Question

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)

Answer 1

(A). $\Delta$H^o_rxn = 243.5 kJ/mol = 243500 J/mol

$\Delta$S^o_rxn = 172.0 J/(mol⋅K)

For reaction to be spontaneous,

$\Delta$G^o_rxn < 0

$\Delta$G^o_rxn =  $\Delta$H^o_rxn - T*$\Delta$S^o_rxn

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) $\rightarrow$ BaO(s) + CO2(g

At T = 25 oC = 298 K

$\Delta$G^o_rxn =  $\Delta$H^o_rxn - T*$\Delta$S^o_rxn

= 243500 - 298 * 172

= 192244 J/mol

$\Delta$G^o_rxn = - RT lnKeq

192244 = - 8.314 * 298 * lnKeq

- 77.59 = lnKeq

Keq = 2.0x10^-34