The enthalpy of vaporization of ethanol is 38.7 kJ/mol at its boiling point (78∘C). Determine △Ssys,△Ssurr,and △Suniv when 1.00 mole of ethanol is vaporized at 78∘C and 1.00atm.

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The enthalpy of vaporization of ethanol is 38.7 kJ/mol at its boiling point (78∘C). Determine △Ssys,△Ssurr,and △Suniv when 1.00 mole of ethanol is vaporized at 78∘C and 1.00atm.

Donald Cheek · Accepted Answer

Here we have the reaction: C2H5OH(1)⇒C2H5OH(g) The values are: △Hvap=38.7kJmol T=78∘C Firstly we will calculate △Ssys: △Ssys=△HvapT =38.7kJmol78+273K =+110.2JmolK Now, we will calculate △Ssurr: △Ssurr=−△HvapT =−38.7kJmol78+273K =−110.2JmolK Therefore, at the end we can calculate △Suniv as the sum of △Ssys and △Ssurr: △Suniv=△Ssys+△Ssurr =+110.2JmolK+(−110.2JmolK) =0

Foreckije · Accepted Answer

Step 1 Given: Enthalpy of vaporization of ethanol =38.7kJmol Boiling point (T)=78∘C=351K Step 2 Calculation Vaporisation of ethanol is given by, C2H5OH(1)⇒C2H5OH(g) △Hvap=38.7kJmol Step 3 Calculation for entropy of change system (△Ssys): △Ssys=+△HvapT △Ssys=38.7×103Jmol351K △Ssys=+110.26Jmol.K Calculation for entropy change of surroundings (△Ssurr): △Ssurr=−△HvapT △Ssurr=−38.7×103Jmol351K △Ssurr=−110.26Jmol.K Step 4 Calculation for entropy change of universe (△Suniv): △Suniv=△Ssys+△Ssurr △Suniv=+110.26Jmol.K−110.26Jmol.K △Suniv=0

The enthalpy of vaporization of ethanol is 38.7 kJ/mol at

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