To determine the free energy change
(∆G) for the self-ionization of water, we can use the relation between the equilibrium constant (
Kw ) and the Gibbs free energy change. The formula is given by:
∆G=−RTlnKwwhere:
R is the universal gas constant, which is approximately
1.987calmol−1K−1.
T is the temperature in Kelvin. At
25∘C, the temperature is 298 K .
Kw is the ionic product of water, given as
10−14.
First, convert the given units of
R to
kcalmol−1K−1.
Now, substitute the values into the Gibbs free energy equation:
∆G=−(0.001987kcalmol−1K−1)×298K×ln(10−14)
Calculate the natural logarithm of
10−14.
ln(10−14)=−14ln(10)≈−14×2.303=−32.242
Substitute this value back into the equation:
∆G=−(0.001987kcalmol−1K−1)×298K×(−32.242)
Now perform the multiplication:
∆G=0.001987×298×32.242∆G≈19.1kcalmol−1 Therefore, the free energy change for the self-ionization of water at
25∘C is close to
19.1kCalmol−1.
The correct answer is:
Option C: 19.1