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.):

Calculate the temperature in kelvins above which this reaction is spontaneous.

Express your answer to four significant figures and include the appropriate units.

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Relationship between free energy and the equilibrium constant

The standard free energy change, G , and the equilibrium constant K for a reaction can be related by the following equation:

G=RTlnK

where T is the Kelvin temperature and R is equal to 8.314 J/(molK) .

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

Question 1.

Find T for spontaneous process

so

dG < 0, in order to be spontaneous

and

dG = dH - T*dS

so

dH - T*dS < 0

dH < T*dS

dH/dS < T

243.5 kJ/mol / 172 J/molK < T

243500/172 K < T

T> 1415.6976 K

B)

solve for K

so

dG = -RT*ln(K)

dG = dH - T*dS = 243500 - 298 * 172 = 192244 J/mol = 192.244 kJ/mol

dG = -RT*ln(K)

192244 = -8.314*298*ln(K)

K = exp(-192244 /(8.314*298))

K =2.002*10^-34

clearly, not favoured at T= 298K

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