Use the data in the question above to determine at what minimum temperature will

Question

Use the data in the question above to determine at what minimum temperature will the reaction become spontaneous under standard –state conditions?

-Calcium carbonate can be converted to quick lime by the following reaction:

CaCO3(s) --> CaO(s) + CO2(g)

Given that:        for CaCO3(s) at 25°C     DHf° = -1206.9 kJ/mol S° = 92.9 J/mol K,

                          for CaO(s)                      DHf° = -635.5 kJ/mol    S° = 39.8 J/mol K

                          for CO2(g))                     DHf° = -393.5 kJ/mol    S° = 213.6 J/mol K

Use these data to calculate DGo for the reaction.

DG was calculated to be 130.1 kJ/mole

       1109 K

       1.5 K   

       298 K   

      1518 K  

      398 K   

       1109 K

       1.5 K   

       298 K   

      1518 K  

      398 K   

Explanation / Answer

Calculating DH and DS we have:

DH = 177.9 kJ/mol

DS = 160.5 J/mol K

I already calculated DG under these conditions and the value obtained was 130.07 kJ/mol (near the one you obtained)

The condition for dG be spontaneus, means that dG must be <0, so the minimun T to begin at this, would be when the system is in equilibrium, ergo, dG = 0. Determine the T when dG = 0 and then, just sum one degree and ou will know the temperature which dG <0 and becomes spontaneus:

dG = dH - TDS

dH = TDS

T = DH/DS

T = 177900 / 160.5

T = 1108.41 K

Therefore, if you raise one degree, the system will become spontaneus, so, option A is the correct option.

Hope this helps.

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