Enthalpy changes

3.2.1. Enthalpy changes If an enthalpy change occurs then energy is transferred between system and surroundings . The system is the chemicals and the surroundings is everything outside the chemicals. In an exothermic change energy is transferred from the system (chemicals) to the surroundings. The products have less energy than the reactants In an exothermic reaction the ∆H is negative In an endothermic change, energy is transferred from the surroundings to the system (chemicals). They require an input of heat energy e.g. thermal decomposition of calcium carbonate The products have more energy than the reactants In an endothermic reaction the ∆H is positive reactants products Activation Energy: EA ∆H Progress of Reaction Energy reactants Activation Energy: EA ∆H Progress of Reaction Energy Common oxidation exothermic processes are the combustion of fuels and the oxidation of carbohydrates such as glucose in respiration The Activation Energy is defined as the minimum energy which particles need to collide to start a reaction Enthalpy changes are normally quoted at standard conditions. Standard conditions are : • 100 kPa pressure • 298 K (room temperature or 25oC) • Solutions at 1mol dm-3 • all substances should have their normal state at 298K When an enthalpy change is measured at standard conditions the symbol is used Eg ∆H Definition: Enthalpy change of reaction ∆rH is the enthalpy change when the number of moles of reactants as specified in the balanced equation react together Standard enthalpy change of formation The standard enthalpy change of formation of a compound is the enthalpy change when 1 mole of the compound is formed from its elements under standard conditions (298K and 100kpa), all reactants and products being in their standard states Symbol ∆fH Mg (s) + Cl2 (g)
MgCl2 (s) 2Fe (s) + 1.5 O2 (g)
Fe2O3 (s) The enthalpy of formation of an element = 0 kJ mol-1. Standard enthalpy change of combustion The standard enthalpy of combustion of a substance is defined as the enthalpy change that occurs when one mole of a substance is combusted completely in oxygen under standard conditions. (298K and 100kPa), all reactants and products being in their standard states Symbol ∆cH CH4 (g) + 2O2 (g)
CO2 (g) + 2 H2O (l) Incomplete combustion will lead to soot (carbon), carbon monoxide and water. It will be less exothermic than complete combustion. 2 N Goalby chemrevise.org Enthalpy change of Neutralisation The standard enthalpy change of neutralisation is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water. Symbol ∆neutH

General method • washes the equipment (cup and pipettes etc) with the solutions to be used • dry the cup after washing • put polystyrene cup in a beaker for insulation and support • clamp thermometer into place making sure the thermometer bulb is immersed in liquid • measure the initial temperatures of the solution or both solutions if 2 are used • transfers reagents to cup. If a solid reagent is used then add the solution to the cup first and then add the solid weighed out on a balance. • stirs mixture • Measures final temperature One type of experiment is one in which substances are mixed in an insulated container and the temperature rise measured. This could be a solid dissolving or reacting in a solution or it could be two solutions reacting together Errors in this method • heat transfer from surroundings (usually loss) • approximation in specific heat capacity of solution. The method assumes all solutions have the heat capacity of water. • neglecting the specific heat capacity of the calorimeter- we ignore any heat absorbed by the