2HIGHER CHEMISTRY REVISION. Unit 1:- EnthalpyAmmonium chloride (NH4Cl) is soluble in water.A student dissolved 10.0 g of ammonium chloride in 200 cm3 of water and found that the temperature of the solution fell from 23.2°C to 19.8 °C.Calculate the enthalpy of solution of ammonium chloride.DH = -cmDT= x 0.2 x –3.4= 2.84 kJ10 g 2.84 kJSo 1 mole of NH4Cl, g 53.5/10 x 2.84 kJ
32. Consider the following potential energy diagram. 12010080604020Potential energykJ mol-1.reactantsproductsReaction pathway(a) (i) +70 kJ(ii) +40 kJ-20 kJExothermic. The products have less energy than reactants so energy is given out to the surroundings.(d) (i) An unstable group of atoms with partly made and partly broken bonds which is mid way between reactants and products.(ii) 110kJ – the activated complex is formed at point shown by X on diagram.XWhat is the value for the activation energy for(i) the un-catalysed forward reaction?(ii) the catalysed forward reaction?(b) What is the value for the enthalpy change for the forward reaction?(c) Is the reaction exothermic or endothermic? Explain your answer.(d) (i) What is meant by the term ‘activated complex’?(ii) What would be the potential energy of the activated complex?
43. When 200 cm3 of 1.0 mol l-1 hydrochloric acid was reacted with 200 cm3 of 1.0 mol l -1 potassium hydroxide the temperature of the mixture rose by 6.8oC.Calculate the enthalpy of neutralisation.DH= -cmDT= x 0.4 x 6.8= kJThe number of moles of acid used = C x V(litres)= 1.0 x 200/ = 0.2There is the same number of moles of KOH used.Equation for the reaction isHCl KOH KCl + H2O1 mol mol molSo 0.2 mol 0.2 mol molWhen 0.2 moles of water is formed DH= kJSo when 1 mole of water is formed DH= x 1.0/ = kJ
5(a) (i) Explain why there a regular increase in the enthalpies of The enthalpies of combustion of methane, ethane, propane are and butane –891, -1560, – 2220 kJ and mol-1 respectively.(a) (i) Explain why there a regular increase in the enthalpies ofcombustion from methane, to ethane to propane to butane?(ii) Estimate the enthalpy of combustion of pentane.(b) Calculate the temperature rise when 0.2g of propane is used toheat 400cm3 of water. Assume there are no heat losses.The value obtained by experiment in the laboratory is much less than the expected answer due to heat losses to the surroundings.Give one other reason why the value in the laboratory is less than the expected answer.(a) (i) There is an extra CH2 group being added each time and the burning of thiswill give out the same additional amount of energy each time.(ii) A value of between 3520 and 3550 kJ mol-1 would be reasonable.Mass of 1 mole of propane, C3H8, = 44gBurning 44g DH= kJSo burning 0.2 g DH= x 0.2/44 = kJDH= -cmDT so DT= -DH/cm = -(-10.1) / x 0.4= 6oC(c) Incomplete combustion of the propane,
65. An experiment using dilute hydrochloric acid and sodium hydroxide solution was carried out to determine the enthalpy of neutralisation. Using the information in the diagram, calculate the enthalpy of neutralisation,in kJ mol-1.No. of moles of HCl = No. of moles of NaOH (same volume and concentration)No of moles = C x V(litres)= 1 x 20/ = molNaOH HCl NaCL H2O1 mol mol molDH = - cmDT= x 0.04 x 6.5= kJWhen 0.02 mol of water is formed DH = kJSo when 1 mol of water is formed DH = kJ