From the data provided in the table for 298 K and standard pressure of 1 bar cal

Question

From the data provided in the table for 298 K and standard pressure of 1 bar calculate Delta H degree_298, Delta S degree_298, Delta G degree_298 for the reaction: CIF_3(g) rightarrow CIF(g) + F_2(g) The standard molar Gibbs free energy of formation (Delta G degree_fm) of CIF_3(g) at 298 K is -118.9 kJ/mol. What is Delta G degree_f, m for CIF(g) at 298 K? Assuming the given heat capacities are independent of temperature and assuming each of the gases behaves ideally calculate Delta G_500 for the reaction at 500 K and 10 bar.

Explanation / Answer

a)

dG = dH - T*dS

dH = -50.3 + 0 - (-163.2) = 112.9 kJ/mol

dS = (217.9 + 202.8) - (281.6) = 139.1 J/mol

dG = dH - T*dS = 112.9 - 298 * 139.1 /1000 = 71.448 kJ/mol

dG = Gprod - Greact

dG = (ClF + F2) - ClF3

71.4482 = ClF + 0 - (-118.9)

ClF = 71.4482 -118.9

ClF = -47.4518 kJ/mol

c)

Calculate dG:

dG = dG° -RT*ln(K)

solve initial K:

dG° = RT*ln(K) = (8.314)(298)*ln(K) = 71.4482 *1000

K = 71.4482 *1000 / (8.314*298) = 28.837

assume this is constant

so

dG = dG° -RT*ln(K)

dG = 71.4482*1000 -8.314*500ln(28.837)

dG = 57473.782 J/mol= 57.47 kJ/mol

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