Calculate $\Delta S$ when 0.5 mol of ethanol evaporates at constant pressure.

Ethanol's boiling point is 78.3°C, and $\Delta H_{\text{vap}}$ is 39.3 kJ/mol.

To calculate ΔS when 0.5 mol of ethanol evaporates at constant pressure, use the equation ΔS = q/T. The heat absorbed during the evaporation of ethanol can be calculated using the equation q = n * ΔHvap. Plugging the values into the equation, we find ΔS = 0.056 J/K.

Explanation:

To calculate ΔS when 0.5 mol of ethanol evaporates at constant pressure, we need to use the equation

ΔS = q/T,

where q is the heat absorbed or released during the process, and T is the temperature in Kelvin.

The heat absorbed during the evaporation of ethanol can be calculated using the equation q = n * ΔHvap,

where n is the number of moles and ΔHvap is the enthalpy of vaporization.

Substituting the given values, we have q = 0.5 mol * 39.3 kJ/mol

= 19.65 kJ.

The temperature can be converted from Celsius to Kelvin by adding 273.15,

so T = 78.3°C + 273.15

= 351.45 K.

Plugging these values into the equation for ΔS, we find ΔS = 19.65 kJ / 351.45 K

= 0.056 J/K.

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Calculate \(\Delta S\) when 0.5 mol of ethanol evaporates at constant pressure. Ethanol's boiling point is 78.3°C, and \(\Delta H_{\text{vap}}\) is 39.3 kJ/mol.

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Calculate \(\Delta S\) when 0.5 mol of ethanol evaporates at constant pressure.

Ethanol's boiling point is 78.3°C, and \(\Delta H_{\text{vap}}\) is 39.3 kJ/mol.