Energetics I PDF

8. Energetics I Definition: Enthalpy change is the amount of heat energy taken in or given out during any change in a system provided the pressure is constant. If an enthalpy change occurs then energy is In an exoth...

8. Energetics I Definition: Enthalpy change is the amount of heat energy taken in or given out during any change in a system provided the pressure is constant. If an enthalpy change occurs then energy is In an exothermic change energy is transferred from transferred between system and the system (chemicals) to the surroundings. surroundings. The system is the chemicals and the surroundings is The products have less energy than the reactants everything outside the chemicals. In an endothermic change, energy is Energy Activation transferred from the surroundings to the Energy: system (chemicals). They require an input of EA heat energy e.g. thermal decomposition of reactants calcium carbonate The products have more energy than the reactants ∆H products Activation Energy: Progress of Reaction EA Energy In an exothermic reaction the ∆H is negative reactants ∆H Common oxidation exothermic processes are the In an endothermic reaction combustion of fuels and the oxidation of the ∆H is positive carbohydrates such as glucose in respiration Progress of Reaction Standard enthalpy change of formation The standard enthalpy change of formation of a compound is the Mg (s) + Cl2 (g)  MgCl2 (s) enthalpy change when 1 mole of the compound is formed from 2Fe (s) + 1.5 O2 (g)  Fe2O3 (s) its elements under standard conditions (298K and 100kpa), all reactants and products being in their standard states The enthalpy of formation Symbol fH of an element = 0 kJ mol-1 Standard enthalpy change of combustion The standard enthalpy of combustion of a substance is defined as CH4 (g) + 2O2 (g) CO2 (g) + 2 H2O (l) the enthalpy change that occurs when one mole of a substance is combusted completely in oxygen under standard conditions. Incomplete combustion will lead to (298K and 100kPa), all reactants and products being in their soot (carbon), carbon monoxide and standard states water. It will be less exothermic than complete combustion. Symbol cH When an enthalpy change is Enthalpy changes are normally quoted at standard conditions. measured at standard Standard conditions are : conditions the symbol is 100 kPa pressure used 298 K (room temperature or 25oC) Eg H Solutions at 1mol dm-3 all substances should have their normal state at 298K N Goalby chemrevise.org 1 Definition: Enthalpy change of reaction is the enthalpy change when the number of moles of reactants as specified in the balanced equation react together 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. Enthalpy changes of neutralisation are always exothermic. For reactions involving strong acids and alkalis, the values are similar, with values between -56 and -58 kJ mol-1 Hess’s Law Hess’s law is a version of the first law Hess’s law states that total enthalpy change for a reaction is of thermodynamics, which is that independent of the route by which the chemical change takes place energy is always conserved. 2H (g) + 2Cl(g) On an energy level diagram the directions of the arrows can show the different routes a reaction can proceed by a In this example one route is arrow ‘a’ H2 + Cl2 b The second route is shown by arrows ΔH plus arrow ‘b’ ΔH So a = ΔH + b 2HCl (g) And rearranged ΔH = a - b ΔH H+ (g) + Br - (g) H+ (aq) + Br - (aq) Interconnecting reactions can also be shown diagrammatically. a d In this example one route is arrow ‘a’ plus ΔH The second route is shown by arrows ‘c’ plus arrow ‘d’ H (g) + Br (g) HBr (g) c So a+ ΔH = c + d And rearranged ΔH = c + d - a N Goalby chemrevise.org 2  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. H reaction This Hess’s law is used to work out the CuSO4 (s) + 5H2O (l) CuSO4.5H2O (s) enthalpy change to form a hydrated salt from an anhydrous salt. -66.1 kJmol-1 + aq + aq +11kJmol-1 This cannot be done experimentally because it is impossible to add the CuSO4 (aq) exact amount of water and it is not easy to measure the temperature change of a solid. H reaction + 11kJmol-1 = -66.1 kJmol-1 Instead both salts