# Calculate the average C-S bond energy

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#1
Done
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9 years ago
#2
8) Use the following data to calculate the average C-S bond energy in CS2(l).

S(s) ---> S(g) Delta H? = 223 kJ mol-1
C(s) ---> C(g) Delta H? = 715 kJ mol-1
Enthalpy of formation of CS2(l) Delta H? = 88 kJ mol-1
CS2(l) ---> CS2(g) Delta H? = 27 kJ mol-1

How would you do this? I thought it is just 27 kJ mol-1 but clearly it is not You must construct a Hess cycle, or manipulate the equations to give you the equation you require.

What is the definition of bond enthalpy?

Write out an equation to represent it and then construct this equation from the ones you are given.
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#3
(Original post by charco)
You must construct a Hess cycle, or manipulate the equations to give you the equation you require.

What is the definition of bond enthalpy?

Write out an equation to represent it and then construct this equation from the ones you are given.
How can you construct a Hess cycle for this though? It would be C + S2 ---> CS2? And the elements at the bottom would be the same: C + S2? I am so confused...
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9 years ago
#4
How can you construct a Hess cycle for this though? It would be C + S2 ---> CS2? And the elements at the bottom would be the same: C + S2? I am so confused...
No, you must be more rigorous and take into account the states of matter...

for example:

C(g) + 2S(g) -->CS2(g)

would represent the formation of two C-S bonds
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#5
(Original post by charco)
No, you must be more rigorous and take into account the states of matter...

for example:

C(g) + 2S(g) -->CS2(g)

would represent the formation of two C-S bonds
But what after that?
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9 years ago
#6
But what after that?
You can't read that post in isolation...
What I'm saying is that you must construct your required equation using the equations given...

OK, so the equation you actually want is:

C(g) + 2S(g) --> CS2(g)

so construct it one step at a time from the other equations.

1. You want C(g). Is there any equation that uses it?

2. Then you want to add 2S(g). Is there any equation that contains S(g)? Well double it and add...

Until you eventually end up with the required equation (doing the same to all of the enthalpy changes)
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#7
(Original post by charco)
You can't read that post in isolation...
What I'm saying is that you must construct your required equation using the equations given...

OK, so the equation you actually want is:

C(g) + 2S(g) --> CS2(g)

so construct it one step at a time from the other equations.

1. You want C(g). Is there any equation that uses it?

2. Then you want to add 2S(g). Is there any equation that contains S(g)? Well double it and add...

Until you eventually end up with the required equation (doing the same to all of the enthalpy changes)
1) Right, and this equation uses it: C(s) --> C(g) (which is 715 kJ mol-1)
2) This one: S(s) --> S(g) (which is 223 kJ mol-1), so 223 x 2 = 446 kJ mol-1.

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9 years ago
#8
1) Right, and this equation uses it: C(s) --> C(g) (which is 715 kJ mol-1)
2) This one: S(s) --> S(g) (which is 223 kJ mol-1), so 223 x 2 = 446 kJ mol-1.

If you add the two you get:

C(s) --> C(g)
S(s) --> S(g)
---------------
C(s) + S(s) --> C(g) + S(g)

is that what you want?

don't you want two sulphur particles?

why not multiply the sulphur equation by 2 and then add them?

But that will give you the reverse of what you want....

OK so reverse the equation and change the sign...

do you get the drift?
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#9
(Original post by charco)
If you add the two you get:

C(s) --> C(g)
S(s) --> S(g)
---------------
C(s) + S(s) --> C(g) + S(g)

is that what you want?

don't you want two sulphur particles?

why not multiply the sulphur equation by 2 and then add them?

But that will give you the reverse of what you want....

OK so reverse the equation and change the sign...

do you get the drift?
Did I not say to x2 to the sulfur? It says in what I just wrote Anyways so you would then reverse the sign to get -1161, and then add 88 to get -1073, then divide by 2 to get -563.5 which is the answer? Still doesn't seem right because I didn't use the value of 27 kJ mol-1... 0
9 years ago
#10
Did I not say to x2 to the sulfur? It says in what I just wrote Anyways so you would then reverse the sign to get -1161, and then add 88 to get -1073, then divide by 2 to get -563.5 which is the answer? Still doesn't seem right because I didn't use the value of 27 kJ mol-1... Because what you are making is CS2(g) and for the enthalpy of formation you want CS2(l) as it is a liquid under standard conditions (and a very smelly one at that!)

