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Problem 73

Ohio State University

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Problem 72

Use $\Delta H^{\circ}$ and $\Delta S^{\circ}$ values to find the temperature at which these sulfur allotropes reach equilibrium at 1 atm:

S(rhombic) $\rightleftharpoons$ S(monoclinic)

Answer

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## Discussion

## Video Transcript

were given a reversible reaction between two different cell for our troops. And we want to find the values for a change in entropy, Delta H and change in entropy. Delta s for this reaction. In order to help us determine at what temperature this reaction comes to equilibrium, we can use thes two equations in order to help us solve four Delta H and Delta s of reaction. We're Delta H. We find the total change in entropy of the products and then subtract the total change in entropy of the reactive. We're Delta s reaction. We find the total entropy of the products and then subtract the total entropy of the reactant. After we find both of those values, he can use this equation in order to find the temperature at which the reaction comes to equilibrium. Since we know that at equilibrium, Delta G is equal to zero and we will have been solved for Delta H in Delta s. So that's in the process that we're going to use work through this problem. We begin by starting with finding Delta H. We use this equation again in order to determine that, and we start with the total change in entropy of the products. And we see the only product of this reversible chemical reaction is s mono clinic. And so we have just a single mole of that and we can look up in the appendix with the change. An entropy of formation is for that substance. And when we multiply those together to cancel out moles and get total energy units of killer jewels for the total change in entropy of the products to get 0.3 killer jewels where the reactant So we just have one mole of sulfur rahm Bic. And when we look up the entropy of formation for that substance, we see that it is zero meaning that the total change in entropy of the reactant zero kill it, Jules. So the total entropy change of the products minus that of the reactant, it's just equal to 0.3 pillage als. And that corresponds to Delta H for this reaction. And now we use this equation in the middle to solve for Delta s reaction. Again, we have one mole of both reactant and product, and we look up the standard Moeller entropy of each one of these substances in the appendix, which has units of jewels per mole, Kelvin. And this time, when we cancel out the units of moles, we're left with total entropy units of jewels per Kelvin so we can get the total entropy of the products and then subtract the total entropy of the reactant to get the overall change in entropy of this reaction to be 0.7 jewels per kelvin. So now we have values or a Delta H reaction and Delta s reaction. And now we can rearrange this equation in order to solve for the temperature when this reaction is at equilibrium, meaning when Delta G is equal to zero. So when we rearrange that reaction, that equation to solve for temperature, we get this expression and recall that at equilibrium, Delta G is equal to zero. So that's what we plug in for Delta G. We just solved for Delta H in Delta s so we can pull again Everything that we know how to solve for that temperature. When Delta G is equal to zero. So temperature equals Delta G, which is zero killer jewels at equilibrium, and then we subtract Delta H, which we found to be 0.3 killer jewels, and then we divide our anything negative of the of the entropy that we calculated, which was 0.7 jewels per kelvin. We should convert this into killer jewels by dividing by 1000 so that the units of killing JAL's and the denominator cancels out with the units of killer jewels in the numerator so that we're left with temperature units of Kelvin. Now when when we solve that out, we should get a temperature or when this reaction equal a briefs to be equal to about 400 29 Kelvin.

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