apparatus. • reaction or dissolving may be incomplete or slow. • Density of solution is taken to be the same as water. C. If the reaction is slow then the exact temperature rise can be difficult to obtain as cooling occurs simultaneously with the reaction To counteract this we take readings at regular time intervals and extrapolate the temperature curve/line back to the time the reactants were added together. We also take the temperature of the reactants for a few minutes before they are added together to get a better average temperature. If the two reactants are solutions then the temperature of both solutions need to be measured before addition and an average temperature is used. • Heat losses from calorimeter • Incomplete combustion of fuel • Incomplete transfer of heat • Evaporation of fuel after weighing • Heat capacity of calorimeter not included • Measurements not carried out under standard conditions as H2O is gas, not liquid, in this experiment

For a reaction in solution we use the following equation energy change = mass of solution x heat capacity x temperature change Q (J) = m (g) x cp (J g-1K-1) x ∆T ( K) This equation will only give the energy for the actual quantities used. Normally this value is converted into the energy change per mole of one of the reactants. (The enthalpy change of reaction, ∆Hr ) Calculating the enthalpy change of reaction, ∆Hr from experimental data General method 1. Using q= m x cp x ∆T calculate energy change for quantities used 2. Work out the moles of the reactants used 3. Divide q by the number of moles of the reactant not in excess to give ∆H 4. Add a sign and unit (divide by a thousand to convert Jmol-1 to kJmol-1 The heat capacity of water is 4.18 J g-1K-1. In any reaction where the reactants are dissolved in water we assume that the heat capacity is the same as pure water. Also assume that the solutions have the density of water, which is 1g cm-3. Eg 25cm3 will weigh 25 g Example 1. Calculate the enthalpy change of reaction for the reaction where 25cm3 of 0.2 M copper sulphate was reacted with 0.01mol (excess of zinc). The temperature increased 7oC . Step 1: Calculate the energy change for the amount of reactants in the test tube. Q = m x cp x ∆T Q = 25 x 4.18 x 7 Q = 731.5 J Step 2 : calculate the number of moles of the reactant not in excess. moles of CuSO4 = conc x vol = 0.2 x 25/1000 = 0.005 mol If you are not told what is in excess, then you need to work out the moles of both reactants and work out using the balanced equation which one is in excess. Step 3 : calculate the enthalpy change per mole which is often called ∆H (the enthalpy change of reaction) ∆H = Q/ no of moles = 731.5/0.005 = 146300 J mol-1 = 146 kJ mol-1 to 3 sf Finally add in the sign to represent the energy change: if temp increases the reaction is exothermic and is given a minus sign e.g. –146 kJ mol-1 Remember in these questions: sign, unit, 3 sig figs. Example 2. 25cm3 of 2M HCl was neutralised by 25cm3 of 2M NaOH. The Temperature increased 13.5oC What was the energy change per mole of HCl? Step 1: Calculate the energy change for the amount of reactants in the test tube. Q = m x cp x ∆T Q = 50 x 4.18 x13.5 Q = 2821.5 J Step 2 : calculate the number of moles of the HCl. moles of HCl = conc x vol = 2 x 25/1000 = 0. 