A Level Chemistry - Average bond Enthalpies

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Silxnt1
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ΔH = H(products) - H(reactants)

ΔrH = ∑(bond enthalpies in reactants) - ∑(bond enthalpies in products)
Why are the total bond enthalpies in the products subtracted from the total bond enthalpies in the reactants, instead of the other way around, when avergae bond enthalpies are used??
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Silxnt1
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And, with experimental results, I've noticed that sometimes the textbook shows all the calculations and all, but then changes the sign from + to -. I don't always get it.
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charco
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(Original post by Silxnt1)
ΔH = H(products) - H(reactants)

ΔrH = ∑(bond enthalpies in reactants) - ∑(bond enthalpies in products)
Why are the total bond enthalpies in the products subtracted from the total bond enthalpies in the reactants, instead of the other way around, when avergae bond enthalpies are used??
They are two different equations.

The first equation refers to the enthalpy of formation. In a reaction you are dismantling the reactants, which is the reverse of formation, so you use the negative formation enthalpy. You are then assembling the products, i.e. forming them, so you use the positive of the formation enthalpy. The words positive and negative here do not refer to the exo or endothermic nature of the formation enthalpy, rather to the mathematical operation performed.

So:

Reaction enthalpy = -∑(ΔHf(reactants) + ∑(ΔHf(products)

Which on rearranging becomes

Reaction enthalpy = ∑(ΔHf(products) -∑(ΔHf(reactants)

But the definition of bond enthalpy is the energy required to break 1 mole of bonds of a specific type measured over a range of similar gaseous molecules.

So breaking the bonds of the reactants (which has to happen) IS the bond enthalpy, which is always a positive (endothermic process)

However making the bonds of the products (which has to happen), which the reverse of the bond enthalpy is always an exothermic process.

So we get:

Reaction enthalpy = ∑(ΔH(bond enthalpy reactants) - ∑(ΔHf(bond enthalpy products)

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Silxnt1
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(Original post by charco)
They are two different equations.

The first equation refers to the enthalpy of formation. In a reaction you are dismantling the reactants, which is the reverse of formation, so you use the negative formation enthalpy. You are then assembling the products, i.e. forming them, so you use the positive of the formation enthalpy. The words positive and negative here do not refer to the exo or endothermic nature of the formation enthalpy, rather to the mathematical operation performed.

So:

Reaction enthalpy = -∑(ΔHf(reactants) + ∑(ΔHf(products)

Which on rearranging becomes

Reaction enthalpy = ∑(ΔHf(products) -∑(ΔHf(reactants)

But the definition of bond enthalpy is the energy required to break 1 mole of bonds of a specific type measured over a range of similar gaseous molecules.

So breaking the bonds of the reactants (which has to happen) IS the bond enthalpy, which is always a positive (endothermic process)

However making the bonds of the products (which has to happen), which the reverse of the bond enthalpy is always an exothermic process.

So we get:

Reaction enthalpy = ∑(ΔH(bond enthalpy reactants) - ∑(ΔHf(bond enthalpy products)

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That makes more sense.


And just to clear something up;

Are ΔrH, ΔfH, ΔcH, ΔneutH relative to the chemical surroundings

and

the reaction enthalpy (change), from using average bond enthalpies, relative to the chemical system?


Or is that wrong and/or ΔrH is the same as the reaction enthalpy (change) from using average bond enthalpies?
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charco
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(Original post by Silxnt1)
That makes more sense.


And just to clear something up;

Are ΔrH, ΔfH, ΔcH, ΔneutH relative to the chemical surroundings

and

the reaction enthalpy (change), from using average bond enthalpies, relative to the chemical system?


Or is that wrong and/or ΔrH is the same as the reaction enthalpy (change) from using average bond enthalpies?
No, nothing is relative to the chemical surroundings, except in the original definitions in which we assign the enthalpy of elements to be zero under standard conditions and therefore the ΔfH of all elements under standard conditions = zero

The enthalpy change can be considered to be the heat lost or gained by interconversion with chemical internal potential energy.

It's not quite as simple as work can be done by or to the system, which affects the amount of chemical potential energy interchanging with the heat.

Enthalpy change values may be different when calculated by different means, but this is simply because of inaccuracies/inappropriate conditions. This typically affects bond enthalpy calculations as the average bond enthalpy may vary considerably from the actual bonds in a molecule.

