School of Chemistry and Physics, University of KwaZulu-Natal, Westville Campus, Durban
CHEMICAL REACTIVITY – CHEM120
Tutorial 4 – 22 and 24 August 2012
1. The vapour pressure of ethanol (C2H5OH) and 1-propanol (C3H7OH) at 35°C are 100
mmHg and 37.6 mmHg, respectively. Assume ideal behaviour and calculate the
partial pressures of ethanol and 1-propanol at 35°C, over a solution of ethanol in 1propanol. The mole fraction of ethanol is 0.300.
Pethanol = xP°ethanol = (0.300)(100 mmHg) = 30.0
P1-propanol = xP°1-propanol = (0.700)(37.6 mmHg) = 26.3
2. As solution is prepared by dissolving 35.0 g of haemoglobin (Hb) in enough water to
make up 1 L in volume. If the osmotic pressure of the solution is found to be 10.0
mmHg at 25°C, calculate the molar mass of haemoglobin.
𝛑 = MRT
M = 𝛑/RT = 10.0 mmHg x 1 atm/760 mmHg
=
0.0821 L atm K-1 mol-1 (298 K)
5.38 x 10-4 M
or
𝛑 = (n/V)RT
n = 𝛑RT/V = 10.0 mmHg x 1 atm/760 mmHg(1 L)
[0.0821 L atm K-1 mol-1 (298 K)]
= 5.38 x 10-4 M
The volume of solution is 1 L, so it must contain 5.38 x 10-4 mol of Hb.
Molar mass of Hb = mass of Hb/ moles of Hb
=
35.0 g/ 5.38 x 10-4 mol
=
6.51 x 104 g mol-1
3.
Calculate the standard entropy changes for the combustion of 1 mol of propane vapour
at 25oC
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
o
-1
-1
[S (J K mol ): C3H8(g) = 269.9, O2(g) = 205.1, CO2(g) = 213.7 and H2O(l) =
69.90].
So =  So(products) -  So(reactants)
= {3So[CO2(g)] + 4So[H2O(l)]} – {So[C3H8(g)] + 5So[O2(g)]}
= {3(213.7 J K-1 mol-1) + 4(69.90 J K-1 mol-1)} – {269.9 J K-1 mol-1) +
5(205.1J K-1 mol-1)} = -374.7 J K-1
4.
(a) Potassium chlorate, a common oxidizing agent in fireworks and match-heads,
undergoes a solid state disproportionation reaction when heated.
4KClO3(s)  3KClO4(s) + KCl(s)
Calculate the standard free energy change, Go, for the reaction at 25oC:
4KClO3(s)
3KClO4(s)
KCl
Hfo (kJ mol-1)
-397.7
-432.8
-436.7
So (J K-1 mol-1)
+143.1
+151.0
+82.6
Horxn = {3Hof [KClO4(s)] + Hof [KCl(s)]} - 4Hof [KClO3(s)]
= [(3)(-432.8 kJ mol-1) + (-436.7 kJ mol-1)] - (4)(-397.7 kJ mol-1)
= -144 kJ
Sorxn = [3(151.0 J K-1 mol-1) + 82.6 J K-1 mol-1] - 4(143.1 J K-1 mol-1)]
= -36.8 J K-1
Gorxn = Horxn -TSorxn = -144 kJ – [(298 K)( -36.8 × 10-3 kJ K-1)] = -133 kJ
(b) Is the reverse reaction which is the formation of KClO3 spontaneous or nonspontaneous?
Non-spontaneous, Gorxn > 0
(c) What is the standard free energy change for 1 mole of KClO4 formed?
Gof = -133 kJ/3 = -44.3 kJ