Energy Changes and Chemical Reactions Very often chemical changes are accompanied by changes in the heat content (enthalpy, H) of the materials which are reacting. This is H. The two diagrams below show the way in which the heat content of a set of reactants changes as they form products. Label one as an exothermic reaction and the other as an endothermic reaction . <ul><ul><li>Label the amount of energy (ΔH or change in heat / change in enthalpy) that would be given out (-) or taken in (+) as the reaction happens. </li></ul></ul><ul><ul><li>Write on each diagram whether the temperature of the reaction surroundings (eg air or water, if the reactants are in solution) goes up or down as the reaction occurs. </li></ul></ul><ul><ul><li>Give two examples of an exothermic reaction that you have met in chemistry – writing balanced equations if you can. </li></ul></ul>
<ul><li>When an exothermic reaction occurs, the products have more/less energy content than the reactants. </li></ul><ul><li>The heat enegry lost / gained by the reactants is given to other reacting particles / the surroundings ( which can be air or the water used to make aqueous solution). f H in this case is positive / negative . </li></ul>Standard Conditions In comparing enthalpy changes it is essential to ensure the conditions of the system are the same before and after the reaction because ΔH is affected by temperature, pressure and concentration of solutions. The standard conditions for temperature and pressure are _______ and _________ respectively. The substances involved must also be in their normal physical states. Any enthalpy change measured under these conditions is described as a standard enthalpy change of reaction, ΔH Θ 298 .
5.2 Calculation of Enthalpy Changes (3h) 5.2.1 Calculate the heat change when the temperature of a pure substance is changed Students should be able to calculate the heat change for a substance given the mass, specific heat and temperature change. 5.2.2 Design suitable experimental procedures for measuring the heat energy change of a reaction. Students should consider in aqueous solution and combustion reactions. Use of the bomb calorimeter and calibration of calorimeters will not be assessed. 5.2.3 Calculate the enthalpy change for a reaction in aqueous solution using experimental data on temperature changes, quantities of reactants and mass of solution Enthalpy change of an acid–base reaction could be investigated. 5.2.4 Evaluate the results of experiments to determine enthalpy changes. Students should be aware of the assumptions made and errors due to heat loss.
Measuring Enthalpy Changes Enthalpy changes are measured in KJ per mole, but experiments on measuring enthalpy changes always measure a temperature change. The temperature change is then converted to enthalpy change using the following formula: Energy = m x c x change in temperature = mc T m = mass in grams c = specific heat capacity of water in Joules/Kelvin/gram = 4.18 J /K / g N.B. 1:If you are carrying out an experiment where you heat up something other than water, you must use the specific heat capacity of that substance. 2. A 1K temperature change is the same as a 1 o C change If an aluminium block weighing 10 g is heated by a chemical reaction from 25 o C to 40 o C what is the amount of heat energy produced by this reaction? (The specific heat capacity of Al is 0.9 J per g per K)
Working Out Enthalpy Change from the Temperature Change 14 grams of sodium hydroxide pellets were dissolved in 100 cm3 of water. The temperature before adding the sodium hydroxide pellets was 25 degrees C, and after adding the pellets it was 35 degrees C. Calculate the enthalpy change in KJ/mole of the reaction (Specific heat capacity of water = 4.18 J/K/g) Step 1: Work out the energy change using the formula energy = m x c x change in temperature m = 100 grams (100 cm 3 of water is the “same” as 100 grams of water) change in temperature = 10 K energy = 100 x 4.18 x 10 J Step 2: Convert the enthalpy change to kJ by dividing by 1000 energy = 100 x 4.18 x 10 KJ 1000
Step 3: Work out how many moles there are …0.1 moles (moles = mass/formula mass of NaOH)) Step 4: Work out the enthalpy change per mole for this exothermic reaction so the enthalpy change for 1 mole = 4.2 / 0.1 = - 42 kJ/mol
