A.) For the decomposition of barium carbonate, consider the following thermodynamic data (Due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.):
ΔH∘rxn | 243.5kJ/mol |
ΔS∘rxn | 172.0J/(mol⋅K) |
Calculate the temperature in kelvins above which this reaction is spontaneous.
B.) The thermodynamic values from part A will be useful as you work through part B
ΔH∘rxn243.5kJ/mol
ΔS∘rxn172.0J/(mol⋅K)
Calculate the equilibrium constant for the following reaction at room temperature, 25 ∘C:
BaCO3(s)→BaO(s)+CO2(g)
(A). Horxn = 243.5 kJ/mol = 243500 J/mol
Sorxn = 172.0 J/(mol⋅K)
For reaction to be spontaneous,
Gorxn < 0
Gorxn = Horxn - T*Sorxn
243500 - T * 172.0 < 0
T * 172.0 > 243500
T > 1415.6 K
Hence, the reaction will be spontaneous above 1415.6 K
(B). BaCO3(s) BaO(s) + CO2(g
At T = 25 oC = 298 K
Gorxn = Horxn - T*Sorxn
= 243500 - 298 * 172
= 192244 J/mol
Gorxn = - RT lnKeq
192244 = - 8.314 * 298 * lnKeq
- 77.59 = lnKeq
Keq = 2.0x10-34
A.) For the decomposition of barium carbonate, consider the following thermodynamic data (Due to variations in thermody...
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