Calculate ∆H°f in kJ/mol for benzene, C6H6, from the following da...

Question

Calculate ∆H°f in kJ/mol for benzene, C6H6, from the following data: 2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(l) ∆H°=-6534 kJ ∆H°f (CO2) = -393.5 kJ/mol ∆H°f(H2O) = -285.8 kJ/mol

Accepted Answer

Welcome back. Everyone were given the following reaction and standard entropy values, we need to find the entropy of formation in kilograms per mole for cyclo hexane C six H 12. Now, based on our given equation, we need to make sure that this is balanced. And so looking at our atoms, we have six moles of carbon on the reacting side and six on the product side. For hydrogen, we have 12 moles of hydrogen on the product side and 12 on the reactant side. And for oxygen, we have 18 moles of oxygen on the reactant side and also 18 moles of oxygen on the product side. So we can confirm that this reaction is balanced. As written Note that we're given the entropy change of our reaction as 39,000 or 3919.6 kg Joel's as well as our entropy of formation of carbon dioxide and water. Now recall that our entropy change for the reaction is found as taking the difference between the some of the moles of our atom multiplied by the entropy of formation of that atom. And this is all applying to our products, which is subtracted from the next bracket where we have the, some of our moles of our reactant multiplied by the entropy of formation of our reactant. And so this gives us our entropy change of the reaction. So with this in mind, we can plug in the known values that we have to find the entropy of formation of our cyclo hexane reactant. We know that delta H of the reaction is given as negative 3909 sorry, negative 3000 919.6 kg joules, which is equal to in our bracket. We have the moles of our products where we have six moles of carbon dioxide multiplied by its entropy of formation. Given in the prompt as -3 93.5 kill jules per mole. We still have our second product being six moles of water. And let's actually move this over so that we have enough room In which we have an entropy of formation of water given in the prompt as -285.8 kill jules per mole. This completes the entropy of formation some of our products. And now we want to subtract from the entropy of formation of our reactant. And so beginning with our reacting cyclo hexane, we have just one mole of that. So one more times its entropy of formation, which we don't know. And so we'll just plug that in as delta H F C six H and I'll actually move this bracket below since we're running out of room. So this is subtracted and we have our second reactant, which is oxygen in which we have nine moles of oxygen multiplied by its entropy of formation, which is not given to us. But recall that oxygen is in its standard state as a di atomic gas, which is, which is written as here in our equation. And so it has an entropy of formation of zero killed jules, Permal recall that any atom in its standard state will always have that Ethiopia formation being zero, killed joules per mole. So this completes the sum of N three P A formation of our reactant. So now we're just going to simplify our equation we have on the left hand side still negative 3919.6 kg joules, which is equal to our right hand side in which our bracket and blue for the sum of entropy of formation of our products results in a value of negative 4075.8 kg joules because our units of moles cancel out. And for our react ints, we have, I'm sorry, we don't need our bracket anymore. But for our reactant since we would have the results being minus one mole times The entropy of formation of psycho heche scene. So we need to isolate for our variable, which is our entropy of formation of cycle hexen, meaning we want to Add one mole of or multiplied by our entropy. A formation of cyclo hexane to both sides. And in that case to isolate it, we would then add 3 90 or 3919.6 kg joules to both sides. And so that would cancel out here on the left hand side in which now we have our simplified equation being one mole of multiplied by our entropy. A formation of cyclo hexane equal to negative. And we'll keep the color blue on the right hand side, negative 40,075 or negative 4075.8 kg joules plus 3919.6 kg jewels. And so we would just take the some on the right hand side. And so we would find that our entropy of formation of Cyclo hexane is equal to the sum of the right hand side which results in a value of negative 156.2 kg jewels. Recall that entropy of formation is in terms of joules per mole of our substance. And so in this case, we would just say that our units are killing jewels Permal. And so what's highlighted in yellow is our final answer for the entropy of formation of Cyclo Hexane corresponding to choice be in the multiple choice. I hope that everything I went through is clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

Calculate ∆H°f in kJ/mol for benzene, C6H6, from the following data: 2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(l) ∆H°=-6534 kJ ∆H°f (CO2) = -393.5 kJ/mol ∆H°f(H2O) = -285.8 kJ/mol

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