05 mol Step 3 : calculate ∆H the enthalpy change per mole which might be called the enthalpy change of neutralisation ∆H = Q/ no of moles = 2821.5/0.05 = 564300 J mol-1 = -56.4 kJ mol-1 to 3 sf Exothermic and so is given a minus sign Remember in these questions: sign, unit, 3 sig figs. Note the mass equals the mass of acid + the mass of alkali, as they are both solutions. Note the mass is the mass of the copper sulphate solution only. Do not include mass of zinc powder. Example 3. Calculate the enthalpy change of combustion for the reaction where 0.65g of propan-1-ol was completely combusted and used to heat up 150g of water from 20.1 to 45.5oC Step 1: Calculate the energy change used to heat up the water. Q = m x cp x ∆T Q = 150 x 4.18 x 25.4 Q = 15925.8 J Step 2 : calculate the number of moles of alcohol combusted. moles of propan-1-ol = mass/ Mr = 0.65 / 60 = 0.01083 mol Step 3 : calculate the enthalpy change per mole which is called ∆Hc (the enthalpy change of combustion) ∆H = Q/ no of moles = 15925.8/0.01083 = 1470073 J mol-1 = 1470 kJ mol-1 to 3 sf Finally add in the sign to represent the energy change: if temp increases the reaction is exothermic and is given a minus sign eg –1470 kJ mol-1 Remember in these questions: sign, unit, 3 sig figs. Note the mass is the mass of water in the calorimeter and not the alcohol Enthalpies of combustion can be calculated by using calorimetry. Generally the fuel is burnt and the flame is used to heat up water in a metal cup. N Goalby chemrevise.org M

Mean Bond energies Definition: The Mean bond enthalpy is the Enthalpy change when one mole of bonds of (gaseous covalent) bonds is broken (averaged over different molecules) These values are positive because energy is required to break a bond. The definition only applies when the substances start and end in the gaseous state. We use values of mean bond energies because every single bond in a compound has a slightly different bond energy. E.g. In CH4 there are 4 CH bonds. Breaking each one will require a different amount of energy. However, we use an average value for the C-H bond for all hydrocarbons. In general (if all substances are gases) ∆H = Σ bond energies broken – Σ bond energies made ∆H values calculated using this method will be less accuate than using formation or combustion data because the mean bond energies are not exact Reaction profile for an EXOTHERMIC reaction Reaction profile for an ENDOTHERMIC reaction In an exothermic reaction more energy is released when making bonds than is absorbed when breaking bonds Gaseous atoms of elements Reactants Products Σbond energies broken in reactants ∆H reaction Σbond energies made in products Energy Progress of reaction Energy breaking bonds products Gaseous atoms Activation Energy Energy making bonds reactants Energy Progress of reaction Energy breaking bonds products Gaseous atoms Activation Energy ∆H ∆H N Goalby chemrevise.org 6 Bond breaking absorbs energy and bond making releases energy In an endothermic reaction more energy is absorbed when breaking bonds than is released when making bonds. Example 4. Using the following mean bond enthalpy data to calculate the heat of combustion of propene Bond Mean enthalpy (kJ mol-1) C=C 612 C-C 348 O=O 496 O=C 743 O-H 463 C-H 412 ∆H = Σ bond energies broken – Σ bond energies made = [E(C=C) + E(C-C) + 6 x E(C-H) + 4.5 x E(O=O)] – [ 6 xE(C=O) + 6 E(O-H)] = [ 612 + 348 + (6 x 412) + (4.5 x 496) ] – [ (6 x 743) + (6 X 463)] = – 1572 kJmol-1. Example 5. Using the following mean bond enthalpy data to calculate the heat of formation of NH3 ½ N2 + 1.5 H2
NH3 (note the balancing is to agree with the definition of heat of formation (i.e. one mole of product) E(N≡N) = 944 kJ mol-1 E(H-H) = 436 kJ mol-1 E(N-H) = 388 kJ mol-1 ∆H = Σ bond energies broken – Σ bond energies made = [0.5 x E(N≡N) + 1.5 x E(H-H)] – [ 3 xE(N-H)] = [ (0.5 x 944) + (1.5 x 436) ] – [ 3 x 388)] = – 38 kJmol-1

