# Enthalpy Change Help!

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#1
Done
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9 years ago
#2
Use hess's law?
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#3
(Original post by soup)
Use hess's law?
I don't know how I would apply Hess' Law in this context, as I have not been given the reactants, but I know the elements will be 2C + 3H2 + Br2. From here I have no idea what to do...
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9 years ago
#4
4) Calculate delta H?f for bromomethane, CH3Br(g), given the bond data in the table and

C(s) ---> C(g) delta H? = 715 kJ mol-1

Br2(l) ---> Br2(g) delta H? = 15 kJ mol-1

Bond data in the table:

Bonds: H–H O=O C–C C=C C–H F–F H–O Br–Br C–Br H–Br C=O
KJ/mol: 436 496 348 612 412 158 463 193 276 366 743

How do I do this? Write out the equation that represents the enthalpy of formation of bromoethane.

But instead of going directly from elements to compound, go VIA the individual atoms:

elements(standard states) ---> gaseous atoms ---> compound

Now try to construct this equation using the data you are given...
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#5
(Original post by charco)
Write out the equation that represents the enthalpy of formation of bromoethane.

But instead of going directly from elements to compound, go VIA the individual atoms:

elements(standard states) ---> gaseous atoms ---> compound

Now try to construct this equation using the data you are given...
The equation that represents the enthalpy of formation of bromomethane would be:
2C + 3H2 + Br2 (all in gaseous states, so using values from table) ---> 3CH3Br ?

And at the bottom of the Hess' cycle for the elements, it would be the same equation as that of the reactants, but using the 715 and 15 kJ mol-1? 0
9 years ago
#6
The equation that represents the enthalpy of formation of bromomethane would be:
2C + 3H2 + Br2 (all in gaseous states, so using values from table) ---> 3CH3Br ?

And at the bottom of the Hess' cycle for the elements, it would be the same equation as that of the reactants, but using the 715 and 15 kJ mol-1? This is not the definition of enthalpy of formation...

It is the energy change when 1 mole of a substnace is formed from its constituent elements IN THEIR STANDARD STATES

carbon is a solid, bromine is a liquid...

Now can you see how to proceed?
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#7
(Original post by charco)
This is not the definition of enthalpy of formation...

It is the energy change when 1 mole of a substnace is formed from its constituent elements IN THEIR STANDARD STATES

carbon is a solid, bromine is a liquid...

Now can you see how to proceed?
I still am unsure what to do So it would then be C(s) + Br2 (l) ---> CH3Br (g)? Also the values of C(s) and Br2(l) are not given, so huh?
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9 years ago
#8
I still am unsure what to do So it would then be C(s) + Br2 (l) ---> CH3Br (g)? Also the values of C(s) and Br2(l) are not given, so huh?
You are not reading all of the posts. I have already told you how to proceed in the post at 18.59. I am not going to write out the same thing again!
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#9
Would you use combustion in Hess' Law? By writing: 2C + 3H2 + Br2 --> 2CH3Br

And at the bottom with: 2CO2 + 3H2O?
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#10
(Original post by charco)
You are not reading all of the posts. I have already told you how to proceed in the post at 18.59. I am not going to write out the same thing again!
I drew a Hess' Cycle and I did 223 + (158x3) = 697, 697 - 1100 = -403, therefore -403/6 = -67.2 which is the answer?
1
9 years ago
#11
I drew a Hess' Cycle and I did 223 + (158x3) = 697, 697 - 1100 = -403, therefore -403/6 = -67.2 which is the answer?
elements(standard states) ---> gaseous atoms ---> compound (CH3Br)

EQU 1: C(s) --> C(g): 715 kJ
Br2(l) --> Br2(g): 15kJ
EQU 2: 1/2Br2(l) --> Br(g): 7.5kJ
H2 --> 2H(g): 436 kJ
EQU 3: 3/2H2 --> 3H(g): 654 kJ

add up equations 1 +2 +3

C(s) --> C(g): 715 kJ
1/2Br2(l) --> Br(g): 7.5kJ
3/2H2 --> 3H(g): 654 kJ
---------------------------------------
C(s) + 1/2Br2(l) + 3/2H2(g) --> C(g) + Br(g) + 3H(g): ΔH = 1376.5 kJ

The next stage is to turn the gaseous atoms into CH3Br

Can you see how to proceed?
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#12
(Original post by charco)
elements(standard states) ---> gaseous atoms ---> compound (CH3Br)

