# Enthalpy of reaction

by susdu
Tags: enthalpy, reaction
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 P: 24 Given some chemical reaction, A+B -> C+D (all are ideal gases) that occures in 298K and 1 atm. why can't I automatically say that ΔH=0 for this reaction? I know that enthalpy is a state function that is dependant on temperature alone (for an ideal gas), and the final and initial temperature are the same.
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 Quote by susdu Given some chemical reaction, A+B -> C+D (all are ideal gases) that occures in 298K and 1 atm. why can't I automatically say that ΔH=0 for this reaction?
Because it's false.
ΔH of a reaction can be zero, greater than zero or less than zero; anyway it's very unprobable that it's exactly zero.
 I know that enthalpy is a state function that is dependant on temperature alone (for an ideal gas), and the final and initial temperature are the same.
No, it's the internal energy U which, for an ideal gas, depends on temperature only; enthalpy doesn't depend on temperature only: H = U - P*V.
 P: 24 isn't the enthalpy change for isothermal proccess in ideal gas 2.5nR(Tf-Ti)
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Enthalpy of reaction

 Quote by susdu Given some chemical reaction, A+B -> C+D (all are ideal gases) that occures in 298K and 1 atm. why can't I automatically say that ΔH=0 for this reaction? I know that enthalpy is a state function that is dependant on temperature alone (for an ideal gas), and the final and initial temperature are the same.
For a solution of ideal gaseous chemical species, enthalpy depends not only on temperature but also on the enthalpies of the individual chemical species at that temperature. If A and B disappear and are replaced by C and D, since the ideal gas enthalpies of pure A, B, C, and D differ at the same temperature, there will be a change in enthalpy (or heat has to be added or removed to maintain the same temperature) when A and B react to form C and D. We have extensive tables of the heats of formation of chemical species at a reference temperature and at ideal gas pressures to enable us to calculate the heats of reactions.
 P: 24 Thanks for the clear explanation, feels like a stone choked down :)
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 Quote by lightarrow H = U - P*V.
Sorry, H = U + P*V
P: 1,520
 Quote by susdu isn't the enthalpy change for isothermal proccess in ideal gas 2.5nR(Tf-Ti)
This is the enthalpy variation for an isobar process, not isothermal. However, as Chestermiller said, the enthalpy *of reaction* is another thing because it doesn't refer simply to the thermodynamical state:

http://en.wikipedia.org/wiki/Standar...py_of_reaction

<<The standard enthalpy of reaction (denoted ΔrH⊖) is the enthalpy change that occurs in a system when one mole of matter is transformed by a chemical reaction under standard conditions.>>

"Standard conditions" means that the initial and final states are the same.

Example: 1 mole of H2 and 1 mole of Cl2 are inside a cylynder with a piston at P = 101325 Pa and T = 25°C. Then they are ignited with a spark and react:

H2 + Cl2 → 2HCl

then you let the gas expand at constant pressure and temperature, and you measure the heat exchanged with the environment; you notice the system has given heat to the environment. Since, at constant pressure, the enthalpy variation equals the heat given to the system, it means the system's entalpy has decreased. Since you also have kept the temperature constant, it means that you have kept standard conditions, so the enthalpy variation is exactly the enthalpy of the reaction, which is negative in this case.

For this reason enthalpy in general doesn't depend on the state only (the initial and the final states are the same by definition, in this case) but also on the chemical species present and the number of moles of them.

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