are dissolved in excess H reaction = -66.1 - 11 water to form a solution of copper sulfate. = -77.1 kJmol-1 The temperature changes can be measured for these reactions. H reaction This Hess’s law is used to work out the CaCO3 (s) CaO (s) + CO2 (g) enthalpy change for the thermal decomposition of calcium carbonate. ΔH1 ΔH2 (+2HCl) (+2HCl) This cannot be done experimentally because it is impossible to add the heat CaCl2 (aq) + H2O (l) + CO2 (g) required to decompose the solid and to measure the temperature change of a solid at the same time. H reaction + ΔH2 = ΔH1 H reaction = ΔH1 - ΔH2 Instead both calcium carbonate and calcium oxide are reacted with hydrochloric acid to form a solution of calcium chloride. The temperature changes can be measured for these reactions. Using Hess’s law to determine enthalpy changes from enthalpy changes of formation. H reaction Reactants Products H reaction = Σ fH products - Σ fH reactants Σ fH reactants Σ fH products Elements in standard states Example 1. Calculate the enthalpy change for this reaction Al2O3 + 3 Mg  3 MgO + 2 Al H reaction Remember Al2O3 (s) + 3 Mg (s) 3 MgO (s) + 2 Al (s) H = ΣfH products – Σ fH reactants elements have f H = 0 fH(Al2O3) 3 x fH (MgO) H = 3 x fH (MgO) – fH (Al2O3) H = (3 x –601.7) – –1675.7 fH(MgO)= -601.7 kJ mol-1 2Al (s) + 3 Mg (s) + 1.5O2 (g) = -129.4 kJ mol-1 fH(Al2O3) = -1675.7 kJ mol-1 N Goalby chemrevise.org 3  Example 2. Use the following data to calculate the enthalpy of combustion of propene fH C3H6(g) = +20 kJ mol-1 fH CO2(g)= –394 kJ mol-1 fH H2O(g)= –242 kJ mol-1 C3H6 + 4.5 O2  3CO2 + 3H2O  cH C3H6 (g) + 4.5 O2 (g) 3 CO2 (g) + 3 H2O (g) cH = Σ fH products – Σ fH reactants cH = [3 x fH (CO2) + 3 x fH (H2O)] - fH (C3H6) 3 x fH (CO2) fH(C3H6) 3 x fH (H2O) cH = [(3 x –394) + (3 x –242)] – 20 = -1928 kJ mol-1 3C (s) + 3 H2 (g) + 4.5 O2 (g) Using Hess’s law to determine enthalpy changes from enthalpy changes of combustion. H reaction Reactants Products H reaction = Σ cH reactants - Σ cH products Σ cH reactants +O2 +O2 Σ cH products Combustion Products Example 3. Use the following combustion data to calculate the enthalpy of reaction CO (g) + 2H2 (g)  CH3OH (g) cH CO(g) = -283 kJ mol cH H2 (g)= –286 kJ mol-1 cH CH3OH(g)= –671 kJ mol-1 -1 H reaction H reaction = Σ cH reactants - Σ cH products CO (g) + 2H2(g) CH3OH(g) H = cH (CO) + 2 x cH (H2) - cH (CH3OH) +O2 +O2 cH(CO) + cH(CH3OH) H = -283+ 2x –286 - -671 2x cH (H2) CO2 (g) + 2 H2O (l) = -184 kJ mol-1 Example 4. Use the following combustion data to calculate the enthalpy 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 f H H = Σ cH reactants - Σ cH products 3C (s) + 3H2 (g) C3H6 (g) fH = 3 x cH (C) + 3 x cH (H2 ) - cH (C3H6) 3 x cH (C) + cH(C3H6) fH = 3x -393+ 3x –286 - -2058 3 x cH (H2 ) = +21 kJ mol-1 3 CO2 (g) + 3 H2O (l) N Goalby chemrevise.org 4 Measuring the enthalpy change for a reaction experimentally Calorimetric method This equation will only give the energy for the actual quantities used. Normally this value is For a reaction in solution we use the following equation converted into the energy energy change = mass of solution x heat capacity x temperature change change per mole of one of the Q (J) = m (g) x cp (J g-1K-1) x T ( K) reactants. (The enthalpy change of reaction, rH) Calorimetric method Practical One type of experiment is one in which substances are This could be a solid dissolving or reacting mixed in an insulated container and the temperature rise in a solution or it could be two solutions measured. reacting together 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  Measure out desired volumes of solutions with volumetric pipettes and transfer to insulated cup  clamp thermometer into place making sure the thermometer bulb is immersed in solution  measure the initial temperatures of the solution or both solutions if 2 are used. Do this every minute for 2-3 minutes  At minute 3 transfer second reagent 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.  