Question

Given the following equations, calculate the ΔH of reaction of ethylene gas (C2H4) with fluorine gas to make carbon tetrafluoride (CF4) gas and hydrogen fluoride gas. Show your work.
H2 (g) + F2 (g) → 2HF (g) ΔH = –537 kJ
C (s) + 2F2 (g) → CF4 (g) ΔH = –680 kJ
2C (s) + 2H2 (g) → C2H4 (g) ΔH = +52.3 kJ

Answer 1

You may want to post in the Chemistry section next time.

Answer 2

I thought this was chemistry.

Answer 3

I have no idea how to do this one though, sorry.

Answer 4

Okay. Thanks anyway. :)

Answer 5

And you posted in the right section, my computer was spazzing and said this was in math. Sorry!

Answer 6

That's okay!

Answer 7

I have been working on it and was wondering if this looks correct.

C2H4 + 6F2 ------> 2CF4 + 4HF

H2 + F2 ---> 2HF ..ΔH = -537 kj ...(1)

C + 2F2 -----> CF4 ...ΔH = -680 kj....(2)

2C + 2H2 -----> C2H4 ...ΔH = 52.3 kj ....(3)

Multiply (2) by 2

2C + 4F2 ----> 2CF4 ...ΔH = -680X2 = -1360 kj ....(4)

Subtract (3) from (4)

2C + 4F2 - (2C + 2H2) -----> 2CF4 - C2H4 .....ΔH = -1360 - 52.3 = -1412.3 kj

2C + 4F2 - 2C -2H2 -----> 2CF4 - C2H4 ...ΔH = -1412.3 kj

4F2 + C2H4 ------> 2CF4 + 2H2 ....ΔH = -1412.3 kj ....(5)

Multiply (1) by 2

2H2 + 2F2 -----> 4HF ....ΔH = -537X2 = -1074 kj ....(6)

Add (5) and (6)

4F2 + C2H4 + 2H2 + 2F2 ------> 2CF4 + 2H2 + 4HF ....ΔH = -1412.3 - 1074 = -2486.3 kj

C2H4 + 6F2 ------> 2CF4 + 4HF ....ΔH = -2486.3 kj

Answer 8

@aaronq @thomaster

Answer 9

If you did the math right, yea I think so.

Answer 10

I'm a bit unclear as to how you came up with (5)...

Answer 11

I don't remember — can you really just move stuff back and forth across the arrow like that?

Answer 12

Hmm now I'm not sure. Okay if I am at this step what is next.

C2H4(g) + 6 F2(g) --> 4 HF(g) + 2 CF4(g).. ΔH = -52.3 kJ - 1360kJ - 1074kJ

Answer 13

These are the steps for Hess's Law:

1. Write out the balanced thermochemical equations for the step reactions.

2. Write a balanced chemical equation for the target reaction.

3. Reverse any step reactions so that products/reactants match the target reaction (remember to reverse the signs of ΔH).

4. Scale the step reactions so products/reactants that don't appear in the target reaction will cancel out (multiply the ΔH by the scale value)

5. Add the step reactions (add the ΔH values).

6. Scale the resulting reaction so that it matches the target reaction. If necessary, multiply or divide the summed ΔH by the scale factor.

Answer 14

Yeah, I think your (5) violates #3 here....

Answer 15

Could you help me rewrite it?

Answer 16

sometimes it is easier to spot a mistake than to correct it :-)

Answer 17

I'm not sure how to fix my mistake.

Answer 18

Is this first part correct?

C2H4 + 6F2 ------> 2CF4 + 4HF

H2 + F2 ---> 2HF ..ΔH = -537 kj ...(1)

C + 2F2 -----> CF4 ...ΔH = -680 kj....(2)

2C + 2H2 -----> C2H4 ...ΔH = 52.3 kj ....(3)

Multiply (2) by 2

2C + 4F2 ----> 2CF4 ...ΔH = -680X2 = -1360 kj ....(4)

Subtract (3) from (4)

2C + 4F2 - (2C + 2H2) -----> 2CF4 - C2H4 .....ΔH = -1360 - 52.3 = -1412.3 kj

2C + 4F2 - 2C -2H2 -----> 2CF4 - C2H4 ...ΔH = -1412.3 kj

Answer 19

I believe what you had prior to (5) was correct in its working, although I'm not positive it was the right route yet.

