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# chemistry watch

1. http://pmt.physicsandmathstutor.com/...%20Entropy.pdf

question 6b,

it says use data from part aii
I don't understand which ones from part aii we have to use and why??
2. (Original post by chemquestion)
http://pmt.physicsandmathstutor.com/...%20Entropy.pdf

question 6b,

it says use data from part aii
I don't understand which ones from part aii we have to use and why??
YOu need to work out the enthalpy change for this reaction and then substitute into Gibbs free energy equation.

To work out the enthalpy change you have to break the equation down into a series of energetic steps:

BaCl2 --> Ba2+(g) + 2Cl-(g) ..............is the first step (the endothermic lattice enthalpy)

... can you take it from here?
3. (Original post by charco)
YOu need to work out the enthalpy change for this reaction and then substitute into Gibbs free energy equation.

To work out the enthalpy change you have to break the equation down into a series of energetic steps:

BaCl2 --> Ba2+(g) + 2Cl-(g) ..............is the first step (the endothermic lattice enthalpy)

... can you take it from here?
No not sure
How do we calculate the enthalpy change?
4. (Original post by chemquestion)
No not sure
How do we calculate the enthalpy change?
The sum of the individual steps.

BaCl2(s) --> Ba(s) + Cl2(g)

I showed you the first step:

BaCl2 --> Ba2+(g) + 2Cl-(g)

Now the next step is:

Ba2+(g) + 2Cl-(g) --> Ba(g) + 2Cl(g) (actually two steps combined)

Then

Ba(g) + 2Cl(g) --> Ba(s) + Cl2(g) (once again two steps combined)

You add up the energy changes for each of these steps by consulting the data table
5. (Original post by charco)
The sum of the individual steps.

BaCl2(s) --> Ba(s) + Cl2(g)

I showed you the first step:

BaCl2 --> Ba2+(g) + 2Cl-(g)

Now the next step is:

Ba2+(g) + 2Cl-(g) --> Ba(g) + 2Cl(g) (actually two steps combined)

Then

Ba(g) + 2Cl(g) --> Ba(s) + Cl2(g) (once again two steps combined)

You add up the energy changes for each of these steps by consulting the data table
How are we supposed to know that we have to use these individual steps and combine? Can't remember learning content on this (combing steps).
6. (Original post by chemquestion)
How are we supposed to know that we have to use these individual steps and combine? Can't remember learning content on this (combing steps).
It is precisely what you do when you use a Born-Haber cycle!
7. (Original post by charco)
It is precisely what you do when you use a Born-Haber cycle!
I'm stuck on this question.
Why is it ΔH = 859 × 1000
8. (Original post by chemquestion)
I'm stuck on this question.
Why is it ΔH = 859 × 1000
Why is what ΔH = 859 × 1000?
9. Maybe you should check out the Born-Haber cycle for barium chloride
10. (Original post by charco)
Maybe you should check out the Born-Haber cycle for barium chloride
Can you show me the numbers we are adding up?
Also isn't Enthalpy of formation of barium chloride, the enthalpy change we need to calculate temperature?
11. (Original post by chemquestion)
Can you show me the numbers we are adding up?
Also isn't Enthalpy of formation of barium chloride, the enthalpy change we need to calculate temperature?
Calculate a value for the electron affinity of chlorine.

Data

Enthalpy of atomisation of barium +180 kJ mol–1
Enthalpy of atomisation of chlorine +122 kJ mol–1
Enthalpy of formation of barium chloride –859 kJ mol–1
First ionisation enthalpy of barium +503 kJ mol–1
Second ionisation enthalpy of barium +965 kJ mol–1
Lattice formation enthalpy of barium chloride –2056 kJ mol–1

Ba(s) + Cl2(g) --> BaCl2(s)............. -859 kJ

Ba(s) --> Ba(g)......................... +180kJ

Cl2(g) --> 2Cl(g)....................... +244kJ

Ba(g) --> Ba2+(g)..................... +1468 kJ

2Cl(g) + 2e --> 2Cl-(g)................. 2 x electron affinity

Ba2+(g) + 2Cl-(g) --> BaCl2(s)......... -2056 kJ

Calculation:

180 + 244 + 1468 + 2EA -2056 = -859

gather and rearrange

2EA = -695

Electron affinity of chlorine = -347.5 kJ mol-1

(b) The equation given is the reverse of the formation enthalpy = +859 kJ

ΔG = ΔH – TΔS

etc.

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