Ethanol is manufactured in industry by the hydration of ethylene:...

Question

Ethanol is manufactured in industry by the hydration of ethylene:  Using the data in Appendix B, calculate ∆G° and show that this reaction is spontaneous at 25 °C. Why does this reaction become nonspontaneous at higher temperatures? Estimate the temperature at which the reaction becomes nonspontaneous. <QUESTION REFERENCES APPENDIX B>

Accepted Answer

Hello everyone today. We have the following problem. Ethylene reacts with chlorine to yield di chloral ethane. And we have the following reaction here. Use the following gibbs free formation values to calculate the gibbs free energy for this reaction and show that this reaction is spontaneous at 25 degrees Celsius. And then we have the following information here and it says to explain why this addition reaction is non spontaneous and higher temperatures, calculate the temperature where the spontaneity of the reaction changes from non spontaneous to spontaneous to non spontaneous. So first we have to calculate the gibbs free energy of the reaction. In that formula is going to be the gibbs free energy of our reaction is equal to the sum of our gifts. Free energy of our products minus the sum of the gibbs free energy of our reactant of our reactant, our products. It's going to be this dike low ethane. And we have the gibbs free energy formation value here And there's only one mole of that. So we're just going to say that negative 79.6 kg per mole. And then we're going to subtract by the some of the reactant and we have our chlorine gas And we have our ethylene so far ethylene and chlorine gas. We both have one mole. So we're just simply going to have our 68.1 kg joules per mole. And we'll leave out the chlorine gas because it is simply going to be zero. So it won't add to the equation. And to our total gifts for energy for this is going to be negative 47.7 kg jewels. So we have to show that it is spontaneous. So since we have a negative gibbs free energy, this is again to note that this reaction is spontaneous. Now we have to see if the reaction is non spontaneous at higher temperatures. And so how are we going to do that? Well, we have to calculate the entropy change and the entropy change. So the entropy changes can be found in the reference text. And if we were to look up the entropy changes of each of our values, we have our ethylene, we have our chlorine and we have our di chloral ethane. We would see that they are 52.3 kg per mole zero kg per mole And - 65.2 kg Primal respectively. And so to calculate this is gonna be a similar process to calculating the gibbs free energy for the reaction. So the entropy change for our reaction is going to be equal to the sum of the entropy change for our products minus the entropy change of our reactant. So that is simply going to be our product which is that Dick local ethic which is negative 1 65.2 kg joules per mole minus. We're just gonna have our 52.3 kg per mole for our reactant because zero is not going to change the outcome of the reaction, that's going to give us negative 2 17.5 kg joules. Then we can calculate the entropy change for the reaction. And those values can also be found in a reference text. So we'll have the entropy change here a formation for ethylene. We will have it for our chlorine and then we will have it for our chloral ethane. And those values are going to be 219 .5 jewels per mole Kelvin, 2 23 jewels per mole kelvin And jewels primal kelvin. And so that process is going to be very similar to how you solve for entropy. And Gibbs free energy is going to be in trippy for the reaction is gonna be equal to the total entropy change for our products minus our total entropy change for our reactors. Our product was the decline of the things. That's 2 08.5 joules per mole Kelvin, That's gonna be minus for our reactant. We're going to have the ethylene which is 2 .5. And I left out the units to save on time and space and we're going to add that to our chlorine which was 2 23 jewels per mole kelvin. And that's going to give us a total entropy change of negative 234 jewels per kelvin. So now that we have these values, we can solve for our give us free energy. And that's going to be the gibbs free energy is equal to the entropy change or the entropy change minus temperature times our entropy change. So let's solve for that. We have to assume that an equilibrium are Gibbs free energy is going to be equal to zero. We have our entropy changes was negative to 17.5 kg joules. We do not have the temperature but we do have the change in entropy which was negative 34 joules per kelvin. And they were just going to convert it from jewels to kill a jewels in one step by using the conversion factor, that one kg jewels equal to 10 to the third jewels. And when those units cancel out, we're going to have a Temperature that is equal to 929.49 Kelvin. However, we must see, We must determine if our temperature matches up with our standard. So we have our 25°C here. If we were to convert that to Kelvin, it would be our 25°C plus our or 25°C is equal to our 2 kelvin at standard temperature. So what we're gonna do is we're gonna take our temperature that we calculated our 929.49 kelvin and subtract that from our 73 kelvin to get 656.3 degrees Celsius. And so what does this number mean here? So this number means that the reaction is actually gonna be non spontaneous at higher temperatures, and this because this is positive. So when this temperature is positive, this reaction is going to be non spontaneous at high temperatures. And with that we have answered the question overall, I hope this helped, and until next time.

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