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Using values of delta H degree and S degree, calculate delta G degree for each o

ID: 1073197 • Letter: U

Question

Using values of delta H degree and S degree, calculate delta G degree for each of the following reaction at 25 degree C Which of these reactions is predicted to be product favored at equilibrium? Are the reactions enthalpy or entropy ? The standard free energy change delta G degree for the formation of O_3(g) from O_2 is + 163.2 kJ/mol at 25 degree C, calculate k_p at this temperature for the equilibrium Comment on the sign of delta G degree and the magnitude of k_p. Calculate delta G degree at 25 degree for the formation of 1.00 mol of C_2H_6(g) from C_2H_4(g) and H_2(g). Use this value to calculate k_p for the equilibrium. Comment on the sign of delta G and the magnitude of k_p. The standard free energy change delta G degree for the formation of MO(g) from its element + 86.58 kJ/mol at 25 degree C, calculate k_p at this temperature for the equilibrium Comment on the sign of delta G degree and the magnitude of k_p.

Explanation / Answer

1)

a) for this reaction,

dHo = (-468.15) - (2 x -285.8) = 103.45 kJ/mol

dSo = (52.3 + 130.5) - (51 + 2 x 70) = -8.2 J/K.mol

dGo = dHo - TdSo = 105.90 kJ/mol

The reaction is entropy gavorued at equilibrium

b) for this reaction

dHo = 49.2 kJ/mol

dSo = 173.3 - (6 x 5.9 + 3 x 130.5) = -253.6 J/K.mol

dGo = 124.77 kJ/mol

The reaction is enthalpy favoured at equilirium

28) dGo = 163.2 kJ/mol

dGo = -RTlnKp

163200 = -8.314 x 298 lnKp

Kp = 2.5 x 10^-29

So, dGo is +ve that is reaction non-sponatenous with very small value of equilibrium constant Kp

29) C2H4(g) + H2(g) --> C2H6(g)

dGo = (-31.95) - (68.43) = -100.38 kJ/mol

dGo = -RTlnKp

-100380 = -8.314 x 298 lnKp

Kp = 3.92 x 10^17

So dGo is -ve, the reaction is sponatenous as written and the equilibrium constant Kp value for the reaction is very high.

27) dGo = 86.58 kJ/mol

dGo = -RTlnKp

86580 = -8.314 x 298 lnKp

Kp = 6.68 x 10^-16

So, dGo is +ve, that is reaction is non-spontaneous as written and equilibrium constant is very small, is reactant favoured