December 29, 2024, 10:05:53 AM
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Topic: How does enthalpy of formation differ for conditions other than Standard?  (Read 4998 times)

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Offline Flyndaran

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This is my first time on this site, so I apologize if I'm asking in the wrong section.

I have had very little formal education in chemistry, so my questions are likely to wobble between advanced and high school level.

Everything I can find assumes standard conditions.  But how are they likely to change for higher/lower pressures and temperatures that don't alter the state?

Offline Arkcon

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It really would depend on what you're talking about, so a specific example would be helpful.  For example, for reactions involving gasses, pressure changes are important.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Flyndaran

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Well, I'm coming up with the biochemistry for an alien world for fun.  But with all the standard conditions assumed, I don't know if my reactions would work there the way they would here.

Their world's evolved biome is 7 atmospheres pressure, averaging 170 degrees Fahrenheit, low humidity mega-desert.  I don't know if those conditions are different enough to mess with the equations.

Specifically, their anaerobic metabolisms for C6H12O6 = 3CH4 + 3CO2.

I'm still trying to work out how they metabolize acetic acid / acetate.

Thank you for helping.

Offline Big-Daddy

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They definitely change but AFAIK the change is usually applied to the enthalpy/internal energy/entropy/Gibbs' free energy change of the reaction as opposed to all of the individual formation or molar (or combustion etc.) values. For temperature, this is done by using heat capacities and integrating over the desired temperature range for the change (from 298.15K to your desired temperature). Don't know about pressure.

Offline Flyndaran

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Any idea where I would go about looking for that information and instructions on how to do that?  Weird that standard state is so warm at 77 degrees F.

Offline Big-Daddy

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By the way, I feel you'll have to start working in K rather than F.

Sure - look up "enthalpy temperature dependence", "heat capacity integration", anything like that. But first make sure you know the basics of writing enthalpy changes for reactions from formation or molar enthalpy values (and know that the same applies for all other state variables - U, S, G, etc.).

If anyone else knows how to deal with the pressure aspect, I'm sure I'd like to know as much as you!

Offline Flyndaran

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Okay, so I've got some examples to work with.  But Integrating heat capacity uses a, b, c, and d.  I have no idea what those variables mean.

Integrating heat capacity over the temperature change from 298 k to my alien world of 350 k is (a +bt + ct^-2 + dt^0.5)

I'm feeling that the answer is so obvious that that's why it's not defined on any pages giving examples.

Offline Flyndaran

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Heat capacity of CO2 at 298 K = 0.90 kJ/kg mol;  at 350 k = 0.84 kJ/kg mol
CH4 at 298 k = 2.23 kJ/kg mol; at 350 K = 2.37 kJ/kg mol
C6H12O6 at 298 k = 1.21 kJ/kg mol; at 350 K = 1.42 kJ/kg mol

Offline Big-Daddy

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They are coefficients in the same way as ax^n+bx^(n-1)+...+k are coefficients in a polynomial. They can take any numerical value, that's what's meant by using a variable. Each compound will have a different polynomial representing its heat capacity; any polynomial heat capacity except the theoretical infinite-length power series is an approximation and what you've just quoted are the crudest result of all: polynomials taking every coefficient to be 0 except the last one, hence the heat capacity itself has no temperature dependence.

Offline Flyndaran

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So the variables stay in the equation until the end?  Do you know if they disappear ever?

Offline Flyndaran

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C \int^{298}_{350} /Delta CpdT  :Testing to see if this works.

Offline Flyndaran

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I want to find the Heat of formation and eventually gibbs free energy of a reaction.  To do that I have to Integrate delta Cpdt over the range of 298 K to 350 K for each of the chemicals.

Even though I want to learn how to do this, I hope there's an easier way if I have the specific heats of these chemicals for 298 K and 350 K

Offline Flyndaran

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Calculating fH° at Temperatures Other Than 298 K
...
We do this by calculating rCP for the reaction that forms the mineral of interest: •rCP = rCPproducts - rCPreactants
and then integrating. Thus, for example if we want to know fH° for quartz at 1000 K, we get coefficients for the heat capacities of Si, O2 and SiO2:

    a         b                c              d
Si 31.778 5.3878E-4 -1.4654E5 -1.7864E2
O2 48.318 -6.9132E-4 4.9923E5 -4.2066E2
SiO2 104.35 6.07E-3 3.4E-4 -1070

for the reaction •Si + O2 = SiO2
and we calculate
a = 24.254
b = 6.2225E-3
c = -3.5E5
d = -470.7

I cannot find where these numbers come from as they aren't included on any site of Google for any of the chemicals I want.  Please help.  I'm almost there.

Offline Flyndaran

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I forgot to mention that I copy pasted an example of the equation I want to solve using three chemicals other than the ones I want.

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