(a) Using data in Appendix C, estimate the temperature at which the free-energy change for the transformation from I₂(s) to I₂(g) is zero. (b) Use a reference source, such as Web Elements (www.webelements.com), to find the experimental melting and boiling points of I₂. (c) Which of the values in part (b) is closer to the value you obtained in part (a)?

Video transcript

Hello everyone. So for this question we have two parts of part A. To predict the temperature at which the change of free energy for the conversion of our in in a solid state to gaseous phase is zero then for part B, we're gonna go ahead and see which of the two experimental values is closer to the one that we predicted. So for the melting, I'm just gonna go ahead route the reaction. So melting is going to be the iron on solid state, going to the iron in its liquid state for our boiling. What that means for iron is the iron in its liquid state will go to iron in its gaseous phase. And then we have kind of the some of those two which is going to be the sublimation. Sublimation is assistance going from the solid state straight to its gashes face. So that's when of course our iron is going from solid to gas. Okay, let's go ahead and start off this calculation by solving for the delta H. Or the change of entropy of sublimation, it's our delta H of sublimation is equal to the delta H. Of the ARN gas minus the delta age of our iron in a solid state. So, plugging in the values then that's given to me over to the left side, right over here, Let's see. So we have one mole of the iron gas and that is 415.5. So that's again of the delta H. For my iron gas. I'm gonna head and subtract that with, let's see. So we have one mole of our iron solid corresponding delta H. Is zero kg joules per mole. Okay so putting that to my calculator I get that the change of entropy in my sublimation Is equal to 415. units being killed, Jules per mole. I'm not calculated for my delta G. We're using of course the gibbs free energy equation. So delta G of sublimation is equal to the delta H. Or change of entropy of sublimation minus the temperature multiplied by the change of entropy of sublimation. Iron and assaults and liquid states. How they delta G. Of zero. We're also gonna go ahead and assume that the delta H. Of sublimation and delta S of sublimation doesn't vary with temperature. So then the Delta U. become zero. And we just have the repeating equation. Of course we're solving for R. T. Value our temperature. We see that T. Is equal to then our delta age of sublimation over my delta S. Of sublimation. Of course we calculate those values already. So let's go ahead and plug that in. So for my delta H of sublimation we have 415.5 killer jewels. And then for my delta s of sublimation that's going to equal to 0.15335 killer jewels per kelvin. Alright so putting those values into the calculator we get that my T. My temperature is 27 oh 9.5 kelvin's let's go ahead and convert that into Celsius. For a better and easier comparison how we do this is just to subtract this by 2 73.15. And then we get that the temperature is equal to degrees Celsius. So that's part A. Is answer Of our part B. It says to compare to the boiling point and melting point and which is the experimental values is closer to that that we predicted or calculated for. So we can see here that this 2,436 value is definitely closer to the boiling point value. So for part B. So this is for A. And then for B. You say that the predictive value is closer two, the boiling point. And that's going to be my second answer for or the answer for part B. And my second answer for this problem. Alright, so again, of course we have a temperature equaling 2, 2436 degrees Celsius. And for part B we say that this calculated value for party is closer to the boiling point of our iron