Exercise 17.62 Consider the following reaction: CaCO3( s )CaO( s )+CO2( g ). Est

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Exercise 17.62

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

315 K

129

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Correct

Part B

1090 K

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Part C

1440 K

53.1

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Part D

Predict whether or not the reaction in part A will be spontaneous at 315 K .

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Part E

Predict whether or not the reaction in part B will be spontaneous at 1090 K .

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Part F

Predict whether or not the reaction in part C will be spontaneous at 1440 K .

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Chemistry 106 Summer 2016

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Homework Set #6

Exercise 17.62

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Exercise 17.62

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

315 K

G =

129

  kJ  

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Correct

Part B

1090 K

G =   kJ  

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Part C

1440 K

G =

53.1

  kJ  

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Incorrect; Try Again; 4 attempts remaining

Part D

Predict whether or not the reaction in part A will be spontaneous at 315 K .

Predict whether or not the reaction in part A will be spontaneous at 315  . spontaneous nonspontaneous

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Part E

Predict whether or not the reaction in part B will be spontaneous at 1090 K .

Predict whether or not the reaction in part B will be spontaneous at 1090  . spontaneous nonspontaneous

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Part F

Predict whether or not the reaction in part C will be spontaneous at 1440 K .

Predict whether or not the reaction in part C will be spontaneous at 1440  . spontaneous nonspontaneous

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Explanation / Answer

For the reaction,

dHo = (-635.13 - 393.51) - (-1207.13) = 178.49 kJ

dSo = (38.20 + 213.68) - (88.70) = 163.18 J/K

dGo = dHo - TdSo

Part A: T = 315 K

dGo = 178.49 - 315 x 0.16318 = 127.10 kJ

Part B: T = 1090 K

dGo = 178.49 - 1090 x 0.16318 = 0.6238 kJ

Part C: T = 1440 K

dGo = 178.49 - 1440 x 0.16318 = -56.49 kJ

Part D : At T = 315 K, reaction is non-spontaneous

Part E : At T = 1090 K, reaction is non-spontaneous

Part F : At T = 1440 K, reaction is spontaneous

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