# Chem Energy Cycles

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#1
Does anyone have any time to explain how to construct energy cycles for deltaH etc? I'm having trouble understanding them...
0
15 years ago
#2
(Original post by littlemisshala)
Does anyone have any time to explain how to construct energy cycles for deltaH etc? I'm having trouble understanding them...
Hmm... it's hard to explain without an example.
Let's see the easy reaction CH4 + O2 -> CO2 + 2H2O. (enthalpy change: dH)
Suppose we knew enthalpy change of the reactions below:
C(s) + 2H2 -> CH4 (dH1)
C(s) + O2 -> CO2 (dH2)
H2 + 1/2O2 -> H2O (dH3)
Remember enthalpy of formation of an element like O2 is 0.
Draw a cycle with an arrow of the reactions.
Code:
```(A)           dH            (B)
CH4    +    O2     ----->      CO2   +      2H2O
^                            ^          ^
|(dH1)                       |(dH2)     |(2.dH3)
C + 2H2 + O2 ________________|__________|
(C)```
So if u go from system A to B, u can go by the 1st reaction. But u can also go around, from A to C, then C to B, but u should follow the arrows to give correct value for dH.
As u see, from C -> A, enthalpy is dH1, so from A -> C, enthalpy is -dH1. From C-> B, it follows the arrows.
So dH = -dH1 + dH2 + 2.dH3.
Other reactions are quite similar.
Does it make sense now?
0
15 years ago
#3
yep.... dont worry... I'll try my best...

ok... First you would need to understand the concept of thermochemical change (delta H)... it is the CHANGE OF ENERGY accompanied by a chemical reaction...

Now its time for the signs....
a POSITIVE delta H indicates an ENDOthermic reaction as energy is PUT INTO the chemical bonds for reaction which is taken from the environment (that why it feels cold)...
a NEGATIVE delta H indicates an EXOthermic reaction as energy is TAKEN OUT of chemical bonds that is dissapated into the surroundings (that is why it feels hot)...

OK... Hess' Law states that the energy change is independent on the route taken... so whether path A or path B does not make a difference to delta H so long as the products and reactants are the same... even though different routes are taken...

eg. take the two pathways A --> B --> C --> D and A --> E --> F --> D, the total delta H is equal for the whole reaction of A --> D regarless of which pathway is taken...

So how do this help us then?
Well... if you have a reaction that only occurs in THEORY... such as determining Enthalpy of formation... (it is virtually impossible to do)... you can construct a side route to get the delta H....

eg. take C + 2 H2 --> CH4 delta H= say A
it is impossible to do experimentally so we burn the mixture of carbon and hydrogen gas in pure oxygen...
C + O2 --> CO2 delta H = say B
2 H2 + 2 O2 --> 4 H2O (notice the 2 for the reaction... so DOUBLE delta H) delta H = say C

Then we burn Methane...
CH4 + 3 O2 --> CO2 + 4 H2O delta H = say D

Now... as we are following the direction of the arrow from carbon and hydrogen to the combustion products.. we take the quoted values... B+C

As we follow the OPPOSITE direction of the arrow from methane to the products... we take the NEGATIVE of the quoted values... -D

So.. A = B+C-D

feel free to ask me anymore questions... email me or PM me....
0
15 years ago
#4
Woa, u took an e.g ... such similar like mine 0
#5
(Original post by BCHL85)
Hmm... it's hard to explain without an example.
Let's see the easy reaction CH4 + O2 -> CO2 + 2H2O. (enthalpy change: dH)
Suppose we knew enthalpy change of the reactions below:
C(s) + 2H2 -> CH4 (dH1)
C(s) + O2 -> CO2 (dH2)
H2 + 1/2O2 -> H2O (dH3)
Remember enthalpy of formation of an element like O2 is 0.
Draw a cycle with an arrow of the reactions.
Code:
```(A)           dH            (B)
CH4    +    O2     ----->      CO2   +      2H2O
^                            ^          ^
|(dH1)                       |(dH2)     |(2.dH3)
C + 2H2 + O2  ________________|__________|
(C)```
So if u go from system A to B, u can go by the 1st reaction. But u can also go around, from A to C, then C to B, but u should follow the arrows to give correct value for dH.
As u see, from C -> A, enthalpy is dH1, so from A -> C, enthalpy is -dH1. From C-> B, it follows the arrows.
So dH = -dH1 + dH2 + 2.dH3.
Other reactions are quite similar.
Does it make sense now?
Thank you so much to both of you for your help. I didn't undertand where you got the bit in bold BCHL85...
0
15 years ago
#6
Sorry, it should be 2O2, I forgot to balance both sides.
0
#7
(Original post by BCHL85)
Sorry, it should be 2O2, I forgot to balance both sides.
But where did you get those from?
0
15 years ago
#8
(Original post by littlemisshala)
But where did you get those from?
Look at the 1st equation., you have CH4, and O2, right?
Look at the next equation they give: CH4 made from C and H2, right?
Look at the next ones, CO2 made from C and O2, H2O from H2 and O2.
So you just draw a cycle.
0
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