2SO4 yields „x‟ kJ x = 13.46 kJ The specific heat capacity of water is 4.184 J. C-1.g-1 To increase the temperature of 1 litre of water by 1 C requires 4.184 x 1000 = 4184 J 13,460 Increase in temperature = = 3.22 C 4184 Example 11.6 1.82g of a biscuit was dried and then burnt in a bomb calorimeter in the presence of excess pure oxygen. The temperature of the water bath surrounding the reaction chamber rose from 22.085 C to 23.117 C. To calibrate the calorimeter, a current of 3.77A at a potential difference of 5.29V was passed through the heating coil for a period of 72.1 seconds; a rise in temperature of 3.326 C being recorded. Calculate the heat content of the biscuit. Solution First, we must calculate the calorimeter constant: E VIt 5.29 377 721 . . Calorimeter constant = 432.3 J.C-1 T T . 3326 The change in temperature caused by the combustion of the biscuit T = 23.117 - 22.085 = 1.032 C Energy released by 1.82g of biscuit = 432.3 x 1.032 = 446.2 J 446.2 Heat content of biscuit = = 245 J g-1 . 182 11.7 Summary/ Objectives At the end of this Chapter you should - understand that chemical reactions involve a change in the energy of the reactants as they change into products and that this change is known as the „enthalpy‟ of the reaction - appreciate that it is much easier to consider a change in enthalpy as a reaction proceeds, such that H = Hproducts - H reactants - recognise that heat change diagrams are most useful devices for representing changes in enthalpy and the effect of catalysts on the activation energy of a reaction - be able to interpret accurately and fully these heat change diagrams - recall that endothermic reactions have a negative H value as they release heat into the environment during the course of a reaction - recall that exothermic reactions have a positive H value as they absorb heat from the environment during the course of a reaction - understand the role of a catalyst as providing an alternative pathway for a reaction of lower activation energy - be able to display catalytic function on a heat change diagram and/or interpret such a diagram - recall the definition of „specific heat capacity‟ of a substance as „the amount of heat energy required to raise the temperature of 1 gram of a substance by 1 C or 1K‟ - recall that water has a very high specific heat capacity and understand the reasons for this (in terms of hydrogen bonding) - be familiar with the components of both „solution‟ and „bomb‟ calorimeters and appreciate the function of these components in terms of the overall purpose of a calorimeter - recall how to calibrate a calorimeter from both a theoretical standpoint and in practice, having been supplied with the relevant data - recognise the experimental errors inherent in calorimetric determinations and discuss how these errors may be minimised - appreciate how to use thermochemical equations in stoichiometric calculations to determine energy changes and/or H values, having been supplied with the relevant data Chapter 11 Questions Q1. Draw and clearly label heat change diagrams to represent (i) an exothermic reaction with a relatively small activation energy (ii) an endothermic reaction with a relatively large activation energy (iii) the effect of a catalyst on the activation energy of both of the above reactions Q2. State which of the following reactions are exothermic and which are endothermic: (a) I2 (s) + Cl2 (g) ----> 2ICl (g) H = +35 kJ mol-1 (b) N2 (g) + 3H2 (g) ----> 2NH3 (g) H = -92.3 kJ mol-1 (c) BCl3 (l) + 6H2 (g) ----> B2H6 (g) + 6HCl (g) H = +315 kJ mol-1 (d) Ag+ (aq) + 2NH3 (aq) ----> Ag(NH3)2+ (aq) H = -111 kJ mol-1 (e) N2 (g) ----> 2N (g) H = +946 kJ mol-1 Q3. Calculate the amount of heat energy absorbed from the surroundings when 25.0 g of sodium sulfate decahydrate crystallises from water, given the equation: Na2SO4.10H2O (s) water Na2SO4 (aq) + 10H2O (l) H = +79.0 kJ mol-1 Q4. How much energy is required to convert 1 litre of boiling water into steam? The heat of vaporisation of water is given below: H2O (l) ----> H2O (g) H = + 44.0 kJ mol-1 Q5. Anhydrous copper sulfate is a hydroscopic substance - it absorbs water from the surroundings according to the equation: CuSO4 (s) water CuSO4 (aq) When 5.73 g of this substance dissolves in water it produces 2.388 kJ of thermal energy. Calculate the H for the dissolution reaction above. Q6. Calculate the energy released when 10.0 g of carbon reacts with excess oxygen according to the equation: C (s) + O2 (g) ----> CO2 (g) H = -393.4 kJ mol-1 Q7. What mass of brown coal, with an average specific heat capacity of 25 kJ g-1 would be required to heat 600 ml of water from 20 C to 100 C? [specific heat capacity of water = 4.184 J C-1g-1] Q8. The heat of combustion of hexane is 4158 kJ mol-1. Calculate H for the reaction: 2C6H14 (l) + 19O2 (g) ----> 12CO2 (g) + 14H2O (l) Q9. Calculate the heat of combustion of butan-1-ol given the following: 2C4H9OH (l) + 12O2 (g) ----> 8CO2 (g) + 10H2O (l) H = -5354 kJ mol-1 