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Question about enthalpy of reaction

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gokuls
#1
Dec26-12, 02:27 PM
P: 35
I think enthalpy of reaction is the amount of energy released when that reaction takes place (correct me if I'm wrong), but enthalpy of reaction is often written as per mole of something. As in kJ released per mol. What does the per mol represent? There are probably many reactants and products, so what is the "per mol" referring to?

Thanks!
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Mandelbroth
#2
Dec26-12, 02:35 PM
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Quote Quote by gokuls View Post
I think enthalpy of reaction is the amount of energy released when that reaction takes place (correct me if I'm wrong), but enthalpy of reaction is often written as per mole of something. As in kJ released per mol. What does the per mol represent? There are probably many reactants and products, so what is the "per mol" referring to?

Thanks!
Per mole of reaction.

If we have the reaction

[itex]H_2O (s) \rightarrow H_2O (l) \ \Delta H = 6.01 kJ/mol[/itex]

the reaction happening for one molecule of water is obviously not going to cause an enthalpy change of 6.01 kJ. However, if a mole of water molecules undergoes this, the change in enthalpy will be 6.01 kJ.
gokuls
#3
Dec26-12, 02:37 PM
P: 35
Quote Quote by Mandelbroth View Post
Per mole of reaction.

If we have the reaction

[itex]H_2O (s) \rightarrow H_2O (l) \ \Delta H = 6.01 kJ/mol[/itex]

the reaction happening for one molecule of water is obviously not going to cause an enthalpy change of 6.01 kJ. However, if a mole of water molecules undergoes this, the change in enthalpy will be 6.01 kJ.
Say there were multiple reactants and products, then which one would it be referring to?

Mandelbroth
#4
Dec26-12, 02:39 PM
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Question about enthalpy of reaction

Quote Quote by gokuls View Post
Say there were multiple reactants and products, then which one would it be referring to?
The reaction happening ≈6.022 X 1023 times.
gokuls
#5
Dec26-12, 02:41 PM
P: 35
Quote Quote by Mandelbroth View Post
The reaction happening ≈6.022 X 1023 times.
No, what I'm saying is, say the enthalpy of reaction for the following reaction is x kJ/mol.
HCl + NaOH -> NaCl + H20.

Then what would the per mol refer to? To the HCl, or the NaOH, or something else?
Mandelbroth
#6
Dec26-12, 03:04 PM
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Quote Quote by gokuls View Post
No, what I'm saying is, say the enthalpy of reaction for the following reaction is x kJ/mol.
HCl + NaOH -> NaCl + H20.

Then what would the per mol refer to? To the HCl, or the NaOH, or something else?
Per mol of reaction. Not anything in the equation.

If the reaction were to occur 6.022 X 1023 times, the change in enthalpy would be x kJ.
Studiot
#7
Dec26-12, 03:18 PM
P: 5,462
It is usually clear from the stoichiometry of the reaction.

Here are some examples, all using heats of combustion that may make this clear.

Burning 32g of sulphur, 12g of carbon or 2g of hydrogen have the following reactions and heats of reaction

S + O2= SO2 : ΔH = -297 kJ/mol
1mole of sulphur + 1mole of oxygen = 1mole of sulphur dioxide

C + O2= CO2 : ΔH = -393 kJ/mol

H2 + 1/2O2= H2O : ΔH = -286 kJ/mol
1mole of hydrogen + half mole of oxygen = 1 mole of water
Studiot
#8
Dec26-12, 03:55 PM
P: 5,462
Per mol of reaction. Not anything in the equation.
Not exactly.

You need to specify the reaction more clearly and fully and also specify the state and condition of the products as this also has abearing.

The per mol is per mol of the principal reactant as named in the reaction so for instance in the soltion of ammonium nitrate in water the heat of solution (reaction) does not even specify a particular number of water molecules, just a general number n, so long as n is large.

NH4NO3(solid) + nH2) =NH4NO3.nH2O

ΔH per mole of ammonium nitrate is 29 J/mol

The amount of water is not specified or fixed.
gokuls
#9
Dec26-12, 04:02 PM
P: 35
Quote Quote by Mandelbroth View Post
Per mol of reaction. Not anything in the equation.

If the reaction were to occur 6.022 X 1023 times, the change in enthalpy would be x kJ.
Thanks, I finally kind of get it!
Studiot
#10
Dec26-12, 04:13 PM
P: 5,462
per mol of reaction?

but which reaction?

