For ethanol, [tex]C_2H_5OH[/tex], the heat of vaporization at its normal boiling point of [tex]78^\circ C[/tex] is [tex]38.6 \, \text{kJ/mol}[/tex].

Calculate the entropy change when [tex]1.72[/tex] moles of [tex]C_2H_5OH[/tex] vapor condenses at [tex]78^\circ C[/tex] and [tex]1 \, \text{atm}[/tex].

Express your answer in [tex]\text{J/K}[/tex].

The heat of vaporization at its normal boiling point of 78°C is 38.6 kJ/mol for ethanol, the entropy change when 1.72 moles of C2H5OH vapor condenses at 78°C and 1 atm isΔS = nΔS° = (1.72 mol) (-129 J/K/mol) = -221.88 J/K or -188 J/K (rounded to two significant figures).

C2H5OH. At 78°C and 1 atm, when 1.72 moles of C2H5OH vapor condensate, the entropy change is given by j/k. The entropy change when 1.72 moles of C2H5OH vapor condensate at 78°C and 1 atm is -188 J/K. The formula for the entropy change when a substance transforms from one state to another is S = ΔH_vap/T, where ΔH_vap is the heat of vaporization, and T is the temperature.

The entropy change when 1.72 moles of C2H5OH vapor condenses can be calculated by the following equation:

ΔS = -ΔH_vap/T

= -38.6 kJ/mol / (78 + 273.15) K

= -0.129 kJ/K/mol

= -129 J/K/mol

Therefore, the entropy change when 1.72 moles of C2H5OH vapor condenses at 78°C and 1 atm isΔS = nΔS° = (1.72 mol) (-129 J/K/mol) = -221.88 J/K or -188 J/K (rounded to two significant figures).

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