Part A For the decomposition of barium carbonate, consider the following thermod

Question

Part 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.): Hrxn 243.5kJ/mol Srxn 172.0J/(molK) Calculate the temperature in kelvins above which this reaction is spontaneous. Express your answer to four significant figures and include the appropriate units.

Part B

The thermodynamic values from part A will be useful as you work through part B:

Calculate the equilibrium constant for the following reaction at room temperature, 25 C:

BaCO3(s)BaO(s)+CO2(g)

Express your answer numerically to three significant figures.

Explanation / Answer

Part A :

Hrxn = 243.5kJ/mol

Srxn = 172.0J/(molK) = 0.172 kJ/mol.K

G = H - T S

if the reaction is sponteneous the G should be less than zero or negative.

G < H - T S

0 < H - T S

H < T S

243.5 < T x 0.172

1416 K < T

the reaction is spontenous greater than 1416 K.

Part B )

G = H - T S

        = 243.5 - 298 x 0.172

        = 192.2 kJ/mol

G = - R T ln Keq

192.2 = - 8.314 x 10^-3 x 298 x ln Keq

equilibrium constant Keq = 2.05 x 10^-34

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