Answer 20

Okay so for (5) is this correct?

4F2 + 2H2 ------> 2CF4 + C2H4 ....ΔH = -1412.3 kj ....(5)

Answer 21

Yes, I think that is correct.

Answer 22

Is (6) correct?

Answer 23

wait...I take that back — it isn't balanced! there's no carbon on the left side

Answer 24

Oh so was it balanced before or are both wrong?

Answer 25

when you subtract (3) from (4) you eliminate the carbon from the left hand side. are you sure you can do that? where does the carbon on the right side come from if you do?

Answer 26

No I'm not sure I can do that.

Answer 27

@#$#@% OpenStudy — I had the whole thing typed in except the last line, and poof!

Answer 28

Have a look at at this:

Answer 29

Sorry about that. I hate when stuff like that happens.

That makes more sense. Is there still a final number or is that it?

Answer 30

I thought you wanted this reaction though.

C2H4 + 6F2 --> 2CF4 + 4HF
@whpalmer4

Answer 31

ah, okay, yes I didn't finish balancing the multiples of each equation to get the overall equation right. well, I need to leave something for you to do :-)

Answer 32

But I think that gives you the general outline, and you just need to scale something in there

Answer 33

Also I think all of the signs are supposed to be switched.

ΔH = +537 kJ
ΔH = +680 kJ
ΔH = -52.3 kJ

Answer 34

huh? this is what you have at the top of the page: H2 (g) + F2 (g) → 2HF (g) ΔH = –537 kJ
that's what I used. I did reverse the 3rd one, but I changed the sign as required when I did.

Answer 35

Right that's the original from the problem. I think you have to change the sign once you start.

Answer 36

Okay, what you're saying isn't making any sense to me, so I think my usefulness here is at an end. Time to make dinner anyhow. Good luck!

Answer 37

Okay. Thanks for all your help.

Answer 38

Before you go, do you know anyone else that could help me with this.

Answer 39

did you get your question answered?

Answer 40

Not really. This was a really difficult problem, @aaronq

Answer 41

is it the same one from the top of the page?

Answer 42

Yes

Answer 43

k give me min

Answer 44

Okay thanks

Answer 45

\(H_2 (g) + F_2 (g) → 2HF (g) \) multiply by 2

\(C (s) + 2F_2 (g) → CF_4 (g) \) multiply by 2

\(2C (s) + 2H_2 (g) → 2C_2H_4 (g) \) reverse
----------------------------------------------------
\(6F_2 + C_2H_4 \rightarrow 2CF_4 + 4HF\)

----------------------------------------------------

\(2H_2 (g) + 2F_2 (g) → 4HF (g) \)

\(2C (s) + 4F_2 (g) → 2CF_4 (g) \)

\(C_2H_4 (g) → 2C (s) + 2H_2 (g)\)

----------------------------------------------------

\(6F_2 (g) +C_2H_4 (g) → 2CF_4 (g)+4HF (g) \)

Answer 46

just apply the operations i did to the \(\Delta H^o\)values and add them up.

Answer 47

So now I have to find the delta H for that reaction? I think that is the part I'm having trouble with.

Answer 48

(1) ΔH = –537 kJ (x2)
(2) ΔH = –680 kJ (x2)
(3)ΔH = +52.3 kJ (-1)


(1) ΔH = 2(–537 kJ)
(2) ΔH = 2(–680 kJ)
(3)ΔH = -52.3 kJ

\(\Delta H_{reaction}=-52.3 kJ +2(–680 kJ)+2(–537 kJ)=-2486.3kJ\)

is that what you got?

Answer 49

Yes!

Answer 50

then it should be right. Is this from a textbook?

Answer 51

Yes it is.

Answer 52

do you have the answers available?

Answer 53

No unfortunately not. It only has the problems.

Answer 54

hm well considering we both did it on our own and arrived at the same answer, I would say that its correct.

Answer 55

Great. Thanks for the help! I really appreciate it. :)

Answer 56

no problem, really !