Uranium-235 is an important nuclear fuel. However, the 235U isotope is only a ve

Question

Uranium-235 is an important nuclear fuel. However, the 235U isotope is only a very small amount of naturally occurring uranium. Therefore, taking advantage of gas diffusion rates and molar masses, naturally occurring uranium(IV) oxide is converted to uranium(IV) fluoride and this in turn is converted to uranium(VI) fluoride in a reaction with fluorine gas. The uranium(VI) fluoride is a gas and the different masses of the two uranium(VI) fluoride isotopes can be separated. Using the data below, determine the lattice energy, Delta HIattice Energy, of the uranium(IV) fluoride that is formed in the reactions described above. U(s) + O2(g) rightarrow UO2(s) Delta H = -1084. 9 kj/mol UO2(s) + 4 HF(g) rightarrow UF4(s) + 2 H2O(g) Delta H = -228. 5 kj/mol 1/2 H2(g) + 1/2F2(g) rightarrow HF(g) Delta H = -271. 1 kj/mol H2(g) + 1/2O2(g) rightarrow H2O(g) Delta H = -241. 8 kj/mol U(s) rightarrow U(g) Delta H = 527. 2 kj/mol U(g) rightarrow U4+(g) + 4 e- IE = 11800 kj/mol F2(g) rightarrow 2 F(g) BE = 159 kj/mol F(g) + e rightarrow F-(g) EA = -328 kj/mol U4(g) + 4 F-(g) UF4(s) L. E. = ?

Explanation / Answer

LE = H(rxn1) + H(rxn2) + 4H(rxn3) - 2H(rxn4) - H(rxn5) - H(rxn6) - 2H(rxn7) - 4H(rxn8)

put the values,

LE = -13274.4 KJ

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