The latent heat of melting of ice at 0^{circ} | vedclass

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(A) The entropy change $\Delta S$ for a phase transition is given by the formula $\Delta S = \frac{\Delta H}{T}$.Given,latent heat of fusion $\Delta H

Vedclass · Accepted Answer

$22$

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(A) The entropy change $\Delta S$ for a phase transition is given by the formula $\Delta S = \frac{\Delta H}{T}$.Given,latent heat of fusion $\Delta H = 6 \, kJ \, mol^{-1} = 6000 \, J \, mol^{-1}$.The temperature of melting of ice at $1 \, atm$ is $T = 0^{\circ} C = 273 \, K$.Substituting the values: $\Delta S = \frac{6000 \, J \, mol^{-1}}{273 \, K} \approx 21.97 \, J \, K^{-1} \, mol^{-1}$.Rounding to the nearest integer,we get $22 \, J \, K^{-1} \, mol^{-1}$.Thus,the correct option is $A$.

The latent heat of melting of ice at $0^{\circ} C$ is $6 \, kJ \, mol^{-1}$. The entropy change during the melting in $J \, K^{-1} \, mol^{-1}$ is closest to

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