The equilibrium constant for the reaction, 2 Fe3+(aq) + Hg22+(aq) <=> 2 Fe2+(aq)

Question

The equilibrium constant for the reaction, 2 Fe3+(aq) + Hg22+(aq) <=> 2 Fe2+(aq) + 2 Hg2+(aq) is 9.1 x 10-6(aq) at 298 K. Calculate ?G in kJ when [Fe3+(aq)] = 0.681 [Hg22+(aq)] = 0.0737 [Fe2+(aq)] = 0.0141 [Hg2+(aq)] = 0.0183

Explanation / Answer

?Grxn = ?G°rxn + R·T·ln(Q) ?G°rxn = -R·T·ln(k) ==> ?Grxn = -R·T·ln(K) + R·T·ln(Q) ==> ?Grxn = R·T·{ln(Q) - ln(K)} ==> ?Grxn = R·T·ln(Q/K) ~(eq1) Q = reaction quotient Q is calculated like an equilibrium constant. ----------------------------- To calculate an equilibrium constant: Balanced equation: aA + bB cC + dD Equilibrium constant equation: ......[C]^c x [D]^d K = ----------------- ......[A]^a x [B]^b ----------------------------- Calculating the reaction quotient: 2 Fe3+(aq) + Hg22+(aq) 2 Fe2+(aq) + 2 Hg2+(aq) ......[Fe2+(aq)]^2 x [Hg2+(aq)]^2 Q = ---------------------------------------- ......[Fe3+(aq)]^2 x [Hg22+(aq)] Substituting numbers: ......(0.0895)^2 x (0.0710)^2 Q = -------------------------------- .....,..(0.299)^2 x (0.0206) ==> Q = 0.0219257 Recalling (eq1): ?Grxn = R·T·ln(Q/K) where, R = 8.314 J/(mol·K) T = 298 K Q = 0.0219257 K = 9.1 x 10^-6 Substituting numbers into (eq1) ?Grxn = {8.314 J/(mol·K)}·(298 K)·ln(0.0219257 / 9.1 x 10^-6) = {8.314 J/(mol·K)}·(298 K)·ln(2,409.42) = {8.314 J/(mol·K)}·(298 K)·(7.78714) = 19,293.2 J = 19.2932 kJ *** Answer: 19.29 kJ

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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