HIGHER CHEMISTRY REVISION.

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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 23.2C 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
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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?

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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
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• 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 (from
  page 9 of data book)
• 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,