Heat required to liberate CO2 with one kg-mole CaCO3 1

[SamCheung]

Please advise me what is the required heat to liberate CO2 with one kg-mole of CaCO3, starting from room temperature of 68F, thanks. I know CO2 will start to liberate @932F.

 

[SeanB]

Q = mcpT

m = mass of CaCO3
cp = heat capacity
T = delta T

 

[SamCheung]

Do you know Cp of CaCO3 at 932F?
Do you know Cp at 68F?
Is average/mean Cp_mean=(Cp_932+Cp_68)/2?

Thanks for the response.

 

[SamCheung]

Another question is any latent heat involved with liberation of CO2 at 932F?

Thanks.

 

[Zoobie]

Sam, you are right to question SeanB's response. No offence to SeanB but I don't think his answer is correct. Liberating CO2 from CaCO3 is not just a heating issue. It involves a chemical reaction (CaCO3 -> CaO + CO2). This is the process that occurs in the lime kiln of a Kraft pulp mill. mCdt gets you from 68 to 932 but there is also a heat of reaction that needs to be taken into account. Lets say that you are already at 932....that would mean that you would need zero heat to liberate CO2...obviously this makes no sense.

Sorry that I don't have some actual numbers to help you out with.

 

[SamCheung]

SeanB,
My apologies if my questions are too direct.

Zoobie,
Thanks for the response.


Sam

 

[Shimasu]

I think you can get the heat of reaction out of the heat of formations, remember the hesse equation... you are going to take the heat of formation of your reactives and your products and the difference will be your heat of reaction, I dont know if it is an exothermical or endothermical reaction but you will know with hesse. you can find heats of formation in the "reidz & Proudnitz" properties of gases and liquids. if you can not find the heat of formation for one of your compounds just make the same thing, look for the heat of formation for their elements and make the difference.

you can make it in two steps, first the heat required to get to 932F, then the heat of reaction for the given conversion, if you know your equilibrium composition use it otherwise you should try to know it.

 

[SeanB]

Well I got half of it right!
Forgot about the reaction half.

 

[SamCheung]

Unfortunately I know nothing about the Hesse Equations. I know that CO2 liberation absorbs heat but to quantify it will need a lot of time to explore these equations and associated equations in mult-dimensional analysis.

Is there a simple way of finding the heat quantity absorbed by CaCO3 to liberate the CO2?

 

[Shimasu]

You will need your heat of formations and your chemical reaction

Zoobie told us

CaCO3 -> CaO + CO2

first you will have to balance the equation, then you will use the heat of formations (usually refered to as DeltaF) multiplied by your stechiometric coeffs (remember reactives are negative, products are possitive) and make the addition, the result of that is your heat of reaction.

 

[kenvlach]

SamCheung,
Do you have a solution yet?
This is a very basic equilibrium CaCO₃ = CaO + CO₂(g)
It isn't quite correct to say that CO₂ is liberated at a certain temperature; instead, the equilibrium CO₂ pressure is a function of temperature. Write each [Δ]G_f of each compound as a function of temperature. Of course, [Δ]G_f = [Δ]H_f – T [Δ]S_f.
From [Δ]G_Rxn = [Δ]G_{f, CaO} + [Δ]G_{f, CO2} - [Δ]G_{f, CaCO3}, get [Δ]G_Rxn = [Δ]H_Rxn – T [Δ]S_Rxn.

This [Δ]H_Rxn may be sufficient to answer your question. For a little more understanding,
[Δ]G_Rxn = - RT ln P_CO2
so
P_CO2 = exp [-[Δ]G_Rxn/RT]

and at P_CO2 = 1 atm,* [Δ]G_Rxn = 0

*1 bar if that's the thermochemical standard state being used.

Gaskell's book Introduction to Metallurgical Thermodynamics (most recent edition is titled Introduction to the Thermodynamics of Materials) is pretty thorough at explaining condensed phase-gaseous equilibria (including oxide-carbonate equilibria).

Thermodynamic Properties of Minerals and Related Substances at 298.15 K and 1 Bar (10 Pascals Pressure and at Higher Temperatures) (US Geological Survey Bulletin 1452) by R. A. Robie et al. has all the necessary data.
Note that CaCO₃ has more than one crystal structure; use the data for calcite.

 

[SamCheung]

kenvlach,

Thanks for the reply. Without going to the books you mentioned, I cannot comprehend the equations with lots of unknowns.

 

[kenvlach]

The references mentioned above are convenient but certainly not necessary.

The thermodynamic data are readily available, e.g., see the tables
'Standard Thermodynamic Properties of Chemical Substances' in CRC Handbook of Chemistry & Physics or
'Heats and Free Energies of Formation of Inorganic Compounds' in Perry's Chemical Engineers' Handbook.
Note that the CRC book gives heats of formation in kJ/mol while Perry's uses kcal/mol (& these are gram moles).

 

[SamCheung]

Thanks. I will try to get my hands on these handbooks.

 

[kenvlach]

This being a very well-known reaction, you can find much of the information in Wikipedia:

http://en.wikipedia.org/wiki/Calcination

http://en.wikipedia.org/wiki/Calcium_carbonate#Calcination_Equilibrium

Besides the heat of reaction (which can be taken from above articles), the other factor needed is the heat capacity C_p of CaCO₃, which from Robie et al. is

C_p = 99.715 + 0.026920 T - 2.1576x10^-6 T^-2, J/mole/^oK

To heat CaCO₃ from T₁ to T₂ , integrate: [∫]C_p dT.
Of course, T is in ^oK.

For the reaction to go rapidly, heat to T₂ = 1171 ^oK, which is where the equilibrium P_CO2 = 101 kPa (1 atm).

 

[SamCheung]

This is the answer I am looking for. You solved my problem, thanks. For Cp, I can simply ignore the third term (right or wrong, its insignificant in real world).

 

[kenvlach]

The 3rd term isn't quite insignificant; I mistyped it, so ~3% error. Should be

C_p(CaCO₃) = 99.715 + 0.026920 T - 2.1576x10⁶ T^-2, J/mole/^oK
valid for T = 298 to 1200 K,
from Thermodynamic Properties of Minerals..., Robie et al. (1979).