1. The e ∆Ssys, ∆Ssurr, and ∆Suniv when 1.00 mol of ethanol at 78 o C and 1.00 atm.

Chemistry

1 answer:

Crazy boy [7]2 years ago

6 0

The values based on the enthalpy of vaporization of ethanol illustrated is given below.

<h3>What is enthalpy?</h3>

It should be noted that enthalpy is the sum of the internal energy as well as the product of the pressure and the volume of a thermodynamic system.

It should be noted that enthalpy is a property of a thermodynamic system, and it's the sum of the system's internal energy and the product of its pressure and volume.

It is a state function that's used in many measurements in biological, chemical, and physical systems at a constant pressure.

ΔSsys = ΔH/T = (38700 J/mol) /(78+273) = 110.3 J/Kmol

ΔSsurr = -ΔSsys = -110.3 J/Kmol

ΔSuniverse = ΔSsurr +ΔSsys = 110.3 + -110.3 = 0 J/Kmol

Complete question:

The enthalpy of vaporization of ethanol is 38.7 kJ/mol at its boiling point (78°C).

Determine ΔSsys,ΔSsurr, andΔSuniv when 1.00 mol of ethanol is vaporized at 78°C and 1.00 atm

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