Download Self-Test Worksheet for Thermodynamics Section (Quiz

Self-Test Worksheet for Thermodynamics Section
1.
Substance
C6H12O6 (s)
O2 (g)
CO2 (g)
H2O (l)
Absolute Entropy, S° (J / mol-K)
212.13
205
213.6
69.9
Molecular Weight
180
32
44
18
Energy is released when glucose is oxidized in the following reaction, which is a
metabolism reaction that takes place in the body.
C6H12O6 (s) + 6 O2 (g) 6 CO2 (g) + 6 H2O (l)
The standard enthalpy change, ∆H°, for the reaction is -2801 kJ at 298 K.
(a)
(b)
(c)
Calculate the standard entropy change, ∆S°, for the oxidation of glucose.
Calculate the standard free energy change, ∆G°, for the reaction at 298 K.
How much energy is given off by the oxidation of 1.00 gram of glucose?
2.
N2(g) + 3 H2(g) 2 NH3(g)
The heat of formation, ∆Hf°, of NH3(g) is -46.2 kJ/mol. The free energy of formation, ∆Gf°, of
NH3(g) is -16.7 kJ/mol.
(a)
What are the values of ∆H° and ∆G° for the reaction?
(b)
What is the value of the entropy change, ∆S° for the reaction above at 298 K?
(c)
As the temperature is increased, what is the effect on ∆G° for the reaction? How does this
affect the spontaneity of the reaction?
3. After a winter cold spell, a swimming pool has a surface layer of ice. By measuring the thickness
of the ice, and the area of the pool, and the depth of the water under the ice, the owner figures
there is 20,000 kg of ice in the pool and 600,000 kg of water. After breaking up the ice he
measures the pool temperature to be 0 °C. He then runs the heater to warm the pool up to 30 °C.
The heater costs $1 per min to run and it supplies 200,000 kJ of heat per minute (heat capacity of
water is 4.18 J/g/°C, ∆Hfusion for water = 6.01 kJ/mol).
(a) How much heat energy does the owner use?
(b) How much does it cost the owner to do this?
4.
2 Al(s) + 3 Cl2(g) 2 AlCl3(s)
The reaction above is not spontaneous under standard conditions, but becomes spontaneous as the
temperature decreases towards absolute zero. Are ∆H° and ∆S° positive or negative at standard
conditions?
5.
2 SO3(g) 2 S(s) + 3 O2(g)
2 SO3(g) 2 SO2(g) + O2(g)
∆H = -800 kJ/mol
∆H = -200 kJ/mol
What is the enthalpy change, ∆H for the following reaction?
S(s) + O2(g) SO2(g)
Solutions:
1. (a) Use the entropy values in the table.
∆S = ∑ S
− ∑ S Re ac tan ts
Pr oducts
∆S° = [(6)(213.6) + (6)(69.9)] – [(212.13) + (6)(205)] J/K
∆S° = 259 J/K
(b) Use the following equation. Remember that enthalpy values are given in kJ, and entropy values
are given in J.
∆G° = ∆H° - T∆S°
∆G° = (-2801 kJ) – (298)(0.259 kJ) = -2880 kJ
(c) The enthalpy change of the reaction, ∆H°, is a measure of the energy given off by 1 mole of
glucose.
Moles = mass / mwt
Moles of glucose = 1.00 g / 180 = 0.00556 moles.
So, enthalpy change for 0.00556 moles = (0.00556 mol)(2801 kJ/mol) = 15.6 kJ
2. (a)By definition, ∆Hf° and ∆Gf° for N2(g) and H2(g) are equal to zero.
∆H = ∑ H
− ∑ H Re ac tan ts
Pr oducts
∆H° = [(2 mol)(-46.2 kJ/mol)] – [0] = -92.4 kJ
∆G = ∑ G
Pr oducts
− ∑ G Re ac tan ts
∆G° = [(2 mol)(-16.7 kJ/mol)] – [0] = -33.4 kJ
(b) Use ∆G° = ∆H° - T∆S°
∆S° = (∆H – ∆G) / T = [(-92.4 kJ) - (-33.4 kJ)] / (298 K)
∆S° = -0.198 kJ/K = -198 J/K
(c) Use ∆G° = ∆H° - T∆S°
From (b), ∆S° is negative, so increasing the temperature increases the value of ∆G°, making the
reaction less spontaneous.
3. (a) Use q = (moles)(∆Hfusion) for the heat needed to melt the ice at 0 °C
q = ((20 x 106) / 18) (6.01 kJ/mol) = (1.11 x 106) (6.01 kJ/mol) = 6.68 x 106 kJ
Use q = mc∆T for the heating of all the water (600,000 kg + 20,000 kg from the melted ice)
from 0°C to 30 °C.
q = (620 x 106 g) (4.18 J/g/°C) (30-0) = 7.77 x 1010 J = 7.77 x 107 kJ
Total q = 6.68 x 106 kJ + 7.77 x 107 kJ = 8.44 x 107 kJ
(b) Heater supplies 2.0 x 105 kJ/min, so it supplies 8.44 x 107 kJ in;
(8.44 x 107 kJ) / (2.0 x 105 kJ/min) = 422 minutes
Heater costs $1 per minute to run, so total cost is $422.
4.
If the reaction is spontaneous only when the temperature is very low, then ∆G is only negative
when T is very small. This can only happen when ∆H is negative (which favors spontaneity)
and ∆S is negative (which favors non-spontaneity). A very small value for T will eliminate the
influence of ∆S.
5.
Use Hess’s Law. Reverse the first reaction and divide it by 2, then add this to half the second
reaction;
∆H = +400 kJ/mol
S(s) + 3/2 O2(g) SO3(g)
∆H = -100 kJ/mol
SO3(g) SO2(g) + ½ O2(g)
------------------------------
----------------------
S(s) + O2(g) SO2(g)
∆H = +300 kJ/mol