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)

The equation is given as

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.

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.

The equilibrium constant for the given reaction is 1.0 x 10¹⁴

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the formula for the conversion of temperature from degree celsius to kelvin is:

the sample of water at is converted to kelvin temperature by using the above formula as:

on comparing the two samples of water at or we can say that at and at respectively. the difference between the temperatures of two samples of water is:

hence, one can say that the sample of water at is greater than the sample of water at by .