Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: birdnerd007 on October 17, 2019, 04:51:40 PM

This is the reaction:
Calculate heat of formation for .6C + .3O2 > .6CO at 1300K
I know that I can get heat of formation for standard states on a table. How do I go about solving heat of formation for nonstandard state reactions? Do I have to use an integral with the Cp values and then do products minus reactants? I am trying to solve for Enthalpy of the reaction, but I am stuck how to calculate heat of formations for nonstandard state reactions. I appreciate any and all help to understand this.

[...] Do I have to use an integral with the Cp values and then do products minus reactants? [...]
Yes, that's it.
Here the reactants and products stay in the same state between RT and 1300K so it's easier.
Integrating Cp over T isn't comfortable. Here you're lucky because tables of the enthalpy of these compounds versus T exist, for instance in the CRC Handbook of Chemistry and Physics. So to say, these tables are already integrals of Cp over T. So you just take the differences of enthalpy between RT and 1300K. In case the wanted temperatures are not tabulated, interpolate.

Thanks so much, Enthalpy! I appreciate your response.

This is the reaction:
Calculate heat of formation for .6C + .3O2 > .6CO at 1300K
I know that I can get heat of formation for standard states on a table. How do I go about solving heat of formation for nonstandard state reactions? Do I have to use an integral with the Cp values and then do products minus reactants? I am trying to solve for Enthalpy of the reaction, but I am stuck how to calculate heat of formations for nonstandard state reactions. I appreciate any and all help to understand this.
Yea you have to use an integral with the Cp values and then do products minus reactants. If possible calculate the entropy first.