The equation is given as
[tex] H_{3}O^{+} (aq)+ OH^{-} (aq)\rightarrow 2H_{2}O (l) [/tex]
We have been given the standard free energy , ΔG° = -79.9 kJ
The relationship between standard free energy, ΔG° and equilibrium constant K is given by the following equation.
[tex] \bigtriangleup G^{o}= - R\times T\times ln (Keq) [/tex]
Here ΔG° is standard free energy in J
We have,
ΔG° = -79.9 kJ = -79900 J
T = Temperature in Kelvin = 25 + 273 = 298 K
R = Gas Constant = 8.314 J/mol K
Let us plug in these values in above equation.
[tex] -79900 J = - ( 8.314J/mol) * ( 298K) * ( lnK_{eq}) [/tex]
[tex] 79900 = 2477.6 * (lnK_{eq}) [/tex]
[tex] ln K_{eq} = \frac{79900}{2477.6} [/tex]
[tex] ln K_{eq} = 32.25 [/tex]
[tex] K_{eq} = e^{32.25} [/tex]
[tex] K_{eq} = 1.0 * 10^{14} [/tex]
The equilibrium constant for the given reaction is 1.0 x 10¹⁴
