Welcome to the Free Energy debriefing

Answer the questions below, and click on the Check buttons when you are ready.

1. Calculate DG for the following reaction, which is used to produce urea, a common fertilizer.

CO2(g) + 2NH3(g) CO(NH2)2(s) + H2O()

Correct!
Although numerically correct, your answer has the wrong number of significant digits.
Remember to use the correct units in your answer.
That is incorrect. Try again.
The solution is shown below.

DGº = 1 mol•DGfº(CO(NH2)2(s)) + 1 mol•DGfº(H2O(l)) - 1 mol•DGfº(CO2(g)) - 2 mol•DGfº(NH3(g))
DGº = 1 mol•(-197.4 kJ/mol) + 1 mol•(-237.129 kJ/mol) - 1 mol•(-394.359 kJ/mol) - 2 mol•(-16.45 kJ/mol)
DGº = -7.27 kJ

2. Which of the following reactions, when coupled to the reduction of iron(III) oxide to iron metal with no net production or consumption of oxygen, is exergonic enough to make the overall reaction product-favored?

2Fe2O3(s) 4Fe(s) + 3O2(g)      DGº = 1484.4 kJ

a. 2Al(s) + 3/2O2(g) Al2O3(s) DGfº = -1582.3 kJ/mol
b. 2Cr(s) + 3/2O2(g) Cr2O3(s) DGfº = -1058.1 kJ/mol
c. Cu(s) + 1/2O2(g) CuO(s) DGfº = -129.7 kJ/mol
d. 2Ag(s) + 1/2O2(g) Ag2O(s) DGfº = -11.2 kJ/mol
Correct! The oxidation of aluminum and chromium are exergonic enough to drive the reduction of iron. This sort of reaction is called a thermite reaction. It is often used for welding underwater, since the reaction provides its own oxygen and is exergonic enough to melt metal.

4Al(s) + 2Fe2O3(s) 4Fe(s) + 2Al2O3(s)     DGº = -1680.2 kJ
4Cr(s) + 2Fe2O3(s) 4Fe(s) + 2Cr2O3(s)     DGº = -631.8 kJ

True, those reactions are exergonic, but not all of them release enough free energy when only enough reaction occurs to consume all the oxygen released by the reduction of iron. Try again.
Correct! The oxidation of aluminum releases enough free energy to make the reduction of iron product-favored. But remember, the free energy changes must be multiplied by the same coefficient used to ensure all the oxygen is consumed. Therefore, another reaction will also work. Try again.
The solution is shown below.

Each reaction and its free energy change must be adjusted so that all the oxygen produced by the reduction of iron is consumed:

4Al(s) + 3O2(g) 2Al2O3(s)     DGº = 2 mol•(-1582.3 kJ/mol) = -3164.6 kJ
4Cr(s) + 3O2(g) 2Cr2O3(s)     DGº = 2 mol•(-1058.1 kJ/mol) = -2116.2 kJ
6Cu(s) + 3O2(g) 6CuO(s)     DGº = 6 mol•(-129.7 kJ/mol) = -778.2 kJ
12Ag(s) + 3O2(g) 6Ag2O(s)     DGº = 6 mol•(-11.2 kJ/mol) = -67.2 kJ

Then the reactions with free energy changes whose magnitude are greater than the free energy change for the reduction of iron will do. Thus, the oxidation of aluminum and chromium are exergonic enough to drive the reduction of iron. This sort of reaction is called a thermite reaction. It is often used for welding underwater, since the reaction provides its own oxygen and is exergonic enough to melt metal.

4Al(s) + 2Fe2O3(s) 4Fe(s) + 2Al2O3(s)     DGº = -1680.2 kJ
4Cr(s) + 2Fe2O3(s) 4Fe(s) + 2Cr2O3(s)     DGº = -631.8 kJ

Please make your selection from the list above.

3. What is Kth for the synthesis of ammonia?

N2(g) + 3H2(g) 2NH3(g)

Correct!
Although numerically correct, your answer has the wrong number of significant digits.
Remember, equilibrium constants have no units.
That is incorrect. Try again.
The solution is shown below.

DGº = 2 mol•DGfº(NH3(g)) - 1 mol•DGfº(N2(g)) - 3 mol•DGfº(H2(g))
DGº = 2 mol•(-16.45 kJ/mol) - 1 mol•(0) - 3 mol•(0) = -32.90 kJ

DGº = -RTlnKth
Kth = e-DGº/RT = e-32.90 kJ/8.314 J/mol•K • (273.15 + 25)K • 1000 J/1 kJ
Kth = 1.7213 x 10-6

Congratulations!

You have completed the Free Energy module.

Click on the Thermodynamics icon below to return to the Thermodynamics Gateway page.