Use data from Appendix IIB to calculate ΔS°rxn for each of the reactions. In each case, try to rationalize the sign of ΔS°rxn. b. Cr2O3(s) + 3 CO(g) → 2 Cr(s) + 3 CO2(g)

Video transcript

Hello everyone today. We have the following problem. Using the thermodynamic data below calculate the change in entropy for the reaction for the following reaction and justify the change in entropy. So the first thing we're gonna do is we're gonna make note of what the equation looks like where they change an entropy. It's going to be the some of the change in our entropy for our products minus the sum of our change in entropy for our reactive so our products are going to be this solid here and this gas here. So for our solid for our sodium chloride, we note that we have two of them. So we're going to just multiply That change in entropy by two and that change in entropy is 72.1 jewels per mole kelvin. And then we're gonna add that to the one mole of our fluoride gas which is which has an entropy of 202. jewels per mole kelvin. And then we're going to subtract that by our reactant, which is sodium fluoride and chloride gas. We have two moles of that sodium fluoride. So we're gonna do two times are 51.1 jewels per mole kelvin. And then we're gonna add that to the one mole of chloride which is 223.1 jewels per mole kelvin. And in doing this we're going to get a change in our entropy for the reaction of positive 27.1 jules per mole and so why is this change so small, this is because there is no net change in the moles of gas. If we look at our reactions, we have three moles of gas, or we have one mole of gas, chloride. And if we look at our products, we have one mole of gas as well. So that's going to be 1-1. So we're not changing at must as much. So we're going to have a very small change in our entropy. And with that we've answered the question overall, I hope this helped. And until next time.