QuestionJune 17, 2025

1. Calculate the change in entropy in J/K for A. the vaporization of 50.0 g of benzene (C_(6)H_(6)) at its boiling point of 80^circ C.Delta H_(vap)=31.0kJ/mol B. the melting of 50.0 g of benzene (C_(6)H_(6)) at its freezing point of 5.5^circ C.Delta H_(fus)=9.92kJ/mol

1. Calculate the change in entropy in J/K for A. the vaporization of 50.0 g of benzene (C_(6)H_(6)) at its boiling point of 80^circ C.Delta H_(vap)=31.0kJ/mol B. the melting of 50.0 g of benzene (C_(6)H_(6)) at its freezing point of 5.5^circ C.Delta H_(fus)=9.92kJ/mol
1. Calculate the change in entropy in J/K for A. the vaporization of 50.0 g of benzene (C_(6)H_(6)) at its boiling point of 80^circ C.Delta H_(vap)=31.0kJ/mol B. the melting of 50.0 g of benzene (C_(6)H_(6)) at its freezing point of 5.5^circ C.Delta H_(fus)=9.92kJ/mol

Solution
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Answer

A. \Delta S_{vap} = 5.64 \, J/K ### B. \Delta S_{fus} = 2.27 \, J/K Explanation 1. Calculate moles of benzene Molar mass of C_{6}H_{6} is 78.11 g/mol. Moles = \frac{50.0 \text{ g}}{78.11 \text{ g/mol}}. 2. Calculate entropy change for vaporization Use formula \Delta S = \frac{\Delta H_{vap}}{T}. Convert temperature to Kelvin: 80^{\circ}C = 353.15 K. \Delta S_{vap} = \frac{31.0 \text{ kJ/mol}}{353.15 \text{ K}} \times \text{moles}. 3. Calculate entropy change for melting Use formula \Delta S = \frac{\Delta H_{fus}}{T}. Convert temperature to Kelvin: 5.5^{\circ}C = 278.65 K. \Delta S_{fus} = \frac{9.92 \text{ kJ/mol}}{278.65 \text{ K}} \times \text{moles}.

Explanation

1. Calculate moles of benzene<br /> Molar mass of $C_{6}H_{6}$ is 78.11 g/mol. Moles = $\frac{50.0 \text{ g}}{78.11 \text{ g/mol}}$.<br /><br />2. Calculate entropy change for vaporization<br /> Use formula $\Delta S = \frac{\Delta H_{vap}}{T}$. Convert temperature to Kelvin: $80^{\circ}C = 353.15 K$. $\Delta S_{vap} = \frac{31.0 \text{ kJ/mol}}{353.15 \text{ K}} \times \text{moles}$.<br /><br />3. Calculate entropy change for melting<br /> Use formula $\Delta S = \frac{\Delta H_{fus}}{T}$. Convert temperature to Kelvin: $5.5^{\circ}C = 278.65 K$. $\Delta S_{fus} = \frac{9.92 \text{ kJ/mol}}{278.65 \text{ K}} \times \text{moles}$.
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