contradiction-enthalpy change from bond enthalpy?? Watch

debbie394
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1) i use hess' law cycle to find bond enthalpy changes
this means that to find enthalpy change i do reactants - products

2) this question contradicts what i know about bond enthalpies: first question 1a)i)

http://www.chemhume.co.uk/ASCHEM/Exa.../U3June.05.PDF

the bond enthalpy of reactants is 2220
the bond enthalpy of products is 2797
the mark scheme states that bonds broken is +ve and bonds formed is -ve
so the answer is 577

But if i were to calculate the answer using the drawn cycle i would get 2220-2797= -577
would i always have to add a minus sign for bonds formed? even if did this the answer would then turn out to be 2220--2797= 5017?
for some questions i didn't have to consider this?
how would i know when to use this?


this question completely contradicts what i know ??
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KeepItUnreal
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(Original post by esmeralda123)
1) i use hess' law cycle to find bond enthalpy changes
this means that to find enthalpy change i do reactants - products

2) this question contradicts what i know about bond enthalpies: first question 1a)i)

http://www.chemhume.co.uk/ASCHEM/Exa.../U3June.05.PDF

the bond enthalpy of reactants is 2220
the bond enthalpy of products is 2797
the mark scheme states that bonds broken is +ve and bonds formed is -ve
so the answer is 577

But if i were to calculate the answer using the drawn cycle i would get 2220-2797= -577
would i always have to add a minus sign for bonds formed? even if did this the answer would then turn out to be 2220--2797= 5017?
for some questions i didn't have to consider this?
how would i know when to use this?


this question completely contradicts what i know ??
The answer is +577 kJmol-1.
The only thing you need to know is that unless its a combustion reaction, the method of calculating enthalpy change will be Products-Reactants.
For combustion reactions, you do Reactants-Products, as energy is given out as the reaction tends to be exothermic.
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charco
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(Original post by esmeralda123)
1) i use hess' law cycle to find bond enthalpy changes
this means that to find enthalpy change i do reactants - products

2) this question contradicts what i know about bond enthalpies: first question 1a)i)

http://www.chemhume.co.uk/ASCHEM/Exa.../U3June.05.PDF

the bond enthalpy of reactants is 2220
the bond enthalpy of products is 2797
the mark scheme states that bonds broken is +ve and bonds formed is -ve
so the answer is 577

But if i were to calculate the answer using the drawn cycle i would get 2220-2797= -577
would i always have to add a minus sign for bonds formed? even if did this the answer would then turn out to be 2220--2797= 5017?
for some questions i didn't have to consider this?
how would i know when to use this?


this question completely contradicts what i know ??
To break the bonds in the reactants requires 2220 kJ
BUT
making the bonds in the products releases -2797 kJ

Hence the overall change = 2220 - 2797 kJ = -557 kJ
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debbie394
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(Original post by charco)
To break the bonds in the reactants requires 2220 kJ
BUT
making the bonds in the products releases -2797 kJ

Hence the overall change = 2220 - 2797 kJ = -557 kJ
the mark scheme states the answer is +557 kJ
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charco
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(Original post by esmeralda123)
the mark scheme states the answer is +557 kJ
The MS is wrong!
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debbie394
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(Original post by KeepItUnreal)
The answer is +577 kJmol-1.
The only thing you need to know is that unless its a combustion reaction, the method of calculating enthalpy change will be Products-Reactants.
For combustion reactions, you do Reactants-Products, as energy is given out as the reaction tends to be exothermic.
But what about the answer to this question on page 9 of Task1- AS Energetics Revision Q3
it uses : bond broken- bonds formed (reactants- products) to get the answer
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Pigster
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(Original post by charco)
The MS is wrong!


It turns out your statement is incorrect:

1(a)(i) bonds broken
N N + (O=O) + 4(N-H) = 163 + 497 + 4(390)= 2220 (kJ mol -1) (1)
bonds made (N-N) + 4(O-H) = 945 + 4(463) = 2797 (kJ mol-1) (1)
broken ΔH is +ve and made ΔH is –ve (1)
enthalpy of reaction = -577 (kJ mol-1) (1) [4]
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charco
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(Original post by Pigster)

It turns out your statement is incorrect:

1(a)(i) bonds broken
N N + (O=O) + 4(N-H) = 163 + 497 + 4(390)= 2220 (kJ mol -1) (1)
bonds made (N-N) + 4(O-H) = 945 + 4(463) = 2797 (kJ mol-1) (1)
broken ΔH is +ve and made ΔH is –ve (1)
enthalpy of reaction = -577 (kJ mol-1) (1) [4]
Dammit, I can only work with the information given...

I've been onto the website and to be fair the copy of the mark scheme is awful and it is easy to see where the OP misread it:

Attachment 716840

Here it is.

http://www.chemhume.co.uk/ASCHEM/Exa.../U3An06.05.PDF
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KeepItUnreal
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(Original post by esmeralda123)
But what about the answer to this question on page 9 of Task1- AS Energetics Revision Q3
it uses : bond broken- bonds formed (reactants- products) to get the answer
Actually come to think of it, I believe that the mark scheme is wrong as well. Considering its a combustion reaction as hydrazine is used as a fuel and is reacted with Oxygen to release its energy. So technically the enthalphy change should be -577kjmol-1.
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Pigster
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(Original post by charco)
To break the bonds in the reactants requires 2220 kJ
BUT
making the bonds in the products releases -2797 kJ

Hence the overall change = 2220 - 2797 kJ = -557 kJ
I can forgive your inability to read a shoddy pdf file, but getting -557 I can't forgive

The curious thing is, though, it how infectious that 5 turned out to be...

(Original post by esmeralda123)
the mark scheme states the answer is +557 kJ
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charco
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(Original post by Pigster)
I can forgive your inability to read a shoddy pdf file, but getting -557 I can't forgive

The curious thing is, though, it how infectious that 5 turned out to be...
I confess, I didn't bother to do the sums ...
:getmecoat:
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