Unit 5 worksheets 5 Entropy & spontaneity
1. Given that the standard entropy of gaseous water is 189 J K-1 mol-1 and the standard entropy of liquid
water is 69.9 J K-1 mol-1calculate the standard entropy change when one mol of gaseous water
condenses to the liquid state, H2O(g) → H2O(l).
2. (a) Predict the sign (positive or negative) for the entropy change for the reaction:
2NO2(g) → N2O4(g)
(b) Explain why the entropy change for the above reaction is considerably more than for the reaction:
H2(g) + I2(g) → 2HI(g)
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3. Use the information below to calculate the entropy change of formation of copper(I) oxide.
Substance
Entropy / J K-1 mol-1
Cu(s)
33.3
O2(g)
205
Cu2O(s)
101
4. You are unlikely to use orange crystals of ammonium dichromate,
(NH4)2Cr2O7(s), in a school laboratory as it is a highly toxic substance.
It is also carcinogenic and can explode when heated.
When it thermally decomposes it is both oxidised and reduced according to the equation:
(NH4)2Cr2O7(s) → Cr2O3 (s) + N2 (g) + 4 H2O (g)
(a) State what has been oxidized and what has been reduced.
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(b) Clearly the thermal decomposition of ammonium dichromate is very exothermic (ΔH = −1794 kJ
mol-1).
What can you also deduce about the change in entropy?
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5. In a blast furnace limestone (calcium carbonate) is decomposed into calcium oxide and carbon dioxide
according to the equation:
CaCO3(s) → CaO(s) + CO2(g)
(a) The standard enthalpies of formation of CaCO3(s), CaO(s) and CO2(g) are – 1207 kJ mol-1, – 635 kJ
mol-1 and – 394 kJ mol-1respectively.
Calculate the standard enthalpy change, ∆H , for the thermal decomposition of calcium carbonate.
(b) The standard entropy values for CaCO3(s), CaO(s) and CO2(g) are 92.9 J K-1 mol-1,
40.0 J K-1 mol-1 and 214 J K-1 mol-1 respectively.
Calculate the change in entropy, ∆S , for the thermal decomposition of calcium carbonate.
(c) Calculate the change in free energy, ∆G , for the thermal decomposition of calcium carbonate at 298
K.
(d) State whether the thermal decomposition of calcium carbonate is spontaneous or non-spontaneous
at 298K.
(e) Explain why limestone cliffs are thermodynamically stable whereas limestone decomposes readily in
a blast furnace.
6. Propene reacts with hydrogen at 180oC in the presence of a nickel catalyst to give propane.
C3H6(g) + H2(g) → C3H8(g)
(a) Use the information below to determine the value for the free enthalpy change for this reaction at
180oC.
(b) At 180oC this reaction is spontaneous.
Determine the temperature above which the reaction becomes non-spontaneous.
Answers
1. ∆S = S
products
–S
reactants
= 69.9 – 189 = – 119.1 J K-1 mol-1.
2. (a) Negative. It is becoming more ordered as two moles of gas are converted into one mole of gas.
(b) In the second reaction there is no change in state or in the amount of moles present (two moles of
reactants and products) so there is little change in disorder, whereas the amount of moles is reduced in
the first reaction resulting in the system becoming less disordered.
3. 2Cu(s) + ½ O2(g) → Cu2O(s)
S
reactants=
S
products
(2 x 33.3) + (½ x 205) = 169.1 J K-1 mol-1
= 101 J K-1 mol-1
∆S (Cu2O) = 101 – 169.1 = – 68.1 J K-1 mol-1
4. (a) Nitrogen has been oxidised as its oxidation number increases from – 3 in NH4+ to 0 in N2.
Chromium has been reduced as its oxidation number decreases from + 6 in Cr2O72- to +3 in Cr2O3.
(b) One mole of solid produces five moles of gas (and another one mole of solid) so the system has
become much more disordered therefore the change in entropy, ∆S , will have a high positive value.
5. (a)
Using Hess’s law
– 1207 + ∆H = – 635 + (– 394)
∆H = + 178 kJ mol-1
(b)
∆S = S
products–
S
reactants
∆S = (40.0 + 214) – 92.9 = + 161 J K-1 mol-1
(c) ∆G = ∆H – T∆S = + 178 – (298 x 161/1000) = + 130 kJ mol-1
(d) Since ∆G has a positive value (∆G > 0) the thermal decomposition of calcium carbonate is nonspontaneous at 298 K.
(e) At normal atmosphere temperatures ∆G > 0 so limestone cliffs will be thermally stable as the
reaction is non-spontaneous.
∆G = 0 when ∆H = T∆S . This is when T = 178 x 1000/161 = 1106 K (or 833 oC). Hence at
temperatures higher than 833 oC in the blast furnace the decomposition proceeds readily as the reaction
is spontaneous at this temperature.
6. (a)
Using Hess’s law
∆H = –104 – (20.4 + 0) = – 124.4 kJ mol-1
∆S = S
products–
S
reactants
∆S = 270 – (267 + 131) = – 128 J K-1 mol-1
∆G = ∆H – T∆S = – 124.4 – [(180 + 273) x –128/1000] = – 66.4 kJ mol-1
(b) The reaction becomes non-spontaneous above the temperature when ∆G = 0
∆G = 0 when – 124.4 = – T(128/1000)
T = – 124.4 x (1000/128) = 972 K
The reaction becomes non-spontaneous above 972 K (or 699 oC).