Question

Calculate the standard free-energy change and the equilibrium constant Kc for the following reaction at 25°C. See Appendix C for data. Fe(s) + Cu2+ (aq) = Fe2+ (aq) + Cu(s) Find equilibrium constan + K at 25°C

Answer 1

We have relation, $\Delta$G ⁰ reaction = - 2.303 R T log K  

Where, $\Delta$G ⁰ is standard Gibbs free energy change of reaction and K is equilibrium constant of reaction.

To calculate K first calculate $\Delta$G ⁰ reaction

Consider reaction, Fe (s) + Cu ²⁺ (aq) $\rightleftharpoons$ Fe ²⁺ (aq) + Cu (S)

For above reaction, G ⁰ reaction = sum of G ⁰ (products) - sum of G ⁰ (reactants)

G ⁰ reaction = [ G ⁰_f Fe ²⁺ (aq) + G ⁰_f Cu (S) ] - [ G ⁰_f Cu ²⁺ (aq) + G ⁰_f Fe (S) ]

= [ - 78.87 kJ / mol + 0 kJ /mol ] - [ 65.52 kJ / mol + 0 kJ / mol ]

= - 144.39 kJ / mol

= - 144390 J / mol

We have, $\Delta$G ⁰ reaction = - 2.303 R T log K  

$\therefore$ - 144390 J / mol = - 2.303 $\times$ 8.314 J / K mol $\times$ 298 K $\times$ log K

log K = - 144390 J / mol / - 2.303 $\times$ 8.314 J / K mol $\times$ 298 K

log K = 25.3

K = 10 ^25.3 = 1.995 $\times$ 10 ²⁵

ANSWER : K = 1.995 $\times$ 10 ²⁵