Question

The standard free energy change for the formation of two moles of h2o(l) in a strong acid–strong base neutralization reaction at 25°c is -79.9kj. calculate the equilibrium constant for the reaction. see equation 11.1. h3o+(aq) + oh-(aq) = 2 h2o (l)

Answer 1

The equation is given as

`H_(3)O^(+) (aq)+ OH^(-) (aq)\rightarrow 2H_(2)O (l)`

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.

`\bigtriangleup G^(o)= - R* T* ln (Keq)`

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.

`-79900 J = - ( 8.314J/mol) * ( 298K) * ( lnK_(eq))`

`79900 = 2477.6 * (lnK_(eq))`

`ln K_(eq) = (79900)/(2477.6)`

`ln K_(eq) = 32.25`

`K_(eq) = e^(32.25)`

`K_(eq) = 1.0 * 10^(14)`

The equilibrium constant for the given reaction is 1.0 x 10¹⁴