So you must change the gas to a liquid using the value given...
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#11
(Original post by charco)
Because what you are making is CS2(g) and for the enthalpy of formation you want CS2(l) as it is a liquid under standard conditions (and a very smelly one at that!)

So you must change the gas to a liquid using the value given...
But how do you use that value given to convert gas to a liquid?
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9 years ago
#12
But how do you use that value given to convert gas to a liquid?
!!!!

you are given:

CS2(l) ---> CS2(g) ΔHº = 27 kJ mol-1

So to reverse the process simply reverse the sign

CS2(g) ---> CS2(l) ΔHº = -27 kJ mol-1
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#13
(Original post by charco)
!!!!

you are given:

CS2(l) ---> CS2(g) ?Hº = 27 kJ mol-1

So to reverse the process simply reverse the sign

CS2(g) ---> CS2(l) ?Hº = -27 kJ mol-1
So then wouldn't the answer be -1161 - 27 = -1188 kJ?
0
9 years ago
#14
So then wouldn't the answer be -1161 - 27 = -1188 kJ?
Use the following data to calculate the average C-S bond energy in CS2(l).

S(s) ---> S(g) Delta H? = 223 kJ mol-1
C(s) ---> C(g) Delta H? = 715 kJ mol-1
Enthalpy of formation of CS2(l) Delta H? = 88 kJ mol-1
CS2(l) ---> CS2(g) Delta H? = 27 kJ mol-1

The average bond energy can be obtained by dividing the following equation by 2

CS2(g) --> C(g) + 2S(g)

So we must construct it stepwise.
-------------------------------------

equ1: C(s) ---> C(g) ΔH = 715 kJ mol-1
equ2: 2S(s) ---> 2S(g) ΔH = 446 kJ
equ3: C(s) + 2S(s) --> C(g) + 2S(g) ΔH = 1161 kJ
equ4: C(s) + 2S(s) --> CS2(l) ΔH = 88 kJ
-------------------------------------- --------- subtract 3 from 4
equ5: C(g) + 2S(g) --> CS2(l) ΔH = -1073 kJ
equ6: CS2(l) ---> CS2(g) ΔH = 27 kJ
C(g) + 2S(g) --> CS2(g) ΔH = -1046 kJ

Reverse and divide by 2 for C-S bond enthalpy = +523 kJ
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5 years ago
#15
(Original post by charco)
Use the following data to calculate the average C-S bond energy in CS2(l).

S(s) ---> S(g) Delta H? = 223 kJ mol-1
C(s) ---> C(g) Delta H? = 715 kJ mol-1
Enthalpy of formation of CS2(l) Delta H? = 88 kJ mol-1
CS2(l) ---> CS2(g) Delta H? = 27 kJ mol-1

The average bond energy can be obtained by dividing the following equation by 2

CS2(g) --> C(g) + 2S(g)

So we must construct it stepwise.
-------------------------------------

equ1: C(s) ---> C(g) ΔH = 715 kJ mol-1
equ2: 2S(s) ---> 2S(g) ΔH = 446 kJ
equ3: C(s) + 2S(s) --> C(g) + 2S(g) ΔH = 1161 kJ
equ4: C(s) + 2S(s) --> CS2(l) ΔH = 88 kJ
-------------------------------------- --------- subtract 3 from 4
equ5: C(g) + 2S(g) --> CS2(l) ΔH = -1073 kJ
equ6: CS2(l) ---> CS2(g) ΔH = 27 kJ
C(g) + 2S(g) --> CS2(g) ΔH = -1046 kJ

Reverse and divide by 2 for C-S bond enthalpy = +523 kJ
Hi, I get the rest of your work and did the same myself, however I dont understand why you have subtracted equation 3 from 4 could you please explain yourself thanks !!
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5 years ago
#16
(Original post by Hudl)
Hi, I get the rest of your work and did the same myself, however I dont understand why you have subtracted equation 3 from 4 could you please explain yourself thanks !!
To make sure that the carbon disulphide(l) appears on the RHS of equation 5 ...

so that is subsequently gets cancelled when added to equation 6
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5 years ago
#17
(Original post by charco)
To make sure that the carbon disulphide(l) appears on the RHS of equation 5 ...

so that is subsequently gets cancelled when added to equation 6

I kinda get you but its still not perfectly clear in my head

& why did you do equ 4 - 3 instead of equ 3 - equ 4 thanks in advance
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5 years ago
#18
(Original post by Hudl)
I kinda get you but its still not perfectly clear in my head

& why did you do equ 4 - 3 instead of equ 3 - equ 4 thanks in advance
3-4 puts liquid carbon disulfide on the LHS
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