[TA1LGUNN3R]

Hi all-

I have a problem in an example, which seems to have a number which isn't right.  I was wondering if it was a typo of if I'm missing something.  The book is 4th ed. Physical Chemistry by Silbey, Alberty, and Bawendi, section 5.8.

The lesson deals with equilibrium constants in heterogeneous reactions.  Their example reaction is:

CaCO₃(s)=CaO(s)+CO₂(g)

and they go on to say that, for ideal gases, the equilibrium constant expression for this reaction is:

[tex]K=\frac {P_{CO_2}}{P°}[/tex]

Okay, that's fine.  There is an example after this:

Example 5.16, Calculation of reaction properties at high temperatures assuming [tex]\Delta_rC^{°}_{P}[/tex] is constant.

Calculate standard Gibbs energy of formation, standard enthalpy of formation, and standard entropy of formation for CaCO₃(s)=CaO(s)+CO₂(g) at 1000 K using data in the table. 

The table has some temperature and P_CO2/P° values, where at 1000 K P_CO2/P°=3.820.  Their solution is:

[tex]\Delta_{r}G^{°}=-RTlnK=-(8.3145 J K^{-1} mol^{-1})(1000 K)(-3.00)=24.9 kJ mol^{-1}[/tex] 

My question is: where did the -3.00 come from?  If the equilibrium constant is equal to the partial pressure of CO₂ over the standard pressure, shouldn't the lnK factor in the equation be ln(3.28), which does not equal -3.00?

Thanks for any guidance.

-TG

[mjc123]

I suspect their value of 3.82 applies to 1000°C, not 1000K. (Assuming P° = 1 atm)

Look at the table of values in the Wikipedia article http://en.wikipedia.org/wiki/Calcium_carbonate (see under Calcination equilibrium)
It states that the equilibrium CO₂ pressure does not exceed 1 atm until 898°C. The value at 727°C (1000K) is 5.9 kPa, so ln K is -2.84 - approximately, but not exactly, -3.00. Perhaps there's a typo in the table, or different sources give slightly different values.
Check your table again, does it give T in °C and expect you to convert it to K?

[TA1LGUNN3R]

I suspect their value of 3.82 applies to 1000°C, not 1000K. (Assuming P° = 1 atm)

Look at the table of values in the Wikipedia article http://en.wikipedia.org/wiki/Calcium_carbonate (see under Calcination equilibrium)
It states that the equilibrium CO₂ pressure does not exceed 1 atm until 898°C. The value at 727°C (1000K) is 5.9 kPa, so ln K is -2.84 - approximately, but not exactly, -3.00. Perhaps there's a typo in the table, or different sources give slightly different values.
Check your table again, does it give T in °C and expect you to convert it to K?

Ah, right in front of my face.  Yep, that was it, I read it as K for some reason.  Thank you.

-TG