Medikj
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To find the enthalpy of formation of KBr, I should sum all the other values (d and e with a negative sign in front). However, shouldn’t I halve the value for reaction ‘c’, since it is given in kJ mol^-1
and the equation only uses half a mole of Br?
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I’m confused because according to the given solutions i shouldn’t do that:
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Thank you!😅
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Roha125
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Reaction c,
Yes it is .5 mole but it is also Br2 so overall 1 Br (2×0.5=1)
That 1Br (liquid) turns to 1Br (gas) and 112 is required to do it.

You don't need to half it.
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Medikj
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(Original post by Roha125)
Reaction c,
Yes it is .5 mole but it is also Br2 so overall 1 Br (2×0.5=1)
That 1Br (liquid) turns to 1Br (gas) and 112 is required to do it.

You don't need to half it.
I thought so too, but then why do they halve it here for O2?
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(Top left corner reaction)
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Roha125
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because the values give '498' '496' and '107' are for 2 moles

if you look at the data booklet/chart where this given values are then you will see the the equations doesn't have '0.5'mol instead one mole whereas born-haber chains have '0.5',thats when we ÷2/×2
the 'chart equation' isn't similar to 'bornhaber chain equation'

in your earlier question the 'c' value of 112 is for that specfic equation

O2-->2O is 498 (chart equation)
so
0.5 O2 --> O is 498÷2 (born haber chain equation) these are different equation so we divide or multiple

your earlier
0.5 Br2 --> Br is 112 ,the bornhaber chain value is already given so no need to divide or multiple (no chart value)
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HecticG
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(Original post by Roha125)
because the values give '498' '496' and '107' are for 2 moles

if you look at the data booklet/chart where this given values are then you will see the the equations doesn't have '0.5'mol instead one mole whereas born-haber chains have '0.5',thats when we ÷2/×2
the 'chart equation' isn't similar to 'bornhaber chain equation'

in your earlier question the 'c' value of 112 is for that specfic equation

O2-->2O is 498 (chart equation)
so
0.5 O2 --> O is 498÷2 (born haber chain equation) these are different equation so we divide or multiple

your earlier
0.5 Br2 --> Br is 112 ,the bornhaber chain value is already given so no need to divide or multiple (no chart value)
So why is it that for oxygen (diatomic molecule) the enthalpy of atomisation you produce 2 moles of O, but for Br2 which is also a diatomic molecule, that value is actually for 1Br and so you don't have to divide by 2? Where did you get this equation from "O2-->2O is 498 (chart equation)" ? How do you know?
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Medikj
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(Original post by Roha125)
because the values give '498' '496' and '107' are for 2 moles

if you look at the data booklet/chart where this given values are then you will see the the equations doesn't have '0.5'mol instead one mole whereas born-haber chains have '0.5',thats when we ÷2/×2
the 'chart equation' isn't similar to 'bornhaber chain equation'

in your earlier question the 'c' value of 112 is for that specfic equation

O2-->2O is 498 (chart equation)
so
0.5 O2 --> O is 498÷2 (born haber chain equation) these are different equation so we divide or multiple

your earlier
0.5 Br2 --> Br is 112 ,the bornhaber chain value is already given so no need to divide or multiple (no chart value)
The problem is, as I said in the first post, the values are given in kJ mol^-1, so the value for c (112) is for one mole not for half a mole. That’s why I don’t get it😅. Maybe the solutions are wrong or there’s a typo
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Roha125
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(Original post by HecticG)
but for Br2 which is also a diatomic molecule, that value is actually for 1Br
Because the 'c' is given by the arrow so c is value is specifically for that equation which is
0.5Br2 --> Br (atomisation reaction)
Bond energy of Br2 is not given instead directly atomisation value of 0.5Br2 is given

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You will get this kind of chart in every problem and I am pretty sure the above bornhaber was also solved using a similar chart

O2 -->2O 498 ( Bond energy)not atomisation
So
0.5O2 --> O 249 (atomisation reaction)(born haber chain equation)

See in top left, bond energy of O2÷2

Even I had similar trouble understanding born-haber,when to multiple and when to not multiple
Do more born haber maths yourselve and compare.

If both chartequation and chain equation matches by moles then don't have to divide or multiple
If diffenrent then do
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HecticG
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(Original post by Roha125)
Because the 'c' is given by the arrow so c is value is specifically for that equation which is
0.5Br2 --> Br (atomisation reaction)
Bond energy of Br2 is not given instead directly atomisation value of 0.5Br2 is given

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You will get this kind of chart in every problem and I am pretty sure the above bornhaber was also solved using a similar chart

O2 -->2O 498 ( Bond energy)not atomisation
So
0.5O2 --> O 249 (atomisation reaction)(born haber chain equation)

See in top left, bond energy of O2÷2

Even I had similar trouble understanding born-haber,when to multiple and when to not multiple
Do more born haber maths yourselve and compare.

If both chartequation and chain equation matches by moles then don't have to divide or multiple
If diffenrent then do
In the table where in brackets they put (1/2O2---> O) and then its enthalpy is very basic and easy to understand- you can't go wrong there. However in the exam they wouldn't give you the bracket. Hence why I was confused
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Roha125
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(Original post by HecticG)
In the table where in brackets they put (1/2O2---> O) and then its enthalpy is very basic and easy to understand- you can't go wrong there. However in the exam they wouldn't give you the bracket. Hence why I was confused
understandable
do as many born-haber as possible
tips-be calm and don't confuse them.
KNOW ALL definition by heart of all terms (atomisation,affinity etc) these definations are critical is writing equation.

write chart equation and compare to what is in born-haber and then decide to multiple or not
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HecticG
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(Original post by Roha125)
understandable
do as many born-haber as possible
tips-be calm and don't confuse them.
KNOW ALL definition by heart of all terms (atomisation,affinity etc) these definations are critical is writing equation.

write chart equation and compare to what is in born-haber and then decide to multiple or not
So if they give me the bond dissociation of oxygen by saying "the bond dissociation enthalpy of oxygen is 500 kjmol-1" and I have a compound like Na2O (so 2 sodium ions for one oxygen ion), then I will keep the value of oxygen as 500 right- I don't need to x2/divide by 2 ?
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Roha125
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(Original post by HecticG)
So if they give me the bond dissociation of oxygen by saying "the bond dissociation enthalpy of oxygen is 500 kjmol-1" and I have a compound like Na2O (so 2 sodium ions for one oxygen ion), then I will keep the value of oxygen as 500 right- I don't need to x2/divide by 2 ?
Yes but
It also depends on how you/question writes the equations in born-haber
If Na2O is broken (in other steps) and they write 0.5O2 then ÷2
But if equation if question is written in another way and O2 is given then no need to divide
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HecticG
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(Original post by Roha125)
Yes but
It also depends on how you/question writes the equations in born-haber
If Na2O is broken (in other steps) and they write 0.5O2 then ÷2
But if equation if question is written in another way and O2 is given then no need to divide
Sorry but it's kinda hard understanding born-haber through text- I didn't understand "If Na2O is broken (in other steps) and they write 0.5O2 then ÷2". But isn't 0.5O2 just O ? Hence wouldn't the value still be "500"- why would you have to divide by 2 ? And what do you mean by if Na2O is broken?
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Roha125
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"bond dissociation enthalpy of oxygen is 500"
oxygen is normally O2 so this value of 500 is for O2 actually not 1 O

For 1O or 0.5O2 we have to divide
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