Experiment to measure some energy changes of reactions occurring in solutions and to use the results to calculate enthalpy changes The experiment is done by simple calorimeter experiments which involve carrying out the reaction in a expanded polystyrene cup (a good insulator) and measuring the change in temperature. This is also a common practical assessment, and you should be aware of the sources of error in your measurements such as heat loss to surroundings, incorrect readings, inaccurate thermometers etc. Apparatus: Thermometer ? range and sensitivity? Lid + Expanded polystyrene cup used as a calorimeter Method: The reaction between copper (II) sulphate and zinc Caution: Zn dust is flammable 1. Write the balanced ionic equation for the reaction: 2. Measure out 25cc of 0.2M copper (II) sulphate solution into the calorimeter. 3. Measure the temperature of the solution. 4. Add 0.1mol of Zn. 5. Stir gently and continuously and note the highest temperature reached. 6. Work out the temperature change to the nearest 0.1 o C 7. Now try to work out: a) the energy given out by the reaction b) the enthalpy change per mol of Cu 2+ 8. Can you show the zinc was added in excess? 9. What key assumptions have you made in these calculations
<ul><li>Discussion: Getting more accurate values of an enthalpy change from a simple experiment with an expanded polystyrene cup. </li></ul><ul><li>There are some simple improvements that you could make to the previous experiment. After a short class discussion, try to give details of three things that you could do: </li></ul><ul><li>1. </li></ul><ul><li>2. </li></ul><ul><li>3. </li></ul><ul><li>There is one more complicated improvement that you could make to help you minimise the uncertainty due to heat loss to the surroundings – the “bit” you can’t measure – you can plot a heating (exothermic reaction) or cooling (endothermic reaction) curve. </li></ul><ul><li>After a class chat, draw a real heating curve for an exothermic reaction in the space below and use it to show how you would calculate a more accurate value for the temperature change that takes into account some of the heat lost to the surroundings. </li></ul>
<ul><ul><ul><li>If the above experiment were done in the following ways, what would happen to the temperature rise observed and hence the energy change? </li></ul></ul></ul><ul><ul><li>Double both the volume of copper sulphate solution and the number of moles of zinc. </li></ul></ul><ul><ul><li>Double the concentration of copper but use the same volume of solution and the same number of moles of zinc. </li></ul></ul><ul><ul><li>Halve the concentration of copper and leave the volume of solution and the number of moles of zinc unchanged. </li></ul></ul><ul><ul><li>What happens to the enthalpy change (energy / number of moles of limiting reactant) in each of the above cases? </li></ul></ul>
STANDARD ENTHALPY CHANGES AND ENTHALPY CHANGES OF COMBUSTION <ul><li>Information needed </li></ul><ul><li>Mass of water heated </li></ul><ul><li>Temperature change of water </li></ul><ul><li>Mass of fuel burnt </li></ul><ul><li>Molar mass of fuel </li></ul><ul><li>Specific heat capacity of water </li></ul>
<ul><li>COMBUSTION OF ALCOHOLS </li></ul><ul><li>Sample Calculation: </li></ul><ul><li>Propan-1-ol calibration calculation: </li></ul><ul><ul><li>Say that 0.090 g of propan-1-ol raised the temperature of your apparatus and water by 12.5 o C </li></ul></ul><ul><ul><li>The enthalpy change of combustion of propan-1-ol tells you that the complete combustion of 60 g of propan-1-ol provides 2021 kJ. </li></ul></ul><ul><ul><li>Work out the number of moles of propan-1-ol burnt in this experiment, using the Periodic table as a source of data. </li></ul></ul><ul><ul><li>Now use the expression H = Q/n to calculate the amount of energy given out by the burning fuel. </li></ul></ul><ul><ul><li>If this amount of energy, in kJ, raised the temperature of the apparatus by 12.5 o C, work out how much energy is needed to raise the temperature of the apparatus by 1 o C </li></ul></ul><ul><ul><li>This is the specific heat capacity of your apparatus in kJ per o C or kJ K -1 . </li></ul></ul>
5.3 Hess’s Law (2h) 5.3.1 Determine the enthalpy change of a reaction which is the sum of two or more reactions with known enthalpy changes. Students should be able to use simple enthalpy cycles and enthalpy level diagrams and to manipulate equations. Students will not be required to state Hess’s law. 5.4 Bond Enthalpies (2h) 5.4.1 Define the term average bond enthalpy Bond enthalpies are quoted for the gaseous state and should be recognized as average values obtained from a number of similar compounds. 5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic
