Chemical Reactions and Energy

Transcription

1 Topic 9 Chemical Reactions and Energy Unit 34 Energy changes in chemical reactions Unit 35 Hess s Law and its applications

2 Key C o ncepts Energy changes in chemical reactions Nature of energy and internal energy of a system Enthalpy changes of reactions Exothermic and endothermic reactions Standard enthalpy changes of formation, combustion, neutralization and solution Calorimetry Chemical Reactions and Energy Hess s Law and its applications Hess s Law Enthalpy change cycle and enthalpy level diagram Determining enthalpy changes which cannot be easily obtained by experiment

3 Topic 9 Chemical Reactions and Energy Unit 34 Energy changes in chemical reactions Unit 34 Energy changes in chemical reactions 34.1 Heat-packs and cold-packs 34.2 What is energy? 34.3 Specific heat capacity 34.4 The system and the surroundings 34.5 Internal energy of a system 34.6 Comparing the enthalpy change (ΔH) and internal energy change (ΔE) of a system during a reaction 34.7 Enthalpy change of an exothermic reaction 34.8 Enthalpy change of an endothermic reaction 34.9 Enthalpy changes during physical and chemical changes Thermochemical equations Explaining energy changes breakage and formation of chemical bonds Standard conditions for measuring enthalpy changes Standard enthalpy change of reaction Standard enthalpy change of formation Standard enthalpy change of combustion Standard enthalpy change of neutralization Standard enthalpy change of solution Determining enthalpy changes of neutralization Determining enthalpy changes of combustion An improved apparatus: the flame calorimeter Summary 1 The specific heat capacity (symbol: c) of a substance is the amount of heat required to raise the temperature of 1 g of the substance by 1 K (or 1 C). Amount of heat required = specific heat capacity x mass x rise in temperature = c x m x T where T is the temperature change. 2 An enthalpy change, H, is the heat released or taken in during any change in a system, provided that the system is kept at constant pressure. 3 In an exothermic reaction, heat is released to the surroundings. Enthalpy (kj mol 1 ) reactants H < 0 (negative) products 4 In an endothermic reaction, heat is taken in from the surroundings. Enthalpy (kj mol 1 ) products H > 0 (positive) reactants 5 The standard enthalpy change of reaction, H O r, is the enthalpy change when molar quantities of reactants as stated in the specified equation react together under standard conditions. 6 The standard enthalpy change of formation of a substance, H O f, is the enthalpy change when one mole of the substance is formed from its elements in their standard states. 7 The standard enthalpy change of combustion of a substance, H O c, is the enthalpy change when one mole of the substance is completely burnt in oxygen under standard conditions. 8 The standard enthalpy change of neutralization, H O n, is the enthalpy change when an acid reacts with an alkali to form one mole of water under standard conditions. 9 The standard enthalpy change of solution of a substance, H O s, is the enthalpy change when one mole of the substance dissolves in an infinite volume of solvent (or enough solvent so that further dilution has no additional effect) under standard conditions.

4 Topic 9 Chemical Reactions and Energy Unit 35 Hess s Law and its applications Exam tips Example The enthalpy change of an exothermic reaction has a ve sign, while that of an endothermic reaction has a +ve sign. DO NOT confuse these two. Students should be able to provide a concise definition for the term standard enthalpy change of formation. In calculations, read the question carefully to see whether it asks about the enthalpy change of reaction or the enthalpy change involved when a certain amount of the reactants undergoes reaction. An experiment was carried out to determine the enthalpy change of the following reaction: Mg(s) + Cu 2+ (aq) Cu(s) + Mg 2+ (aq) 50.0 cm 3 of mol dm 3 CuSO 4 (aq) were transferred to a polystyrene cup with negligible heat capacity, and the temperature of the solution was recorded g of magnesium were added to the cup. The mixture was stirred and its highest temperature was recorded. There was a temperature rise of 63.0 C. a) Show, by calculation, that CuSO 4 is the limiting reactant. (2 marks) b) Assuming that the specific heat capacity and the density of the mixture are 4.18 J g 1 K 1 and 1.00 g cm 3 respectively, calculate the enthalpy change of this reaction, in kj. (Relative atomic mass: Mg = 24.3) (2 marks) Unit 35 Hess s Law and its applications 35.1 Hess s Law 35.2 Enthalpy level diagram and enthalpy change cycle 35.3 Using Hess s Law to determine enthalpy changes that cannot be easily obtained by experiment 35.4 Determining the enthalpy change of formation of magnesium carbonate from enthalpy changes of other reactions 35.5 Determining the standard enthalpy change of formation of a compound from standard enthalpy changes of combustion 35.6 Determining the standard enthalpy change of a reaction from standard enthalpy changes of formation Answer a) Number of moles of CuSO 4 = mol dm 3 x dm3 = 2.50 x 10 2 mol (0.5) Number of moles of Mg = 1.00 g 24.3 g mol 1 = 4.12 x 10 2 mol (0.5) According to the equation, 1 mole of Mg reacts with 1 mole of CuSO 4. During the reaction, 2.50 x 10 2 mole of CuSO 4 reacted with 2.50 x 10 2 mole of Mg. CuSO 4 is the limiting reagent. (1) b) Amount of heat released during the reaction = 50.0 g x 4.18 J g 1 K 1 x 63.0 K = J = 13.2 kj (1) Enthalpy change of reaction = 13.2 kj 2.5 x 10 2 = 528 kj (1) the enthalpy change of the reaction is 528 kj. Remarks* Questions may NOT tell you which of the reagents is the limiting reactant. Using the given number of moles of Mg when carrying out the calculation in (b) will get a wrong answer. So be careful when the question gives the quantities of both reagents.