Hess’s Law Hess’s law states that total enthalpy change for a reaction is independent of the route by which the chemical change takes place Hess’s law is a version of the first law of thermodynamics, which is that energy is always conserved. 2H (g) + 2Cl(g) H2 + Cl2 2HCl (g) a b ΔH On an energy level diagram the directions of the arrows can show the different routes a reaction can proceed by. In this example one route is arrow ‘a’ The second route is shown by arrows ΔH plus arrow ‘b’ So a = ΔH + b And rearranged ΔH = a – b H+ (g) + Br – (g) H+ (aq) + Br – (aq) H (g) + Br (g) HBr (g) a c d ΔH Interconnecting reactions can also be shown diagrammatically. In this example one route is arrow ‘a’ plus ΔH The second route is shown by arrows ‘c’ plus arrow ‘d’ So a+ ΔH = c + d And rearranged ΔH = c + d – a CuSO4 (aq) CuSO4 (s) + 5H2O (l) CuSO4 .5H2O (s) + 11kJmol-1 = -66.1 kJmol-1 ∆H reaction + aq + aq ∆H reaction +11kJmol-1 -66.1 kJmol-1 ∆H reaction = -66.1 – 11 = -77.1 kJmol-1 This Hess’s law is used to work out the enthalpy change to form a hydrated salt from an anhydrous salt. This cannot be done experimentally because it is impossible to add the exact amount of water and it is not easy to measure the temperature change of a solid. Often Hess’s law cycles are used to measure the enthalpy change for a reaction that cannot be measured directly by experiments. Instead alternative reactions are carried out that can be measured experimentally. Instead both salts are dissolved in excess water to form a solution of copper sulphate. The temperature changes can be measured for these reactions. Using Hess’s law to determine enthalpy changes from enthalpy changes of formation. ∆H reaction = Σ ∆fH products – Σ ∆fH reactants Example 6. What is the enthalpy change for this reaction ? Al2O3 + 3 Mg
3 MgO + 2 Al ∆H = Σ∆fH products – Σ ∆fH reactants ∆H = 3 x ∆fH (MgO) – ∆fH (Al2O3 ) ∆H = (3 x –601.7) – –1675.7 = -129.4 kJ mol-1 Remember elements have ∆Hf = 0 ∆fH(MgO)= -601.7 kJ mol-1 ∆fH(Al2O3 ) = -1675.7 kJ mol-1 Example 7. Using the following data to calculate the heat of combustion of propene ∆Hf C3H6 (g) = +20 kJ mol-1 ∆Hf CO2 (g)= –394 kJ mol-1 ∆Hf H2O(g)= –242 kJ mol-1 C3H6 + 4.5 O2
3CO2 + 3H2O ∆cH = Σ ∆fH products – Σ ∆fH reactants ∆cH = [3 x ∆fH (CO2 ) + 3 x ∆fH (H2O)] – ∆fH (C3H6 ) ∆cH = [(3 x –394) + (3 x –242)] – 20 = -1928 kJ mol-1 Using Hess’s law to determine enthalpy changes from enthalpy changes of combustion. ∆H reaction = Σ ∆cH reactants – Σ ∆cH products Example 8. Using the following combustion data to calculate the heat of reaction CO (g) + 2H2 (g)
CH3OH (g) ∆cH CO(g) = -283 kJ mol-1 ∆cH H2 (g)= –286 kJ mol-1 ∆cH CH3OH(g)= –671 kJ mol-1 ∆H reaction = Σ ∆cH reactants – Σ ∆cH products ∆H = ∆cH (CO) + 2 x ∆cH (H2 ) – ∆cH (CH3OH) ∆H = -283+ 2x –286 – -671 = -184 kJ mol-1 Elements in standard states Reactants Products Σ ∆ Σ ∆fH products fH reactants ∆H reaction 2Al (s) + 3 Mg (s) + 1.5O2 (g) Al2O3 (s) + 3 Mg (s) 3 MgO (s) + 2 Al (s) ∆ 3 x ∆fH (MgO) fH(Al2O3 ) ∆H reaction 3C (s) + 3 H2 (g) + 4.5 O2 (g) C3H6 (g) + 4.5 O2 (g) 3 CO2 (g) + 3 H2O (g) 3 x ∆fH (CO2 ) ∆fH(C3H6 ) ∆cH 3 x ∆fH (H2O) Combustion Products Reactants Products Σ ∆cH reactants Σ ∆cH products ∆H reaction CO2 (g) + 2 H2O (l) CO (g) + 2H2 (g) CH3OH(g) ∆cH(CO) + ∆cH(CH3OH) Example 9. Using the following combustion data to calculate the heat of formation of propene 3C (s) + 3H2 (g)
C3H6 (g) ∆cH C (s) = -393kJ mol-1 ∆cH H2 (g)= –286 kJ mol-1 ∆cH C3H6 (g)= –-2058 kJ mol-1 ∆H = Σ ∆cH reactants – Σ ∆cH products ∆fH = 3 x ∆cH (C) + 3 x ∆cH (H2 ) – ∆cH (C3H6 ) ∆fH = 3x -393+ 3x –286 – -2058 = +21 kJ mol-1 3 CO2 (g) + 3 H2O (l) 3C (s) + 3H2 (g) C3H6 (g) ∆cH(C3H6 3 x ∆cH (C) + )