EQU 1: C(s) --> C(g): 715 kJ
Br2(l) --> Br2(g): 15kJ
EQU 2: 1/2Br2(l) --> Br(g): 7.5kJ
H2 --> 2H(g): 436 kJ
EQU 3: 3/2H2 --> 3H(g): 654 kJ

add up equations 1 +2 +3

C(s) --> C(g): 715 kJ
1/2Br2(l) --> Br(g): 7.5kJ
3/2H2 --> 3H(g): 654 kJ
---------------------------------------
C(s) + 1/2Br2(l) + 3/2H2(g) --> C(g) + Br(g) + 3H(g): ?H = 1376.5 kJ

The next stage is to turn the gaseous atoms into CH3Br

Can you see how to proceed?
Alright I see what you have done, so then you reverse the sign of that right? And then add that to (412x3) + (0.5x193) = 1332.5 kJ, to get an answer of -44 kJ?
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9 years ago
#13
Alright I see what you have done, so then you reverse the sign of that right? And then add that to (412x3) + (0.5x193) = 1332.5 kJ, to get an answer of -44 kJ?
No you don't reverse the sign...

The stage is endothermic because it required energy to turn the elements into gaseous atoms.

The second stage is exothermic as you are forming bonds from gaseous atoms.

You just add the two stages together.

elements -- stage 1---> gaseous atoms -- stage 2---> compound

elements ---> compound = stage 1 + stage 2
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#14
(Original post by charco)
No you don't reverse the sign...

The stage is endothermic because it required energy to turn the elements into gaseous atoms.

The second stage is exothermic as you are forming bonds from gaseous atoms.

You just add the two stages together.

elements -- stage 1---> gaseous atoms -- stage 2---> compound

elements ---> compound = stage 1 + stage 2
So it's 1332.5 + 1376.5 = 2709 kJ mol-1 = Enthalpy of Formation for bromomethane?
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9 years ago
#15
So it's 1332.5 + 1376.5 = 2709 kJ mol-1 = Enthalpy of Formation for bromomethane?
noooooooo...

You are clearly more concerned with the actual answer than how to work it out. You are not reading the posts, you are just jumping at what you hope is the 'answer'. The most important thing is the methodology so that you can do this for yourself in the future.

The formation of bonds (second stage) is exothermic ...

It MUST be negative

You just add the two stages together.

elements -- stage 1 [endothermic (positive)] ---> gaseous atoms -- stage 2 [exothermic (negative)] ---> compound

elements ---> compound = stage 1 + stage 2
1
#16
(Original post by charco)
noooooooo...

You are clearly more concerned with the actual answer than how to work it out. You are not reading the posts, you are just jumping at what you hope is the 'answer'. The most important thing is the methodology so that you can do this for yourself in the future.

The formation of bonds (second stage) is exothermic ...

It MUST be negative

You just add the two stages together.

elements -- stage 1 [endothermic (positive)] ---> gaseous atoms -- stage 2 [exothermic (negative)] ---> compound

elements ---> compound = stage 1 + stage 2
I have drawn a Hess' Cycle for this, at the top I have the equation:

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

Now at the bottom I have the elements: C(s) and 2S(s).

So gaseous atoms to elements gives me a negative answer of -1161. Now I am trying to get from elements to CS2 (g), but I don't know how to proceed as I didn't understand what you said is the next step 1
#17
(Original post by charco)
noooooooo...

You are clearly more concerned with the actual answer than how to work it out. You are not reading the posts, you are just jumping at what you hope is the 'answer'. The most important thing is the methodology so that you can do this for yourself in the future.

The formation of bonds (second stage) is exothermic ...

It MUST be negative

You just add the two stages together.

elements -- stage 1 [endothermic (positive)] ---> gaseous atoms -- stage 2 [exothermic (negative)] ---> compound

elements ---> compound = stage 1 + stage 2
I'm sure once you do everything correctly you get -44 kJ mol-1 as the answer...
0
9 years ago
#18
I have drawn a Hess' Cycle for this, at the top I have the equation:

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

Now at the bottom I have the elements: C(s) and 2S(s).

So gaseous atoms to elements gives me a negative answer of -1161. Now I am trying to get from elements to CS2 (g), but I don't know how to proceed as I didn't understand what you said is the next step You're in the wrong thread !

The answers in the other one 0
9 years ago
#19
are u ok with this or still need help?
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#20
(Original post by charco)
You're in the wrong thread !

The answers in the other one Oh sorry! But I meant for this one, you have -1376.5 on one side, which you add to (412x3) + (193/2) = -44 kJ surely?
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