Look at my oxygen/hydrogen example

The heat of reaction of 1 mol of oxygen with hydrogen is twice that of the heat of reaction of 1 mol of hydrogen with oxygen.
gokuls
#11
Dec26-12, 04:20 PM
P: 35
Quote Quote by Studiot View Post
per mol of reaction?

but which reaction?

Look at my oxygen/hydrogen example

The heat of reaction of 1 mol of oxygen with hydrogen is twice that of the heat of reaction of 1 mol of hydrogen with oxygen.
I see your argument, but then I still don't understand how to interpret the mol part. Is it just the mol of the main product.
Borek
#12
Dec26-12, 05:16 PM
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"per mol" in this context is ambiguous, if the source of the data is serious it should clarify what convention is used.
Studiot
#13
Dec26-12, 06:04 PM
P: 5,462
"per mol" in this context is ambiguous, if the source of the data is serious it should clarify what convention is used.
That's exactly what I was trying to put over. The source will (should) tell you more than just
per mol.

So the reaction of (1mol) hydrogen with oxygen requires 1/2 mol of oxygen to produce 1 mol of water, at the enthalpy of reaction I stated earlier per mol (of hydrogen)

The reaction of (1mol) of oxygen with hydrogen requires 2mols of hydrogen and produces 2 moles of water and twice the enthalpy of reaction per mol (of oxygen)
Mandelbroth
#14
Dec27-12, 03:41 PM
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Quote Quote by Studiot View Post
Not exactly. [...]
Quoting from the 9th edition of Chemistry from Raymond Chang, on page 236.
"The 'per mole' in the unit for ΔH means that this is the enthalpy change per mole of the reaction (or process) as it is written [italics are in the text, but emphasize my point all the same], that is, when 1 mole of ice is converted to 1 mole of liquid water [referring to the example I gave]."

So now, I need clarification too, because I thought I was right.
Borek
#15
Dec27-12, 04:16 PM
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Quote Quote by Mandelbroth View Post
Quoting from the 9th edition of Chemistry from Raymond Chang, on page 236.
"The 'per mole' in the unit for ΔH means that this is the enthalpy change per mole of the reaction (or process) as it is written [italics are in the text, but emphasize my point all the same], that is, when 1 mole of ice is converted to 1 mole of liquid water [referring to the example I gave]."

So now, I need clarification too, because I thought I was right.
The book precisely explains what they mean by 1 mole of the reaction. You take the reaction as it is WRITTEN. So, if they wrote H2+1/2O2->H2O that's what they mean, if they write 2H2+O2->2H2O - that's what they mean.

Note that in both cases we deal with 6.02x1023 reactions.
Studiot
#16
Dec27-12, 04:18 PM
P: 5,462
So now, I need clarification too, because I thought I was right
Well you were right, and the comment about Avogadro's number of repeats is very good.

However, it was incomplete. 'The equation as written'

Going back to the reaction between oxygen and hydrogen.

As written

H2 + 1/2O2 = H2O

or

O2 + 2 H2 = 2H2O

Does this clear things up?

I am used to the convention that we use the first substance mentioned as our basis.

Edit Borek is obviously quicker on the draw than I am, so remind me never to have a shoot out at the OK corral with him
Mandelbroth
#17
Dec27-12, 04:39 PM
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Quote Quote by Studiot View Post
Well you were right, and the comment about Avogadro's number of repeats is very good.

However, it was incomplete. 'The equation as written'

Going back to the reaction between oxygen and hydrogen.

As written

H2 + + 1/2O2 = H2O

or

O2 + 2 H2 = 2H2O

Does this clear things up?

I am used to the convention that we use the first substance mentioned as our basis.

Edit Borek is obviously quicker on the draw than I am, so remind me never to have a shoot out at the OK corral with him
Quicker, yes, but you explained my "mistake" better. Thank you.
gokuls
#18
Dec27-12, 09:42 PM
P: 35
Quote Quote by Borek View Post
The book precisely explains what they mean by 1 mole of the reaction. You take the reaction as it is WRITTEN. So, if they wrote H2+1/2O2->H2O that's what they mean, if they write 2H2+O2->2H2O - that's what they mean.

Note that in both cases we deal with 6.02x1023 reactions.
How is it that in both cases we deal with 1 mol worth of reactions? In the first equation you wrote, it would be 1 mol worth of reactions because you produce 1 mol of water, but in the second equation, you must have 2 mols of reactions, since you produce 2 mol of water, n'est pas?


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