Pretend you are a nuclear scientist and try to answer this bonus question. Uranium -235 (235U) is used as a fuel in nuclear power plants. However, natural uranium contains mostly U -238 (238U) and very little uranium -235. Therefore, it is necessary that the natural uranium be enriched in uranium -235 before it can be used. To do this, the natural uranium is first converted to uranium (IV) oxide, UO2, and then to the gaseous compound, uranium (VI) fluoride, UF6, as shown by the equations below.

(1) U(s) + O2(g) -> UO2(s)

(2) UO2(s) + 4 HF(g) -> UF4(s) + 2H2O(g)

(3) UF4(s) + F2(g) -> UF6(g)

The molecules of UF6 that contain the uranium -235 atoms are somewhat lighter in weight than those with the uranium -238 atoms. The difference in mass of the two uranium isotopes enables the molecules of UF6 to be separated by a gaseous diffusion process.

How much energy (in kilojoules) would be involved in producing 100 tons of UF6(g) from UO2(s) (1 ton = 9.08  105 g)? Use the standard heats (enthalpies) of formation below:

COMPOUND HEAT OF FORMATION (ΔHof)
UO2(s) - 1085 kJ/mol
UF4(s) - 1914 kJ/mol
UF6(g) - 2147 kJ/mol
H2O(g) -241.8 kJ/mol
HF(g) -271.1 kJ/mol

can someone please help me! my test is on tuesday on this stuff and this is the one question that I am confused about!

To calculate the energy involved in producing 100 tons of UF6(g) from UO2(s), we need to calculate the energy change for each reaction and then use stoichiometry to convert the amounts of reactants and products.

Step 1: Calculate the moles of UO2(s) needed:

Given: 100 tons of UF6(g)
Convert tons to grams: 100 tons * (9.08 × 105 g/ton) = 9.08 × 107 g
From equation (2), the molar ratio between UO2(s) and UF6(g) is 1:1.
Molar mass of UO2 = 238.03 g/mol
Moles of UO2(s) = (9.08 × 107 g) / (238.03 g/mol) = 3.82 × 105 mol

Step 2: Calculate the energy change for each reaction:

For reaction (1), U(s) + O2(g) → UO2(s), the enthalpy change (ΔH) is -1085 kJ/mol.
The stoichiometric coefficient of UO2 is 1. Therefore, the energy change for this reaction is -1085 kJ/mol.

For reaction (2), UO2(s) + 4HF(g) → UF4(s) + 2H2O(g), the enthalpy change (ΔH) is -1914 kJ/mol.
The stoichiometric coefficient of UO2 is 1. Therefore, the energy change for this reaction is -1914 kJ/mol.

For reaction (3), UF4(s) + F2(g) → UF6(g), the enthalpy change (ΔH) is -2147 kJ/mol.
The stoichiometric coefficient of UF6 is 1. Therefore, the energy change for this reaction is -2147 kJ/mol.

Step 3: Calculate the total energy change for the production of UF6(g):

Multiply the energy change for each reaction by the number of moles of UO2(s) to determine the energy change for that reaction.

Energy change for reaction (1): (-1085 kJ/mol) * (3.82 × 105 mol) = -4.14 × 108 kJ
Energy change for reaction (2): (-1914 kJ/mol) * (3.82 × 105 mol) = -7.32 × 108 kJ
Energy change for reaction (3): (-2147 kJ/mol) * (3.82 × 105 mol) = -8.20 × 108 kJ

The total energy change is the sum of all the energy changes:

Total energy change = -4.14 × 108 kJ + (-7.32 × 108 kJ) + (-8.20 × 108 kJ)
Total energy change = -19.66 × 108 kJ

Step 4: Convert the energy change to kilojoules:

Divide the total energy change by 1000 to convert it to kilojoules:

Total energy change = -19.66 × 108 kJ / 1000 = -1.966 × 107 kJ

Therefore, the energy involved in producing 100 tons of UF6(g) from UO2(s) is approximately -1.966 × 107 kJ.