Consider the following reaction: CaCO3( s )CaO( s )+CO2( g ). Estimate G for thi

Question

Consider the following reaction:
CaCO3(s)CaO(s)+CO2(g).
Estimate G for this reaction at each of the following temperatures. (Assume that H and S do not change too much within the given temperature range.)

Part A

320 K

Part B

1095 K

Part C

1445 K

Part D

Predict whether or not the reaction in part A will be spontaneous at 320 K .

Part E

Predict whether or not the reaction in part B will be spontaneous at 1095 K .

Part F

Predict whether or not the reaction in part C will be spontaneous at 1445 K .

Consider the following reaction:
CaCO3(s)CaO(s)+CO2(g).
Estimate G for this reaction at each of the following temperatures. (Assume that H and S do not change too much within the given temperature range.)

Part A

320 K

G =   kJ  

Part B

1095 K

G =   kJ  

Part C

1445 K

G =   kJ  

Part D

Predict whether or not the reaction in part A will be spontaneous at 320 K .

-spontaneous -nonspontaneous

Part E

Predict whether or not the reaction in part B will be spontaneous at 1095 K .

-spontaneous -nonspontaneous

Part F

Predict whether or not the reaction in part C will be spontaneous at 1445 K .

-spontaneous -nonspontaneous

Explanation / Answer

deltaHf (CaCO3) = -1207 KJ/mol
deltaHf (CaO) = -635.09 KJ/mol
deltaHf (CO2) = -393.51 KJ/mol
delta Hrxn = deltaHf(CO2) + deltaHf(CaO) - deltaHf(CaCO3)
= -393.51 -635.09 + 1207
= 178.4 KJ/mol
= 178400 J/mol

deltaSf(CaCO3) = 93 J/molK
deltaSf (CaO) = 39.75 J/molK
deltaSf (CO2) = 213.74 J/molK

delta Srxn = deltaSf(CO2) + deltaSf(CaO) - deltaSf(CaCO3)
= 213.74 + 39.75 -93
= 160.49 J/molK

part A) G0 = H0 - TS0

  G0 = 178400 - 320 x 160.49 = 127043.2 J/mol

Part D)  G0 is positive therefore the reaction is nonspontaneous.

Part B) G0 = 178400 - 1095 x 160.49 = 2663.45 J/mol

Part E) G0 is positive therefore the reaction is nonspontaneous.

Part C) G0 = 178400 - 1445 x 160.49 = -54100.5 J/mol

Part F) G0 is negative therefore the reaction is spontaneous.

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