Q10. Octane, the principle component of petrol, undergoes combustion with a plentiful supply of oxygen according to the equation: 2C8H18 (l) + 25O2 (g) ----> 16 CO2 (g) + 18H2O (l) H = -10,900 kJ mol-1 Given the density of octane is 0.698 g cm-3 calculate the thermal energy released when 50 litres of octane undergoes complete combustion. Q11. Much research is currently being undertaken to investigate the feasibility of using ethanol (or an ethanol/petrol mixture) as a substitute for the fossil-fuel based petrol currently employed in most cars. Given that ethanol has a density of 0.785 g cm-3 and that it undergoes combustion according to the equation: C2H5OH (l) + 3O2 (g) ----> 2CO2 (g) + 3H2O (l) H = -1364 kJ mol-1 calculate the thermal energy released when 50 litres of ethanol undergoes complete combustion. By referring to Question 9, which of ethanol or octane (on this basis alone) is the better fuel? What other factors should be taken into account? Q12. When an electric current is passed through water it will decompose to its constituent elements according to the equation: 2H2O (l) ----> 2H2 (g) + O2 (g) H = +572 kJ mol-1 Calculate the amount of electrical energy that would be required to produce 500 cm3 of oxygen gas at S.T.P. Q13. As part of the process of manufacturing sulfuric acid the following reaction takes place: 2SO2 (g) + O2 (g) ----> 2SO3 (g) H = -197 kJ mol-1 Calculate the heat energy released in the production of 1 tonne of SO3 gas. Q14. An important reaction which occurs in the blast furnace is the reduction of carbon dioxide to carbon monoxide, according to the equation: CO2 (g) + C (s) ----> 2CO (g) H = +173 kJ mol-1 Calculate the amount of heat energy absorbed as 5000 litres of CO2 is reduced by excess carbon to CO at 1.226 atm pressure and 1100 C. Q15. By using the specific heat capacity data from Table 11.1 calculate the amount of energy (i) required to raise the temperature of 20 ml of water from 11 C to 85 C (ii) given off as 45 ml of ethanol [d = 0.785 g.cm-3] cools from 37 C to 4 C (iii) absorbed by 1000 kg of iron as it is heated from 25 C to 1000 C Q16. A 1.00 kg block of gold at a temperature of 488 C is plunged into a container filled with 20 L of water at a temperature of 18.5 C. The gold block is removed when its temperature has been lowered to 40 C. What is the temperature of the water now? [specific heat of gold = 0.128 JC-1g-1] Q17. A solution calorimeter with an internal volume of 100 ml has a constant of 419 J C-1. A student carefully adds 100 ml of 2 M HCl to the calorimeter and measures the temperature as 19.60 C. He then adds 0.449g of powdered magnesium, stirs the solution to ensure complete reaction and measures the temperature the solution has reached as 25.42 C. Calculate the heat of reaction for the equation Mg (s) + 2HCl (aq) ----> MgCl2 (aq) + H2 (g) Q18. In order to determine the energy content of a sample of bread a food scientist measures a 1.50g sample, dries it fully and places it into the combustion chamber of a bomb calorimeter. The device is calibrated by passing a current of 2.44 A at a potential difference of 5.08 V for 185 seconds. The temperature within the calorimeter increased by 1.69 C. The chamber is then filled with oxygen and the bread sample is ignited. The temperature of the water surrounding the combustion chamber is found to increase from 17.22 C to 29.16 C. (i) Calculate the energy content of the bread. (ii) State the major sources of error which may have affected the result obtained Q19. A student wishes to determine the heat of reaction for a strong acid reacting to completion with a strong base. The ionic equation for this reaction is: H3O+ (aq) + OH- (aq) ----> 2H2O (l) She first electrically calibrates the calorimeter by filling it with 100.0 ml of deionised water and passing a current of 1.496 A at a potential difference of 4.07 V for 160 seconds. The temperature of the water rose from 16.81 C to 19.58 C. She then carefully added 50 ml of 0.110 M NaOH and 50.0 ml of 0.101 M HCl to the calorimeter and measured the increase in temperature as 0.81 C. (i) Calculate the calorimeter constant (ii) Calculate the heat of neutralisation for the reaction (iii) Suggest a reason for using a higher concentration of the NaOH compared to the HCl. (iv) List the main experimental errors the student may have encountered. (v) Would you expect the heat of neutralisation to have been larger, smaller or the same if 50 ml of 0.101 M acetic acid had been reacted with 50 ml of 0.110 M ammonia? Give an explanation for your answer. Solutions to Chapter 11 questions Q2. a) endothermic b) exothermic c) endothermic d) exothermic e) endothermic Q3. 6.13 kJ Q4. 2440 kJ Q5. 66.5 kJ mol-1 Q6. 327.8 kJ Q7. 8.03g Q9. -8316 kJ mol-1 Q10. 1.669 x 106 kJ Q11. 5.819 x 105 kJ Q12. 12.8 kJ Q13. 1.23 x 106 kJ Q14. 9413 kJ Q15. (i) 6.19 kJ (ii) 1.65 kJ (iii) 4.37 x 105 kJ Q16. 21.24 C Q17. -132 kJ. mol-1 Q18. -10.8 kJ. mol-1 Q19. (i) 351.7 C-1 (ii) -56.1 kJ. mol-1
"CHAPTER 11 - DOC 1"