Actual bond dissociation enthalpies vary from the average. A good example of this is the average bond enthalpy of the C-H bond. This is usually quoted at about 414 kJ mol-1

However, the bond dissociation enthalpies of methane vary from between 460 kJ mol-1 for CH3 ==> CH2 + H

to 339 kJ mol-1 for CH ==> C + H

The values for the four dissociations are: 439 kJ, 460 kJ, 423 kJ and 339 kJ, which for methane gives an average bond enthalpy of 415 kJ mol-1.

So you can see that using average bond enthalpies in calculations may give inaccurate answers. There is also the factors of different bonding types (resonance, delocalisation etc) that can affect the bond enthalpy and it must be remembered that bond energy terms are calculated ONLY for gaseous molecules.
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Silxnt1
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I understood the second part of your most recent reply, but for me to be able to understand the first part properly I think I something clarified.


You said

"The first equation refers to the enthalpy of formation. In a reaction you are dismantling the reactants, which is the reverse of formation, so you use the negative formation enthalpy. You are then assembling the products, i.e. forming them, so you use the positive of the formation enthalpy. The words positive and negative here do not refer to the exo or endothermic nature of the formation enthalpy, rather to the mathematical operation performed."

If you dismantle the reactants aren't you putting energy in (the system) to break the bonds in the reactants, therefore causing the system to gain energy from the surroundings and have a positive enthalpy change (from that alone).
And, similarly, when the products are assembled/formed, energy is released (from the system) through forming the bonds in the products, therefore causing the system to lose energy and have a negative enthalpy change (from that alone).
(I understand that the overall enthalpy change is the net of the two)
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charco
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(Original post by Silxnt1)
I understood the second part of your most recent reply, but for me to be able to understand the first part properly I think I something clarified.


You said

"The first equation refers to the enthalpy of formation. In a reaction you are dismantling the reactants, which is the reverse of formation, so you use the negative formation enthalpy. You are then assembling the products, i.e. forming them, so you use the positive of the formation enthalpy. The words positive and negative here do not refer to the exo or endothermic nature of the formation enthalpy, rather to the mathematical operation performed."

1. If you dismantle the reactants aren't you putting energy in (the system) to break the bonds in the reactants, therefore causing the system to gain energy from the surroundings and have a positive enthalpy change (from that alone).

No.

Don't forget that the definition of enthalpy of formation is "from the elements in their standard states". The elements themselves have bonds, so you are not just breaking bonds, you are also making bonds.

By the law of conservation of matter:

A ==> B .................. ΔH = x kJ mol-1

so

B ==> A .................. ΔH = -x kJ mol-1

Similarly the enthalpy of formation:

Elements needed for compound X ==> Compound X .............. ΔHf = x kJ mol-1

therefore

Compound X ==> Elements needed for compound X .............. -ΔHf = -x kJ mol-1

If the enthalpy of formation of compound X is exothermic then ΔHf will take a negative value and the reverse will be endothermic and positive.

Don't forget that you are using the negative sign in two different ways here. Conventionally an exothermic reaction has a negative value. But mathematically, if you reverse an equation you also reverse the sign.

And, similarly, when the products are assembled/formed, energy is released (from the system) through forming the bonds in the products, therefore causing the system to lose energy and have a negative enthalpy change (from that alone).
(I understand that the overall enthalpy change is the net of the two)
This is the same issue.
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Silxnt1
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(Original post by charco)
No.

Don't forget that the definition of enthalpy of formation is "from the elements in their standard states". The elements themselves have bonds, so you are not just breaking bonds, you are also making bonds.

By the law of conservation of matter:

A ==> B .................. ΔH = x kJ mol-1

so

B ==> A .................. ΔH = -x kJ mol-1

Similarly the enthalpy of formation:

Elements needed for compound X ==> Compound X .............. ΔHf = x kJ mol-1

therefore

Compound X ==> Elements needed for compound X .............. -ΔHf = -x kJ mol-1

If the enthalpy of formation of compound X is exothermic then ΔHf will take a negative value and the reverse will be endothermic and positive.

Don't forget that you are using the negative sign in two different ways here. Conventionally an exothermic reaction has a negative value. But mathematically, if you reverse an equation you also reverse the sign.



This is the same issue.
That makes sense. Are you a Chemistry teacher/professor by the way
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charco
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(Original post by Silxnt1)
That makes sense. Are you a Chemistry teacher/professor by the way
Amongst other things ...
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Silxnt1
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(Original post by charco)
Amongst other things ...
Like what, if you don't mind me asking
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charco
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(Original post by Silxnt1)
Like what, if you don't mind me asking
I design and build software for chemistry education.
And websites.
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Silxnt1
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(Original post by charco)
I design and build software for chemistry education.
And websites.
Nice
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