A reaction A(aq) + B(aq) C(aq) has a standard free-energy change of-4.35 kJ/mol

Question

A reaction A(aq) + B(aq) C(aq) has a standard free-energy change of-4.35 kJ/mol at 25 degree C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively? How would your answers above change if the reaction had a standard free-energy change of +4.35 kJ/mol? All concentrations would be higher. There would be less A and B but more C. There would be no change to the answers. All concentrations would be lower. There would be more A and B but less C.

Explanation / Answer

Dg0 = -RTlnk

-4.35*10^3 = -8.314*298lnk

k = 5.79

k = [C]/[A][B]

5.79 = x/((0.3-x)(0.4-x))

x = 0.171 M

[C] at equilibrium = 0.171 M

[B] = 0.3-0.171 = 0.129 M

[A]= 0.4-0.171 = 0.229 M

there would be more Aand B but less C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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