Question

What is the standard free energy of the reaction, in kJ, at 298K for:

1 A (aq) + 1 B (aq) <--> 2 C (aq) + 2 D (aq)

if   ΔH° (A)= -3.4134 kJ/mol , ΔH° (B) = 93.8823 kJ/mol, ΔH° (C) = -26.8495 kJ/mol, and ΔH° (D) = -39.4984 kJ/mol

ΔS° (A)= 33.3845 J/(mol K) , ΔS° (B) = 92.5722 J/(mol K) , ΔS° (C) = -59.1781 J/(mol K) , and ΔS° (D) = -12.7602 J/(mol K)

Answer 1

Answer -

Given,

ΔH° (A)= -3.4134 kJ/mol

ΔH° (B) = 93.8823 kJ/mol

ΔH° (C) = -26.8495 kJ/mol

ΔH° (D) = -39.4984 kJ/mol

ΔS° (A)= 33.3845 J/(mol K)

ΔS° (B) = 92.5722 J/(mol K)

ΔS° (C) = -59.1781 J/(mol K)

ΔS° (D) = -12.7602 J/(mol K)

Temperature = 298 K

ΔG°rxn = ?

1 A (aq) + 1 B (aq) 2 C (aq) + 2 D (aq)

We know that,

ΔH°rxn = np * ΔH° (products) - nr * ΔH° (reactants)

ΔH°rxn =  2*ΔH°(C) + 2*ΔH°(D) - 1*ΔH°(A) - 1*ΔH°(B)

ΔH°rxn =  2*(-26.8495 kJ/mol) + 2*(-39.4984 kJ/mol) - 1*(-3.4134 kJ/mol) - 1*(93.8823 kJ/mol)

ΔH°rxn = -2*26.8495 kJ/mol - 2*39.4984 kJ/mol) + 3.4134 kJ/mol) - 93.8823 kJ/mol)

ΔH°rxn = - 223.1647 kJ/mol)

Also,

ΔS°rxn = np * ΔS° (products) - nr * ΔS° (reactants)

ΔS°rxn =  2*(-59.1781 J/(mol K)) + 2*(-12.7602 J/(mol K)) - 1*33.3845 J/(mol K) - 1*92.5722 J/(mol K)

ΔS°rxn = - 269.8333 J/(mol K) or 0.269833 kJ/(molK)

Now,

Go = Ho - TSo

So,

Go = - 223.1647 kJ/mol) - 298 K * (- 0.269833 kJ/(mol K))

Go = -142.75 kJ/mol [ANSWER]