1)Calculate the free energy %u0394G (in kJ/mol) for a reaction 4A + 3 B %u21CC 1

ID: 783691 • Letter: 1

Question

1)Calculate the free energy %u0394G (in kJ/mol) for a reaction 4A + 3 B %u21CC 1 C + 3 D

knowing that the concentrations are: [A] = 0.78 M, [B] = 1.49 M, [C] = 0.5 M, [D] = 1.38 M.

The standard free energy and the temperature of the reaction are respectively

DeltaG=?

2)Calculate the free energy %u0394G (in kJ/mol) for a reaction 4A + 1 B %u21CC 2 C + 2 D

knowing that the concentrations are: [A] = 0.43 M, [B] = 1.04 M, [C] = 1.64 M, [D] = 1.19 M.

The equilibrium constant, Kc , is 6.42e+1 , and T = -17 degree C.

DeltaG=?

(A) %u0394G = %u0394G0 + RT ln C^c . D^d/A^a.B^b

where R = 8.3 J mol-1 K-1 and T = 248 Kelvin

= 95.3 + 8.3*248*ln(1.073)

= 95.3 + 8.3*248*0.07

= 239.38 Kj/mol

(B) At equilibrium, DeltaG = 0 and Q = K

But DeltaG = DeltaG0 + RT ln Q

Hence DeltaG0 = - RT ln K

= -8.314*256*ln 6.42e+1

=-3937.5 Kj/mol

So, %u0394G = -3937.5 + 8.314*256*ln(107.12) [ applying the same formula as above]

= -3937.5+ 9939.5

= 6002.05 kj/mol

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.