Presentation on theme: "HIGHER CHEMISTRY REVISION."— Presentation transcript:
1HIGHER CHEMISTRY REVISION. Unit 3 :- Hess’s Law1. The Thermite Process involves the reaction between aluminium and iron(III) oxide to produce iron and aluminium oxide.This highly exothermic reaction, which generates so much heat that the temperature of the mixture rises to around 3000oC, is used for repairing cracked railway lines as shown in the diagram below.(a) Suggest why this process is suitable for repairing cracked railway lines.(b) The enthalpy changes for the formation of one mole of aluminium oxideand one mole of iron(III) oxide are shown below.2Al(s) ½ O2(g) Al2O3(s) DH = kJ mol-1.2Fe(s) ½ O2(g) Fe2O3(s) DH = -825 kJ mol-1.Use the above information to calculate the enthalpy change for the reaction:2Al(s) Fe2O3(s) Al2O3(s) Fe(s)(a) The iron formed ismolten and it fillsthe crack – it thensolidifies.(b) Leave firstequation andreverse thesecond equation.DH = (-1676) +(+825)= kJ
22. The enthalpy of combustion of hydrogen sulphide is -563 kJ mol-1. Use this value and the enthalpy of combustion values in the databooklet to calculate the enthalpy change for the reaction:H2(g) S(s) H2S(g)(rhombic)H ½ O2 H2O DH1 = kJS O2 SO2 DH2 = kJH2S ½ O2 H2O + SO2 DH3 = kJTo form the equation shown we need DH1 + DH2 - DH3And so DH = (-286) + (-297) + (+563) = kJ
33. A calorimeter, like the one shown, can be used to measure the enthalpy of combustion of ethanol.The ethanol is ignited and burns completely inthe oxygen gas. The heat energy released in thereaction is taken in by the water as the hotproduct gases are drawn through the coiled copperpipe by the pump.(a) Why is the copper pipe coiled as shown in the diagram.(b) The value of enthalpy of combustion of ethanol obtained by the calorimeter methodis higher than the value obtained by the typical school laboratory method.One reason for this is that more heat is lost to the surroundings in the typicalschool laboratory method.Give one other reason for the value being higher with the calorimeter method.(c) In one experiment the burning of g of ethanol resulted in the temperature of400 cm3 of water rising from 14.2oC to 31.6oC.Use this information to calculate the enthalpy of combustion of ethanol.(a) Coiling the pipe increases its surface area and ensures as much heat istransferred into the water as possible.(b) Using oxygen ensures complete combustion of ethanol.DH=-cmDT = x 0.4 x 17.4 = kJ0.98 g of ethanol kJSo 1 mol, 46g, of ethanol kJ
44. Potassium hydroxide can be used in experiments to verify Hess’s Law 4. Potassium hydroxide can be used in experiments to verify Hess’s Law. The reactionsconcerned can be summarised as follows.KOH(s) KCl(aq)KOH(aq)(a) State Hess’s Law.(b) Complete the list of measurements that would have to be made in order to calculateDH2.(i) Mass of potassium hydroxide(ii)(iii)(iv)(c) What solution must be added to the potassium hydroxide solution in order tocalculate DH3?+HCl(aq)DH1DH2DH3+H2O(l)The enthalpy of a reaction is independent of the reaction pathway.Mass of water, initial water temperature, final water temperature.(c) Hydrochloric acid.
55. The equation for the enthalpy of formation of propanone is: 3C(s) + 3H2(g) + ½ O2 (g) C3H6O(l)Use the following information on enthalpies of combustion to calculate the enthalpy offormation of propanone.(1) C(s) + O2(g) CO2 (g) DH1 = –394 kJ mol-1.(2) H2 (s) ½ O2 (g) H2O(l) DH2 = –286 kJ mol -1.(3) C3H6O(l) + 4O2 (g) 3CO2 (g) + 3H2O(l) DH3 = –1804 kJ mol -1.To form the equation we want we need:-3 x Eqn(1) x Eqn (2) + Reverse Eqn(3)DH = (3 x –394) + (3 x –286) + (+1804)= kJ
66. Some rockets have a propellant system which combines dinitrogen tetroxide with methylhydrazine.5N2O CH3NHNH2 xN yH22O zCO2(a) State the values of x, y and z required to balance the above equation.(b) Draw the full structural formula for methylhydrazine. (c) Methylhydrazine burns according to the following equation.CH3NHNH O2 CO H2O N2 DH = –1305 kJ mol-1.Use this information, together with information from page 9 of the data booklet, tcalculate the enthalpy change for the following reaction.C + N H2 CH3NHNH2x = 9, y = 12 and z = 4H H HH-C-N –N-HH(c)(1) CH3NHNH O2 CO H2O N2 DH1 = –1305 kJ mol-1.(2) C + O2 CO2 DH2 = –394 kJ mol-1.(3) H2 + ½ O2 H2O DH3 = –286 kJ mol-1.To form the equation we want we needReverse (1) + (2) + 3 x (3)DH = (1305) + (-394) + (3 x –286)= 53 kJ