Question

To prepare nuclear fuel, U₃O₈ ("yellow cake") is converted to UO₂(NO₃)₂, which is then converted to UO₃ and finally UO₂.The fuel is enriched (the proportion of the ${}^{\mathbf{\text{235}}}\mathbf{\text{U}}$ is increased) by a two-step conversion of UO₂ into UF₆, a volatile solid, followed by a gaseous-diffusion separation of the ${}^{\mathbf{\text{235}}}{\mathbf{\text{Uand}}}^{\mathbf{\text{238}}}\mathbf{\text{U}}$isotopes:

$\begin{array}{l}{\mathbf{\text{UO}}}_{\mathbf{\text{2}}}\mathbf{\text{(s)+4HF(g)}}\mathbf{\to }{\mathbf{\text{UF}}}_{\mathbf{\text{4}}}{\mathbf{\text{(s)+2H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(g)}}\\ {\mathbf{\text{UF}}}_{\mathbf{\text{4}}}{\mathbf{\text{(s)+F}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{UF}}}_{\mathbf{\text{6}}}\mathbf{\text{(s)}}\end{array}$

Calculate ${\mathbf{\text{ΔG}}}^{\mathbf{\text{°}}}$ for the overall process at ${\mathbf{\text{85}}}^{\mathbf{\text{°}}}\mathbf{\text{C}}$.

Short answer

At ${\text{85}}^{\text{°}}\text{C(358K),}$the overall process${\text{ΔG}}_{\text{rxn}}$ is equal to -370.497 kJ/mol.

Step-by-step solution

Concept Introduction

The reactions of the fuel enrichment process are first written down:

$\begin{array}{l}{\text{UO}}_{\text{2}}\text{(s)+4HF(g)}\to {\text{UF}}_{\text{4}}{\text{(s)+2H}}_{\text{2}}\text{O(g)}\\ {\text{UF}}_{\text{4}}{\text{(s)+F}}_{\text{2}}\text{(g)}\to {\text{UF}}_{\text{6}}\text{(s)}\end{array}$

To calculate the Gibb's energy change for the process, we must keep in mind that the process occurs at ${\text{85}}^{\text{°}}\text{C(358K),}$therefore the conventional Gibb's energy values supplied are insufficient.

We may then compute Gibb's energy change for the process at the requisite temperature after calculating the enthalpy and entropy changes of the entire process.

Calculation for ΔXrxn° 

Let us calculate ${\text{ΔX}}_{\text{rxn}}^{\text{°}}$

Both enthalpy and entropy are state functions, which means they solely depend on the system's starting and ultimate states:

$\begin{array}{l}{\text{ΔH}}_{\text{ren}}^{\text{°}}{\text{=ΔH}}_{\text{products}}^{\text{°}}{\text{-ΔH}}_{\text{reactants}}^{\text{°}}\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}{\text{=ΔS}}_{\text{products}}^{\text{°}}{\text{-ΔS}}_{\text{reactants}}^{\text{°}}\end{array}$

The products in this example are all chemicals created by both reactions, and the reactants are also all compounds produced by both reactions.

If is a thermodynamic state function,

${\text{ΔX}}_{\text{rxn}}^{\text{°}}\text{=}\left[{\text{ΔX}}^{\text{°}}\left({\text{UF}}_{\text{4}}\text{(s)}\right){\text{+2×ΔX}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(g)}\right){\text{+ΔX}}^{\text{°}}\left({\text{UF}}_{\text{6}}\text{(s)}\right)\right]\text{-}[{\text{ΔX}}^{\text{°}}\left({\text{UO}}_{\text{2}}\text{(s)}\right){\text{+4×ΔX}}^{\text{°}}\text{(HF(g))+}{\text{ΔX}}^{\text{°}}\left({\text{UF}}_{\text{4}}\text{(s)}\right){\text{+ΔX}}^{\text{°}}\left({\text{F}}_{\text{2}}\text{(g)}\right)]$