If using a solid reagent then use ‘before and after’ weighing method  stirs mixture (ensures that all of the solution is at the same temperature)  Record temperature every minute after addition for several minutes 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. Errors in this method Read question carefully. It may be energy transfer from surroundings (usually loss) necessary to describe: approximation in specific heat capacity of solution. The method assumes all Method solutions have the heat capacity of water. Drawing of graph neglecting the specific heat capacity of the calorimeter- we ignore any with extrapolation energy absorbed by the apparatus. Description of the reaction or dissolving may be incomplete or slow. calculation density of solution is taken to be the same as water. N Goalby chemrevise.org 5 Calculating the enthalpy change of reaction, Hr from experimental data The heat capacity of water is General method 4.18 J g-1K-1. In any reaction where the reactants are 1. Using q= m x cp x T calculate energy change for quantities used dissolved in water we assume that the heat capacity is the 2. Work out the moles of the reactants used same as pure water. 3. Divide q by the number of moles of the reactant not in excess to give H Also assume that the solutions have the density of water, 4. Add a sign and unit (divide by a thousand to convert Jmol-1 to kJmol-1 which is 1g cm-3. Eg 25cm3 will weigh 25 g Example 5. Calculate the enthalpy change of reaction for the reaction where 25cm3 of 0.2 mol dm-3 copper sulfate 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 calorimeter. Q = m x cp x T Note the mass is the mass of the copper sulfate Q = 25 x 4.18 x 7 solution only. Do not include mass of zinc powder. Q = 731.5 J Step 2 : calculate the number of moles of the reactant not in excess. moles of CuSO4 = conc x vol If you are not told what is in excess, then you need to = 0.2 x 25/1000 work out the moles of both reactants and work out = 0.005 mol 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 Remember in these questions: sign, unit, sig Finally add in the sign to represent the energy change: if temp increases figs same as data given. the reaction is exothermic and is given a minus sign e.g. –146 kJ mol-1 Example 6. 25cm3 of 2 mol dm-3 HCl was neutralised by 25cm3 of 2 mol dm-3 NaOH. The temperature increased 13.5oC. Calculate the enthalpy change per mole of HCl? Step 1: Calculate the energy change for the amount of reactants in the calorimeter. Q = m x cp x T Note the mass equals the mass of acid + the Q = 50 x 4.18 x13.5 mass of alkali, as they are both solutions. 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 can be called the enthalpy change of neutralisation H = Q/ no of moles = 2821.5/0.05 = 56430 J mol-1 Remember in these Exothermic and so is given a minus sign questions: sign, unit, sig = -56.4 kJ mol-1 to 3 sf figs same as data given. N Goalby chemrevise.org 6  Measuring enthalpies of combustion using calorimetry 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. Example 7. 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 Note the mass is the mass of water in the Q = 150 x 4.18 x 25.4 calorimeter and not the alcohol 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 cH (the enthalpy change of combustion) H = Q/ no of moles = 15925.8/0.01083 = 1470073 J mol-1 Remember in these = 1470 kJ mol-1 to 3 sf questions: sign, Finally add in the sign to represent the energy change: if temp increases unit, 3 sig figs. the reaction is exothermic and is given a minus sign eg –1470 kJ mol-1 Errors in this method Energy losses from calorimeter Incomplete combustion of fuel Incomplete transfer of energy 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 N Goalby chemrevise.org 7  Mean Bond enthalpies These values are positive because Definition: The mean bond enthalpy is the enthalpy needed to energy is required to break a bond. break the covalent bond into gaseous atoms, averaged over different molecules The definition only applies when the substances start and end in the We use values of mean bond enthalpies because every single gaseous state. bond in a compound has a slightly different bond energy. E.g. In CH4 there are 4 C-H 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. The value for the bond enthalpy for the C-H bond in methane matches this reaction ¼ CH4 (g)  C (g) + H (g) Gaseous