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. c. CO( g) + H2O( g) → H2( g) + CO2( g)

Video transcript

Hello everyone say we have the following problem. Using the thermodynamic data below calculate the change of our entropy for the reaction for the following reaction and justify the change in entropy. So first you wanna recall our formula for the change in entropy of a reaction. That's going to be the sum of the change in our entropy for our products minus the sum of the change in entropy for our reactant. Starting with our products We have one mole of each. So we're simply just going to add their total entropy ease. So we have our 201 For our hydrogen cyanide, 201.8 is it gonna be units of joules per mole kelvin. And we're going to add that to our second one which is one mole of our water which is 188.8 jewels per mole kelvin. And then we're going to subtract that by our reactant. So we have carbon monoxide and we have ammonia. We have one mole of each. So we're simply just going to add their changes in entropy. So that's going to be our 197 . Jewels Per Mole Kelvin For our carbon monoxide 192.8 jewels promote kelvin for our ammonia. And so when we calculate this we're going to get Positive 0.1 jewels per mole kelvin. And so we have to justify this small change in entropy. So this entropy is very small due to the moles of gas in the reaction being the same. So we went from two moles of gas in our reactant, and then we produced two moles of gas as well. So we're gonna have a very, very small change in our entropy levels. And so with that we've answered the question overall, I hope this helped, and until next time.