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Current Question (ID: 8330)
Question:
$\text{The entropy change in the fusion of one mole of a solid melting at } 27^{\circ} \text{C is:}$
$(\text{the latent heat of fusion is } 2930 \text{ J mol}^{-1})$
Options:
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1. $9.77 \text{ JK}^{-1}\text{mol}^{-1}$
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2. $19.73 \text{ JK}^{-1}\text{mol}^{-1}$
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3. $2930 \text{ JK}^{-1}\text{mol}^{-1}$
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4. $108.5 \text{ JK}^{-1}\text{mol}^{-1}$
Solution:
$\text{Hint: } \Delta\text{S} = \frac{\Delta\text{H}_f}{\text{T}}$
$\text{Step 1: Identify the known values}$
$\text{Given:}$
$\text{- Latent heat of fusion (}\Delta\text{H}_f) = 2930 \text{ J mol}^{-1}$
$\text{- Temperature (T) = } 27^{\circ}\text{C}$
$\text{Step 2: Convert the temperature to Kelvin}$
$\text{Temperature in Kelvin = Temperature in } ^{\circ}\text{C } + 273.15$
$\text{T = } 27 + 273.15 = 300.15 \text{ K}$
$\text{For simplicity, we can approximate this to 300 K, which appears to be what was done in the solution image.}$
$\text{Step 3: Calculate the entropy change using the formula}$
$\text{The entropy change during fusion (}\Delta\text{S}_f\text{) can be calculated using:}$
$\Delta\text{S}_f = \frac{\Delta\text{H}_f}{\text{T}}$
$\text{Where:}$
$\Delta\text{S}_f = \text{entropy change of fusion}$
$\Delta\text{H}_f = \text{enthalpy change of fusion (latent heat of fusion)}$
$\text{T = temperature in Kelvin}$
$\text{Substituting the values:}$
$\Delta\text{S}_f = \frac{2930 \text{ J mol}^{-1}}{300 \text{ K}} = 9.77 \text{ J mol}^{-1} \text{ K}^{-1}$
$\text{Therefore, the entropy change in the fusion of one mole of the solid is } 9.77 \text{ J K}^{-1} \text{mol}^{-1}$
$\text{This makes physical sense because fusion (melting) is a process where the solid changes to a liquid, resulting in increased disorder and hence a positive entropy change.}$
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