We can simplify that equation,

${\text{ΔX}}_{\text{rxn}}^{\text{°}}\text{=}\left[\text{2×ΔX°}\left({\text{H}}_{\text{2}}\text{O(g)}\right)\text{+ΔX°}\left({\text{UF}}_{\text{6}}\text{(s)}\right)\right]\text{-}\left[\text{ΔX°}\left({\text{UO}}_{\text{2}}\text{(s)}\right)\text{+4×ΔX°(HF(g))+ΔX°}\left({\text{F}}_{\text{2}}\text{(g)}\right)\right]$

This is the simplified equation for

Calculating for Enthalpy and Entropy

Then, using Appendix B, for enthalpy we can write and calculate that,

$\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\left[{\text{2×ΔH}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(g)}\right){\text{+ΔH}}^{\text{°}}\left({\text{UF}}_{\text{6}}\text{(s)}\right)\right]\text{-}\left[{\text{ΔH}}^{\text{°}}\left({\text{UO}}_{\text{2}}\text{(s)}\right){\text{+4×ΔH}}^{\text{°}}{\text{(HF(g))+ΔH}}^{\text{°}}\left({\text{F}}_{\text{2}}\text{(g)}\right)\right]\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\left(\text{2mol×}\left(\text{-241.826}\frac{\text{kJ}}{\text{mol}}\right)\text{+1mol×}\left(\text{-2197.00}\frac{\text{kJ}}{\text{mol}}\right)\right)\text{-}(\text{1mol×}\left(\text{-1085.00}\frac{\text{kJ}}{\text{mol}}\text{+4mol×}\left(\text{-273.00}\frac{\text{kJ}}{\text{mol}}\right)\text{+1mol×}\left(\text{0.00}\frac{\text{kJ}}{\text{mol}}\right)\right)\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=-503.652}\frac{\text{kJ}}{\text{mol}}\end{array}$

Therefore, the enthalpy is $\text{-503.652}\frac{\text{kJ}}{\text{mol}}$

Similarly, for entropy we can write and calculate that,

$\begin{array}{l}{\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=}\left[{\text{2×ΔS}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(g)}\right){\text{+ΔS}}^{\text{°}}\left({\text{UF}}_{\text{6}}\text{(s)}\right)\right]\text{-}\left[{\text{ΔS}}^{\text{°}}\left({\text{UO}}_{\text{2}}\text{(s)}\right){\text{+4×ΔS}}^{\text{°}}{\text{(HF(g))+ΔS}}^{\text{°}}\left({\text{F}}_{\text{2}}\text{(g)}\right)\right]\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=}\left(\text{2mol×}\left(\text{188.72}\frac{\text{J}}{\text{mol×K}}\right)\text{+1mol×}\left(\text{225.00}\frac{\text{J}}{\text{mol×K}}\right)\right)\text{-}(\text{1mol×}(\text{77.00}\frac{\text{J}}{\text{mol×K}}\text{+4mol×}(\text{173.67}\frac{\text{J}}{\text{mol×K}}\text{+1mol×}\left(\text{202.70}\frac{\text{J}}{\text{mol×K}}\right)\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=-371.94}\frac{\text{J}}{\text{mol×K}}\end{array}$

Thus, the entropy is,$\text{-371.94}\frac{\text{J}}{\text{mol×K}}$

Calculating Gibb’s Free Energy

Finally, Gibb's free energy change can be calculated at ${\text{85}}^{\text{°}}\text{C(358K)}$

$\begin{array}{c}{\text{ΔG}}_{\text{rxn}}{\text{=ΔH}}_{\text{rxn}}^{\text{°}}{\text{-T×ΔS}}_{\text{rxn}}^{\text{°}}\\ {\text{ΔG}}_{\text{rxn}}\text{=-503.652}\frac{\text{kJ}}{\text{mol}}\text{-358K×}\left(\text{-371.94}\frac{\text{J}}{\text{mol×K}}\right){\text{×10}}^{\text{-3}}\frac{\text{kJ}}{\text{J}}\\ {\text{ΔG}}_{\text{rxn}}\text{=-370.497kJ}\end{array}$

Therefore, at${\text{85}}^{\text{°}}{\text{C(358K),ΔG}}_{\text{rxn}}\text{=-370.497kJ/mol}$