Question

part 1 answer is correct. Looking for #2 on this sheet with work please

Verarbon dia Gibbs Free Energy & Predicting Spontaneity 1) The decomposition of calcium carbonate into calcium oxide and carbon dioxide (shown below) is not spontaneous at 298 K. CaCO (8) Cao(8) + Co.g) AH = +179 28 Using the data provided above, calculate the value of Great 298 K. 10- +160.29 179.2151005 AGO - DHOTOS I min hof amor AG° = 179200 Frei- 298(160.23imo) DGO=131460.45 sco=131.4605lmo I 2) Using the data provided in problem 1), determine the temperature at which this entropy- driven reaction becomes spontaneous. You must indicate whether the reaction is spontaneous at temperatures greater than or less than your calculated temperature. ACO- AHO- TAS spontaneous @ higher temps - 131.46ong = 179.2 - ED 0:1602 131460.4 = 179200 - (T160.2) 131460= +179.2 - 4A160.2 - اور t=29814 =819 Page 3 of 8 - 821.72

Answer 1

Solutions of above question is answered by using Equation of Gibbs free energy. In second question for reaction being spontaneous the standard Gibbs free energy should be negative.

D I Given (1) CaCO3(s) CaO(s) + CO₂(g) AHgxn = + 179-2 kJ/mol 45xxn = +160-2 Jl molok T = 298k By using the equation, AGO = AH - TASAN sp= 179-2 x 103 m. - (298x1600) mai = 131460.4 I/mol = 131.46 kJ/mol For entropy of reaction to being spontaneous Alico Be a ton TAS8xn < 0 - Allsyn + T AS on > > T Atlan => T> (179- 2 x 10 ik As pxn 1 160.2 Ty the TY 1118.60k For reaction to be spontaneous the temperature should be greates then 1118.60k