5.3 Hess’s Law (2h) What does Hess’ Law say? Why is Hess’s Law useful? What link is there between the enthalpy change of the forward and the back reaction? Now try to use Hess’ Law to answer the following questions: Using the chemical equations below: C(s) + O 2 (g) CO 2 (g) H = -390 kJ mol -1 Mn(s) + O 2 (g) MnO 2 (s) H = -520 kJ mol -1 Work out the enthalpy change for the following reaction: MnO 2 (s) + C(s) Mn(s) + CO 2 (g)
Using the chemical equations below: Cu(s) + ½ O 2 (g) –CuO(s) H = -156 kJ mol -1 2Cu(s) + ½ O 2 (g) Cu 2 O (s) H = -170 kJ mol -1 work out the value of H (in kJ) for the following reaction: 2CuO (s) Cu 2 O (s) +½ O 2 (g)
Enthalpy Cycles & Hess's Law <ul><li>Instead of manipulating chemical equations, you can also use enthalpy cycles like the one shown in the next slide. H cannot be determined directly by experiment. It is possible, however, to determine the enthalpy changes of combustion of carbon and hydrogen ( H 1 ) and the enthalpy change of methane ( H 2 ). The key idea for this cycle is that the total enthalpy change for one route is the same as the total enthalpy change for an alternative route. </li></ul>
If we can measure ΔH2 and ΔH1, we can find ΔH. Using Hess's Law: ΔH + ΔH2 = ΔH1 Hence: ΔH = ΔH1 - ΔH2 Enthalpy cycles are useful because they enable a value for an enthalpy change to be determined for a reaction which cannot be determined directly from experiment.
Using an Enthalpy Cycle to Determine Enthalpy Change of Reaction <ul><li>It is not possible to determine the enthalpy change for the reaction between silicon tetrachloride and water directly by experiment. </li></ul>ΔH1 = -1212 kJ mol-1 ΔH2 = -1280.1 kJ mol-1 Now use Hess’s law to show ΔH is -68.1kJ/mol
Bond enthalpies and Energy from Fuels <ul><li>Different fuels have different enthalpies of combustion </li></ul>Why do they vary so much? In order to release energy, fuels must combine with oxygen.
<ul><li>The enthalpy change of combustion depends on two things: </li></ul><ul><li>The number of bonds to be made and broken and this depends on the size of the molecule. This is why more energy is released from hexane than methane. </li></ul><ul><li>The type of bonds involved. When methane and methanol burn the same products are formed, carbon dioxide and water, however, methanol already has an O-H bond. In other words, one of the bonds to oxygen is already made, unlike methane where all new bonds with oxygen have to be made. </li></ul>The energy released during combustion comes from the making of bonds to oxygen. If methanol already has one bond made, it will give out less energy when it burns. As a general rule, the more oxygen a fuel has in its molecule, the less energy it will give out when it burns but the rate at which it burns may be greater because oxygen from the air might take time to reach the combustion site. Obviously, efficient explosives have a great deal of oxygen in their molecules so that the combustion reaction is very fast .
Bond Enthalpies <ul><li>The quantity of energy needed to break one mole of a particular bond in a molecule is called the bond dissociation enthalpy , or bond enthalpy for short. </li></ul><ul><li>E.g. H 2 (g) -> 2H (g) ΔH Θ = + 436 kJ mol -1 </li></ul><ul><li>This is a positive ΔH value because bond breaking is an endothermic change. Bond making is an exothermic change. </li></ul><ul><li>What do you think is the relationship between the strength of bonds broken and formed in a chemical reaction and whether that reaction is exo or endothermic. </li></ul>
<ul><li>The stronger a bond, the more difficult it is to break - the higher its bond enthalpy. Tables of bond enthalpies are average values because the exact value depends on the particular compound in which the bond is found. Other atoms and bonds around it influence the value. Double bonds have higher bond enthalpies than single bonds. Triple bonds are higher still. In general the higher the bond enthalpy the shorter the bond. </li></ul>
Bonds and Enthalpy Cycles The bond-breaking and bond-making can be represented in an enthalpy cycle. Now can you use your data book and the information about bond enthalpies to calculate H2 ?
Now use Hess’s Law to show the enthalpy change of combustion is -818 kJ mol-1 <ul><li>This value is a little different from the standard enthalpy change of combustion of methane, -890kJ mol-1. </li></ul><ul><li>Give two reasons why the experimental value of this enthalpy change is different from the one calculated from bond energies. </li></ul><ul><li>………… . </li></ul><ul><li>………… .. </li></ul>