5 Topic 9 Chemical Reactions and Energy Unit 35 Hess s Law and its applications Summary 1 Hess s Law states: The enthalpy change of a reaction depends on the initial and final states of the reaction and is independent of the route by which the reaction may occur. 2 Overall reaction = Step 1 + Step 2 + Step 3 then the enthalpy change for the overall reaction H O = H 1 + H 2 + H 3 This concept enables us to find enthalpy changes that cannot be easily obtained by experiment. 3 Hf O [compound] = Hc O [constituent elements] Hc O [compound] 4 Hr O = Hf O [products] Hf O [reactants] The recordings of temperature are shown in the graph below: Temperature ( C) Exam tips Example The enthalpy changes of formation of many compounds (e.g. ethanol) CANNOT be determined directly. The enthalpy changes of formation of these compounds may be determined from H c of the constituent elements and the compound. Drawing an enthalpy change cycle relating the known enthalpy changes will help you calculate the unknown enthalpy change. Alternatively, you may calculate the required enthalpy change by combining the enthalpy changes of appropriate processes as shown in the next example. The enthalpy change of formation ( H f ) of ZnO(s) can be determined indirectly from the enthalpy change of formation of H 2 O(l) and the enthalpy changes of reactions (1) and (2) below. Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) (1) ZnO(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 O(l) (2) An experiment as outlined below was carried out to determine the enthalpy change of reaction (1): 30.0 cm 3 of 1.20 mol dm 3 HCl(aq) (excess) were placed in an expanded polystyrene cup. The temperature of the acid in the cup was measured with a thermometer at half-minute intervals. Right at the third minute, g of Zn(s) was added to the cup. The mixture in the cup was then stirred with the thermometer and its temperature was measured for an additional 8 minutes Time (min) a) i) Deduce the greatest temperature change of the reaction mixture. (Show your working on the graph.) (2 marks) ii) Calculate the molar enthalpy change of reaction (1) under the conditions of the experiment. (Assume that the heat capacity of the expanded polystyrene cup is negligible, and that the specific heat capacity and density of the solutions are 4.18 J g 1 K 1 and 1.00 g cm 3 respectively.) (3 marks) (Relative atomic mass: Zn = 65.4) b) Given that under the same conditions the molar enthalpy change of reaction (2) is 84.0 kj mol 1 and the molar enthalpy change of formation of H 2 O(l) is 286 kj mol 1, calculate H f of ZnO(s). (3 marks)

6 10 Topic 9 Chemical Reactions and Energy Unit 35 Hess s Law and its applications 11 Answer a) i) (Working on the graph to show temperatures at the third minute.) (1) Reverse the equation (2) to give equation (2). By combining equations as shown below, followed by collecting like terms, we can obtain the equation for the formation of ZnO(s) from its constituent elements. (1) Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) H = 48.5 kj (2) ZnCl 2 (aq) + H 2 O(l) ZnO(s) + 2HCl(aq) H = kj (3) H 2 (g) O 2(g) H 2 O(l) H = 286 kj 31 Zn(s) O 2(g) ZnO(s) Temperature ( C) T = 5.8 C Time (min) By Hess s Law, H f [ZnO(s)] = [( 48.5) + (+84.0) + ( 286)] kj mol 1 (1) = 251 kj mol 1 (1) the enthalpy change of formation of zinc oxide is 251 kj mol 1. Remarks* For an experiment carried out to determine the enthalpy change of a certain reaction, students often need to work out the maximum temperature rise of the reaction mixture from the plot of temperature against time by extrapolation of the curves. Be careful when reading the questions; either the specific heat capacity or the heat capacity of a solution may be provided. DO NOT confuse specific heat capacity with heat capacity. Calculated temperature change T = 5.8 C (1) ii) Amount of heat released when Zn reacted with HCl(aq) = 30.0 g x 4.18 J g 1 K 1 x 5.8 K = 727 J (1) Number of moles of Zn reacted = g 65.4 g mol 1 = mol Enthalpy change of the reaction = 727 J mol (1) = 48.5 kj mol 1 (1) b) H f [ZnO(s)] refers to the enthalpy change of the following process: Zn(s) O 2(g) ZnO(s) (1) H values for the following processes are known: (1) Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) H = 48.5 kj (2) ZnO(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 O(l) H = 84.0 kj (3) H 2 (g) O 2(g) H 2 O(l) H = 286 kj