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Problem 92

To prepare nuclear fuel, $\mathrm{U}_{3} \mathrm{O}_{8}(\text { "yellow cake" ) is converted }$ to $\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2}$, which is then converted to $\mathrm{UO}_{3}$ and finally UO2. The fuel is enriched (the proportion of the 235U is increased) by a two-step conversion of $\mathrm{UO}_{2}$ into$\mathrm{UF}_{6}$ , a volatile solid, followed by a gaseous-diffusion separation of the 235U and 238U isotopes:

$$\begin{aligned} \mathrm{UO}_{2}(s)+4 \mathrm{HF}(g) & \longrightarrow \mathrm{UF}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{UF}_{4}(s)+\mathrm{F}_{2}(g) & \longrightarrow \mathrm{UF}_{6}(s) \end{aligned}$$

Calculate $\Delta G^{\circ}$ for the overall process at $85^{\circ} \mathrm{C} :$

Answer

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## Discussion

## Video Transcript

were given to intermediate reactions for the overall reaction used to prepare nuclear fuel. We want to determine what the changing Gibbs free energy is of the overall reaction at 85 degree Celsius. So the first thing that we need to do is combine these two reactions by adding them together in order to first right out that overall reaction, we can see that you f four cancels out on each side. And so the overall reaction is you. 02 solid was or HF gas was f to gas goes to you have six solid plus two h +20 Yes, and now we need to determine the delta G of this reaction at 85 degrees Celsius because the temperature is not at standard conditions of 298 Kelvin, we cannot directly solve four Delta G by finding the overall change in Gibbs free energy of the products and subtracting that seem value for the reactive. Instead, we need to use this equation since it relates temperature to Delta G through Delta H in Delta S and were given values in the problem or Delta H in Delta s and we can use the appendix in order to look up those values for the ones that aren't given in the problem statement. So these air the two equations that we're going to use to find well to each of reaction in Delta s of reaction so that once we find them, we can plug them into this equation at the temperature of 85 degrees Celsius to find Delta G at that temperature. So for Delta H, we first find the total change in entropy of the products and then subtract the total change in entropy of the reactant for the change in entropy Delta s of the reaction we find the total entropy of the products and subtract the total entropy of the reactive. So starting with Delta H, we can break up the changing entropy of the products in reactant and to do two different expressions. Help us organize our work. We start with the reactant again. We're going off of this overall equation. Now, I reserved the products and we see that we have one mole of us six and two moles of H 20 on the product side. So one mole of U F six and two moles of H 20 were given this value for the delta h of U F six, we can look up this value for the delta, each of each to oh, in the appendix. And again, these correspond to the the standard change in entropy, a formation of each one of those compounds. And so when we cancel off units of moles, we're left with energy units of killer jewels really change in entropy of the products And since we added the total entropy change of each one of the products and add them together, that is the total entropy change of all of the products. And now, for the reactant, we have one mole of uo two formals of HF in one mole of F two. But remember that for Delta H values we used up to each of formation at standard conditions in order to help us determine the overall Delta H and F two is a naturally occurring die atomic gas molecule that has a Delta H information value of zero. So we only include the reactions of one mole of uo two and four moles of H f or the Delta H for the reactant. So one mole of uo two and four moles of H f. We look up there values for Delta H information at standard conditions we multiply them by the number of moles of each and then add them together to get the total change in entropy of the reactant. And now we do the total entropy change of the products minus a total entropy change of the reactant And this is the delta each of reaction that we should find And now we use the same process to So for Delta s of reaction, starting with the products we have one mole of us six and two moles of H 20 So one mole of U F 62 moles of H 20 each multiplied by their standard Moeller entropy values which we can look up And when we can't sof units of moles, this time we're left with jewels per kelvin or units of entropy. And when we add the those overall entropy values together or all of the products, we get the total change, an entropy of the products and then we find the total entropy of the reactant, seeing that we have one mole of uo two formals VHF and one mole of H two or of F two. So one mole of uo two for most of HF and one mole of F two multiply each one of those by their respective standard entropy inter Miller entropy values which we can look up And when we add them all together we get the total entropy of the reactions. Now we take the total entropy of the products minus the total entropy of the reactant. We get the total change. Entropy of the reaction. Now that we have Delta H and Delta s and we want to find Delta G at a given temperature, we can use the equation. Delta G equals still to h minus t Delta s in order to so for that value. So Delta G equals Delta H, which we calculated to be negative. 503 0.625 Village ALS so negative 503 0.6 killer jewels minus the temperature is 85 degrees Celsius. We convert that into Kelvin. It's 358. We multiply that by the delta s, which we just found. But we should convert this into killing jewels by dividing by 1000 so that we can subtract the units of killing jewels from the Delta H Value. And so that's negative. 371 0.9 times 10 Do the negative third killer jewels her Kelvin. We cancel off units of Kelvin after we multiply those two together, we're left with units of killing jewels or Delta G. And when we do the math, we should get a final answer for the change in Gibbs free energy of this reaction at 85 degrees Celsius to be negative 300 70 village ALS.

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