Question

Calculate the standard change in Gibbs free energy, AGn, for the following reaction at 25.0 °C. Standard Gibbs free energy of formation values can be found here. rxn: KCI K (aq)+ CI (ag) Number kJ/ mol Then, determine the concentration of K*(aq) if the change in Gibbs free energy, AGn, for the reaction is -8.45 kJ/mol. Number M

i need the K+ value this question was submitted already but the second answer i got was incorrect.

Answer 1

KCl(s) -------------> K^+ (aq) + Cl^- (aq)     $\Delta$G⁰rxn = -8.45KJ/mole

$\Delta$G⁰rxn = -8.45KJ/mole    = -8450J/mole

T = 25+273 = 298K

$\Delta$G⁰    = -RTlnKc

-8450   = -8.314*298lnKc

lnKc     = -8450/(-8.314*298)

lnK    = 3.41

Kc        = 30.265

Kc      = 30.265

KCl(s) -------------> K^+ (aq) + Cl^- (aq)

at equilibrium [K^+] = [Cl^-] = x

Kc    = [K^+][Cl^-]

30.265   = x*x

30.265   = x^2

x    = 5.5

[K^+] = [Cl^-] = x = 5.5M

[K^+]   = 5.5M >>>>answer