Use standard free energies of formation to calculate ΔG° at 25 °C...

Question

Use standard free energies of formation to calculate ΔG° at 25 °C
for each reaction in Problem 61. How do the values of ΔG°
calculated this way compare to those calculated from ΔH° and
ΔS°? Which of the two methods could be used to determine
how ΔG° changes with temperature?

Accepted Answer

hey everyone today, we're being asked to calculate the delta G. Of the reaction for ammonium, bromide decomposing into ammonia and hydrochloric acid. But we're asked to do this in two different ways. One is using the standard entropy and standard entropy of formation along with the temperature. And the second way is using only the standard free energies of formation. Finally, the the question is asking us in determining how delta G changes with temperature, which method would be the best. So without further ado, let's go ahead and just jump right into it. So I'm going to scroll a little bit and let's write out our first method. So it's asking us to utilize the delta H. The change in entropy, the delta S going to change in entropy of the reaction in order to calculate the delta G. Of the reaction. So to calculate the delta G of the reaction, we need to take the delta each other reaction and subtract the temperature in kelvin, multiplied by the change in entropy of the reaction. That's our formula. But to do this we need to also find out what is the change in entropy of the reaction and the change in entropy of the reaction. We can do this using the standard entropy is and entropy of formation respectively. So starting with the central piece to calculate the change in entropy of the reaction, we need to take the difference between the standard entropy of formation of the products which are P and the standard change in entropy of formation of the reactant which I know are and so therefore delta H of the reaction. And we need to consider the moller values and molar ratios as well. But since everything only has one mole in relation to each other, we have one mole of ammonium bromide, one mole of ammonia and one mole of hydrochloric acid. It's much easier for us. So now the entropy of formation, our experimentally calculated values that we'll find in a table but there'll be as such. So for the products which are ammonia bromide and are sorry ammonia and hydrochloric acid. For one mole of ammonia, the change in entropy the entropy of formation Is negative 45.9 killer jewels, per mole killer joules per mole. We will add this because we're taking the some of the or the sum of all the entropy of formation of the products and subtracting the sum of all the entropy ease of formation of the reactions from it. So for hydro Bernick acid HPR the standard entropy information is negative 36.3 kg per mole. And from this we will subtract one mole of ammonium bromide. NH four br which has a standard entropy of formation of negative 270.8 kg joules per mole Permal which will net us a final delta age of the reaction Of 188.6 kg jewels. So that's our first part. The second part is also pretty easy. We just need to find the delta s of the reaction now which is calculated in much the same way delta S of the reaction. The change in entropy of the reaction is simply equal to the difference between the delta S. Of formation of the products and the standard entropy of formation of the reactant. Using a similar set up as well, we get the delta S. Of the reaction sequel to again we have one mole of ammonia, one mole of Ammonia. It needs three Which has a standard entropy information of 192.8 jewels per mole kelvin. It's multiplied Plus one mole of hydra, birdemic acid HBR Which has a standard entropy information of 198. jules. Per mole kelvin minus The one mole of Ammonia ammonium bromide. My bad ammonium bromide H four b. R. Which has a standard entropy of formation of 113. jewels. Per Mole Kelvin. Don't forget that we need our final answer in kill jules because delta G is calculated in joules. And entropy is already given in kill jules. And this is a very simple conversion remember that we have one kill a jewel for every 10 to the third jewels. So this will end up canceling out with the jewels and we will be left with a final answer of 0.2785 killer jewels per kelvin as our standard entropy or entropy for the reaction. Now our temperature was 25°C which means to convert this to kelvin. I'll just do this on the side here for temperature It'll simply be to 73.15 plus 25. Which will give us 298.15 Kelvin. So that'll be our temperature. So plugging all this back into our original equation we get the delta G. The reaction is equal to the delta H. Of the reaction was 1 88.6 kg joules per mole. Oh sorry. Killer jewels minus the temperature to 98.15 Kelvin multiplied by the Entropy or change in entropy for the reaction 0.2785 kill jules per kelvin Which means our Kelvin's will cancel out and we will be left with a final answer of 106 killer jewels. So that is our estimated delta T. Of the reaction, estimated gibbs free energy of the reaction when calculated using the standard entropy ease of formation and standard entropy ease of formation. Let's take a look at our second method using only the standard delta G. The standard gibbs free energy of formation. This is again done in a very similar way. Let's write this out here. Second method to do this just like how we did to find the standard delta G of form or standard delta H of the reaction. We used the entire piece of formation. We have to do a very similar process. This will simply be the standard delta G of formation of the product products. My band subtracted by the standard delta geo formation of the reactant. So again, these are all experimentally calculated values and can be done in much the same way. So we have delta G. Of the reaction is equal to. We have excuse me, we have one more of ammonia. NH three which has a delta G. Or standard delta G of formation of negative 16.4 kg joules per mole, Add that to the one mole of Hydra Bronek acid, which has a delta G. A formation of negative 53.4 kg joules per mole. And from this we subtract the product or sorry, the reactant, which is ammonium bromide, one more NH four BR. Which has a standard delta G information of negative 1 75. kg joules per mole. Finally adding all this up, we get that. The delta G for this reaction, calculated using the standard delta G's of formation is 105 kg jewels. So using the first method, we got 106 killer jewels as our delta G. And using the second method, using only the standard delta G's of formation, we got one of five killer jewels. So both of the values that we got using these methods are really close. But while the first method was longer, if we want to consider how delta G changes with temperature. We need to use that method because it factors in temperature into its calculations right there. So I hope this helps, and I look forward to seeing you all in the next one.

Use standard free energies of formation to calculate ΔG° at 25 °C for each reaction in Problem 61. How do the values of ΔG° calculated this way compare to those calculated from ΔH° and ΔS°? Which of the two methods could be used to determine how ΔG° changes with temperature?

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