atoms Gaseous atoms Energy Energy Energy Energy breaking Activation In an exothermic reaction Energy breaking Activation bonds Energy Energy making the sum of the bonds in bonds Energy products reactants making bonds bonds the reactant molecules H will be less than the sum H products reactants of the bonds in the product molecules Progress of reaction Progress of reaction Reaction profile for an Reaction profile for an exothermic reaction endothermic reaction In general (if all substances are gases) H =  bond enthalpies broken -  bond enthalpies made H reaction Reactants Products Σbond energies Σbond energies broken in reactants made in products Gaseous atoms of elements H values calculated using this method will be less accuate than using formation or combustion data because the mean bond energies are not exact N Goalby chemrevise.org 8 Example 8. Using the following mean bond enthalpy data to calculate Bond Mean enthalpy the enthalpy of combustion of propene (kJ mol-1) H H O C=C 612 H C C C + 4.5 O O 3 O C O + 3 H H C-C 348 H H H O=O 496 H =  bond enthalpies broken -  bond enthalpies 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)] O=C 743 = [ 612 + 348 + (6 x 412) + (4.5 x 496) ] – [ (6 x 743) + (6 X 463)] O-H 463 = - 1572 kJmol-1 C-H 412 Example 9. Using the following mean bond enthalpy data to calculate the enthalpy of formation of NH3 ½ N2 + 1.5 H2  NH3 (note the balancing is to agree with the definition of enthalpy 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 =  bond enthalpies broken -  bond enthalpies 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 A more complicated example that may occur This is a more complicated Working out fH of a compound using bond energies and other data example of the type in example 9 elements  fH Compound in standard state The H’s can be combinations of H to turn elements H to turn compound into different data into gaseous atoms gaseous atoms Gaseous atoms Can be bond energies E(Cl-Cl) Cl2  2Cl  bond energies of Or atomisation energies (if compound + (H to turn to the substance is not diatomic gas if compound is not gaseous) C(s)  C(g) Example 10 Calculate Δ fH for propane, C3H8(g), given the following data. Bond C–C C–H H–H C(s)  C(g) ΔH = 715 kJ mol-1 kJ mol-1 348 412 436 3C (s) + 4H2 (g)  C3H8(g), Σ H to turn elements - Σ H to turn compound  fH = into gaseous atoms into gaseous atoms fH = (3xHat [C] + 4 x E[H-H] ) – (2 x E[C-C]+ 8 x E[C-H] ) = (3x715 + 4 x 436 ) – (2 x 348+ 8 x 412 ) =-103 kJ mol-1 N Goalby chemrevise.org 9  Enthalpies of combustion in a homologous series When comparing the heats of combustion for successive members of a homologous series such as alkanes or alcohols there is a constant rise in the size of the heats of combustion as the number of carbon atoms increases H H O 1 C-C, 5C-H 1C-O 1O-H and 3 O=O H C C O H + 3 O O 2 O C O + 3 H H bonds are broken H H 4 C=O and 6 O-H bonds are made ethanol ΔHc = -1365 kJ mol-1 H H H 2C-C, 7C-H 1C-O 1O-H and 4.5 O=O O bonds are broken H C C C O H + 4.5 O O 3 O C O + 4 H H 6 C=O and 8 O-H bonds are made H H H ΔHc = -2016 kJ mol-1 Propan-1-ol H H H H 3C-C, 9C-H 1C-O 1O-H and 6 O=O H C C C C O H + 6 O 4 bonds are broken O O O C O+ 5 H H H H H H 8 C=O and 10 O-H bonds are made Butan-1-ol ΔHc = -2677 kJ mol-1 As one goes up the homologous series there is a constant amount and type of extra bonds being broken and made e.g. 1C-C, 2C-H and 1.5 O=O extra bonds broken and 2 C=O and 2 O-H extra bonds made, so the enthalpy of combustion increases by a constant amount calculated If the results are worked out experimentally using a calorimeter the experimental results combustion kJ mol-1 Enthalpy of will be much lower than the calculated ones because there will be significant heat loss. There will also be incomplete combustion experimental which will lead to less energy being released. Mr of alcohol Remember that calculated values of enthalpy of combustions will be more accurate if calculated from enthalpy of formation data than if calculated from average bond enthalpies. This is because average bond enthalpy values are averaged values of the bond enthalpies from various compounds. N Goalby chemrevise.org 10