Finding the Molar Enthalpy of a Reaction with One Solid Product and One Gaseous Product
December 11, 2011 4:35 PM Subscribe
This is a hypothetical chemistry question. I'm tutoring someone, and a question like this came up on their exam. There was no answer provided, so I don't know if the answer we found was correct.
Say you have some reaction that (when the limiting reactant has been exhausted) produces 2.5 moles of some solid and 5.2 moles of a gas. You find that the heat released by this reaction is 32.3 kJ.
How do you find the molar enthalpy of reaction? "Molar enthalpy" usually implies that you are finding the enthalpy per each mole of product that is produced- but what is the procedure with two separate products? Do you divide the heat produced by the moles of the gas, by the moles of the solid, or by the combined moles of both?
I have seen problems in which there was one solid product and one aqueous product; in that case, the aqueous product is ignored and the heat produced is divided by the moles of the solid precipitate. I don't think that the above case is the same as that.
Say you have some reaction that (when the limiting reactant has been exhausted) produces 2.5 moles of some solid and 5.2 moles of a gas. You find that the heat released by this reaction is 32.3 kJ.
How do you find the molar enthalpy of reaction? "Molar enthalpy" usually implies that you are finding the enthalpy per each mole of product that is produced- but what is the procedure with two separate products? Do you divide the heat produced by the moles of the gas, by the moles of the solid, or by the combined moles of both?
I have seen problems in which there was one solid product and one aqueous product; in that case, the aqueous product is ignored and the heat produced is divided by the moles of the solid precipitate. I don't think that the above case is the same as that.
I may be wrong here - but my answer would be to divide by the number of moles of the limiting reactant.
posted by NoDef at 5:39 PM on December 11, 2011
posted by NoDef at 5:39 PM on December 11, 2011
Assuming that there's a little rounding error, this is probably meant to be a 1:2 ratio. Assuming that the reaction would be balanced as such:
Reactants --> A(s) + 2B (g)
Then deltaH= -32.3 kJ/ 2.6 moles or -32.3 / (5.2/2) moles, which is the same regardless of which product you use. Change in enthalpy is usually defined per mole of reaction, not per mole of any one substance, so you have to factor in the coefficients.
posted by Dr.Enormous at 5:44 PM on December 11, 2011 [2 favorites]
Reactants --> A(s) + 2B (g)
Then deltaH= -32.3 kJ/ 2.6 moles or -32.3 / (5.2/2) moles, which is the same regardless of which product you use. Change in enthalpy is usually defined per mole of reaction, not per mole of any one substance, so you have to factor in the coefficients.
posted by Dr.Enormous at 5:44 PM on December 11, 2011 [2 favorites]
I'll also note that it's done this way even if you only had one product or one reagent--you divide the moles by the coefficient in the balanced reaction to get "moles of reaction", then use that as the "per mole" in the molar enthalpy.
This avoids precisely the issue that was throwing you off here--if you do it any other way, the same reaction can have different enthalpies depending on which compound you use, which makes no sense.
posted by Dr.Enormous at 6:43 PM on December 11, 2011
This avoids precisely the issue that was throwing you off here--if you do it any other way, the same reaction can have different enthalpies depending on which compound you use, which makes no sense.
posted by Dr.Enormous at 6:43 PM on December 11, 2011
I'd guess that a gaseous component is included in the products to illustrate the fact that enthalpy is not simply heat, and that you have to add the energy necessary to do the work of expanding the reaction products against the ambient pressure of the environment to the heat when you calculate the enthalpy per mole.
posted by jamjam at 7:04 PM on December 11, 2011
posted by jamjam at 7:04 PM on December 11, 2011
[...] if you do it any other way, the same reaction can have different enthalpies depending on which compound you use, which makes no sense.
This depends on which pedantic academic formalism you are operating under. Sometimes you want, for convenience, to have the enthalpy per mole of limiting reagent or what not. Heats of combustion, for example, are usually tabulated per mole of fuel.
I've personally found that in my chemistry textbooks the enthalpy is per mole of reaction, but in my chemical engineering texts the enthalpy is usually given per mole of one of the reactants.
posted by selenized at 7:13 PM on December 11, 2011
This depends on which pedantic academic formalism you are operating under. Sometimes you want, for convenience, to have the enthalpy per mole of limiting reagent or what not. Heats of combustion, for example, are usually tabulated per mole of fuel.
I've personally found that in my chemistry textbooks the enthalpy is per mole of reaction, but in my chemical engineering texts the enthalpy is usually given per mole of one of the reactants.
posted by selenized at 7:13 PM on December 11, 2011
chemical engineering texts the enthalpy is usually given per mole of one of the reactants.
This is usually done in cases where one reactant is always limiting under normal circumstances (combustion, as you mentioned), and the reaction is simply balanced using that reactant as 1. As such, it's still per mole of reaction, and is identical to doing it the other way.
jamjam: while technically true, I highly doubt that's the intention here. For one, there'd have to be more data given. For two, pretty much all chemistry texts operate under the assumption of work being zero unless something is basically telling you "account for work here" in big bold letters.
posted by Dr.Enormous at 8:51 PM on December 11, 2011
This is usually done in cases where one reactant is always limiting under normal circumstances (combustion, as you mentioned), and the reaction is simply balanced using that reactant as 1. As such, it's still per mole of reaction, and is identical to doing it the other way.
jamjam: while technically true, I highly doubt that's the intention here. For one, there'd have to be more data given. For two, pretty much all chemistry texts operate under the assumption of work being zero unless something is basically telling you "account for work here" in big bold letters.
posted by Dr.Enormous at 8:51 PM on December 11, 2011
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posted by kagredon at 5:39 PM on December 11, 2011