Methyl tert-butyl ether (MTBE) is prepared by reaciton of methanol (l) (ΔH°f = -239.2 kJ/mol) with 2-methyl-propene (g), according to the requation Calculate ΔH°f in kJ/mol for 2-methylpropene

Video transcript

Hello everyone today. We have the following question. The reaction of ice a bottle in with ethanol produces ethyl TERT beetle, What is the change of the entropy of formation for Isil bottling in? Kill eagles crumble. So ultimately we have to calculate our entropy of formation here and so we can start off with the equation that the entropy of formation of our reaction is going to be equal to the change in our entropy of formation of our products minus the change in entropy of formation for our reactant. So this is going to indicate that the change of formation of our entropy for our products, our ethyl ether is going to be our c six H 14 minus our change and our heat of formation for our reactant and our reactions are Isil beetle and ethanol. So now we can plug in the values that we have. So we have the overall Entropy of formation for the reaction which is negative 4044 kill eagles per mole. We're going to go ahead and equal that to the heat of formation for our product which we have here is negative 3 50. kg joules per mole. And then we're going to subtract that by the heat of formation of the entropy the entropy of formation for our reactant. So we actually don't have that value for our s a beauty line. That's what we are trying to find. However, we do have that for our ethanol, which is going to be our -2 69.29 Joules per mole simplifying this? We have negative 4044 joules per mole is equal to -81.51 kg per mole minus our entropy of formation for our so beautiful, simplifying this more. And actually solving for that, we get the entropy of formation for our beauty line is equal to 962.49 kg / Mole. And with that we have answered our question overall. I hope that this helped, and until next time.