15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T65c,

To determine the free energy change ($$\Delta G$$) for the self-ionization of water, we can use the relation between the equilibrium constant ($$K_w$$) and the Gibbs free energy change. The formula is given by:

$$\Delta G = -RT \ln K_w$$

where:

$$R$$ is the universal gas constant, which is approximately 1.987 cal mol$$^{-1}$$ K$$^{-1}$$.
$$T$$ is the temperature in Kelvin. At 25$$^\circ$$C, the temperature is 298 K.
$$K_w$$ is the ionic product of water, given as 10$$^{-14}$$.

First, convert the given units of $$R$$ to kcal mol$$^{-1}$$ K$$^{-1}$$:

$$R = 1.987 \, \text{cal mol}^{-1} \, \text{K}^{-1} \times \frac{1 \, \text{kcal}}{1000 \, \text{cal}} = 0.001987 \, \text{kcal mol}^{-1} \, \text{K}^{-1}$$

Now, substitute the values into the Gibbs free energy equation:

$$\Delta G = - (0.001987 \, \text{kcal mol}^{-1} \, \text{K}^{-1} ) \times 298 \, \text{K} \times \ln(10^{-14})$$

Calculate the natural logarithm of $$10^{-14}$$:

$$\ln(10^{-14}) = -14 \ln(10) \approx -14 \times 2.303 = -32.242$$

Substitute this value back into the equation:

$$\Delta G = - (0.001987 \, \text{kcal mol}^{-1} \, \text{K}^{-1} ) \times 298 \, \text{K} \times (-32.242)$$

Now perform the multiplication:

$$\Delta G = 0.001987 \times 298 \times 32.242$$

$$\Delta G \approx 19.1 \, \text{kcal mol}^{-1}$$

Therefore, the free energy change for the self-ionization of water at 25$$^\circ$$C is close to 1

Ionic Equilibrum

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