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Question about enthalpy of reaction |
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| Dec26-12, 02:27 PM | #1 |
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Question about enthalpy of reaction
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?
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| Dec26-12, 02:35 PM | #2 |
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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. |
| Dec26-12, 02:37 PM | #3 |
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| Dec26-12, 02:39 PM | #4 |
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Question about enthalpy of reaction |
| Dec26-12, 02:41 PM | #5 |
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HCl + NaOH -> NaCl + H20. Then what would the per mol refer to? To the HCl, or the NaOH, or something else? |
| Dec26-12, 03:04 PM | #6 |
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If the reaction were to occur 6.022 X 1023 times, the change in enthalpy would be x kJ. |
| Dec26-12, 03:18 PM | #7 |
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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 |
| Dec26-12, 03:55 PM | #8 |
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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. |
| Dec26-12, 04:02 PM | #9 |
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| Dec26-12, 04:13 PM | #10 |
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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. |
| Dec26-12, 04:20 PM | #11 |
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| Dec26-12, 05:16 PM | #12 |
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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.
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| Dec26-12, 06:04 PM | #13 |
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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) |
| Dec27-12, 03:41 PM | #14 |
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"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.
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| Dec27-12, 04:16 PM | #15 |
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Note that in both cases we deal with 6.02x1023 reactions. |
| Dec27-12, 04:18 PM | #16 |
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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 |
| Dec27-12, 04:39 PM | #17 |
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| enthalpy change, thermochemistry |
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