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WBJEE 2024 Physics & Chemistry Paper

Question : 15 of 80

Marks: +1, -0

25^{\circ}\,\mathrm{C}

10^{-14}

{\mathrm{kCal}}^{-1}\,{\mathrm{mol}}^{-1}

is close to

Solution:

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 = - R T \ln K_w

where:

R

is the universal gas constant, which is approximately

1.987\,\mathrm{cal}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}

T

is the temperature in Kelvin. At

25^{\circ}\,\mathrm{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

{\mathrm{kcal}}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}

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

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

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

Now perform the multiplication:

\Delta G = 0.001987 \times 298 \times 32.242

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

Therefore, the free energy change for the self-ionization of water at

25^{\circ}\,\mathrm{C}

is close to

19.1\,\mathrm{kCal}\,\mathrm{mol}^{-1}

The correct answer is:

Option C: 19.1

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