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HIGHER CHEMISTRY REVISION .

HIGHER CHEMISTRY REVISION. Unit 1:- Enthalpy. Ammonium chloride (NH 4 Cl) is soluble in water. A student dissolved 10.0 g of ammonium chloride in 200 cm 3 of water and found that the temperature of the solution fell from 23.2°C to 19.8 °C.

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HIGHER CHEMISTRY REVISION .

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  1. HIGHER CHEMISTRY REVISION. Unit 1:- Enthalpy • Ammonium 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 = -4.18 x 0.2 x –3.4 = 2.84 kJ 10 g  2.84 kJ So 1 mole of NH4Cl, 53.5 g  53.5/10 x 2.84  15.2 kJ

  2. X 120 100 Potential energy kJ mol-1. 80 • (a) (i) +70 kJ • (ii) +40 kJ • -20 kJ • Exothermic. 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. 60 reactants 40 products 20 0 Reaction pathway 2.Consider the following potential energy diagram. • What 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?

  3. 3. 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 = -4.18 x 0.4 x 6.8 = -11.37 kJ The number of moles of acid used = C x V(litres) = 1.0 x 200/1000 = 0.2 There is the same number of moles of KOH used. Equation for the reaction is HCl + KOH  KCl + H2O 1 mol 1 mol 1 mol So 0.2 mol 0.2 mol 0.2 mol When 0.2 moles of water is formed DH= -11.37 kJ So when 1 mole of water is formed DH= -11.37 x 1.0/0.2 = -56.85 kJ

  4. The enthalpies of combustion of methane, ethane, propane are and butane –891, -1560, – 2220 kJ and 2877 mol-1 respectively. • (a) (i) Explain why there a regular increase in the enthalpies of • combustion 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 to • heat 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 this • will 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, = 44g • Burning 44g  DH= -2220 kJ • So burning 0.2 g  DH= -2220 x 0.2/44 = -10.1 kJ • DH= -cmDT so DT= -DH/cm = -(-10.1) / 4.18 x 0.4 • = 6oC • (c) Incomplete combustion of the propane,

  5. 5. 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 20/1000 = 0.02 mol NaOH + HCl  NaCL + H2O 1 mol 1 mol 1 mol 0.02 0.02 0.02 DH = - cmDT = - 4.18 x 0.04 x 6.5 = -1.0868 kJ When 0.02 mol of water is formed DH = -1.0868 kJ So when 1 mol of water is formed DH = -54.34 kJ

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