From the enthalpies of reaction
H21g2 + F21g2¡ 2 HF1g2 H = -537 k...

Question

From the enthalpies of reaction
H21g2 + F21g2¡ 2 HF1g2 H = -537 kJ
C1s2 + 2 F21g2¡ CF41g2 H = -680 kJ
2 C1s2 + 2 H21g2¡ C2H41g2 H = +52.3 kJ
calculate H for the reaction of ethylene with F2:
C2H41g2 + 6 F21g2¡ 2 CF41g2 + 4 HF1g2

Accepted Answer

Hey everyone. So here it says to calculate the standard entropy of reaction for the following reaction. Here we have two moles of carbon solid reacting with one mole of hydrogen gas to produce one mole of a seedling gas. All right, so basically, we may have to manipulate one or more of these reactions given below to obtain this first one given to us. All right, So, let me just write it down here. So we want to see solid Plus H two gas Giving US C two, H 2. Gas. Doing that will help us find delta H of reaction. Alright, so, we need to find two moles of carbon solid. If we look up above here is carbon solid here, the issue is we need two of them. So, that means they're gonna multiply this entire thing. Times two. When I do that, I'm gonna get to see solid Plus 202. Give Me two c. 02. Multiplying by two. Means I also multiply delta H by two. And when I do that it becomes negative 787 kg joules. Alright, so now that I've done that, this is gone because it's now this new equation. Next I need to find one mole of H two. Alright, so where do we ch. Two? We see H two. Right here, it's they're both reactant. They're both one mole each. So that means I don't touch this equation at all. This equation stays as it is. Alright, so then I'm just gonna bring it over. So, I'm gonna have a church two gas Plus 002 gas gives me Htoo liquid. So the number stays negative to 85.8 killer jewels. Lastly I need one mole of C. Two H. Two. Okay, so if I need one more mole of c. two h. 2 as a product, I have to reverse the top equation. So when I reverse it I'm gonna have to C. 02. Gas plus H. Two. A liquid gives me C. Two H. Two. Gas Plus 5/2 is really just 2.5 02 gas. And when you reverse it you reverse the sign. So it's one positive 1129 9.5 kg. Now let's get rid of intermediates. Here we have water as a liquid in reacting form. And here it's in product form they cancel out two moles of C. 02. Gas as reacting cancel out with two moles of C. 02 as product. Now here we have half 02 plus 202. That's 2.502 as reactant. Those 2.5s cancel out with this 2.5. Remember 5/2 is equal to 2.5. Now, if we look what's left, we have two moles of C. Solid, one mole of H 20. And one mole of C. Two H. Two. Those make up are the equation. We need to find all we do. Now is we add up these three new delta H values, adding them all up together will give us our final answer. So when we add them all up together, I get my final answer as 2-6. killing jewels, so this would be my final answer for the given equation.

From the enthalpies of reaction H21g2 + F21g2¡ 2 HF1g2 H = -537 kJ C1s2 + 2 F21g2¡ CF41g2 H = -680 kJ 2 C1s2 + 2 H21g2¡ C2H41g2 H = +52.3 kJ calculate H for the reaction of ethylene with F2: C2H41g2 + 6 F21g2¡ 2 CF41g2 + 4 HF1g2

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