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Chapter 12: Q12.20P (page 494)

From the data below, calculate the total heat (in J) needed to convert 0.333 mole of gaseous ethanol at 300oC and 1 atm to liquid ethanol at role="math" localid="1656934322575" 25.0oCand 1 atm: B.P. at 1 atm: ,78.5oC;螖贬yap: 40.5 kJ/mol;: Cgas1.43 J/role="math" localid="1656934528476" goC;: 2.45 J/goC.

Short Answer

Expert verified

The heat required in the process is 20356.06 J

Step by step solution

01

Heat needed to convert 0.333 mol liquid at 25.0 oCto 78.5oC 

The heat required to convert 0.333 mol (15.34 g) liquid at 25.0oCto 78.5oCis calculated as:

螖H1=Csolid(J/goC)*Weight(g)*螖T(oC)

螖H1=2.45*15.34*53.5=2010.69J

02

Heat needed to convert 0.333 mol liquid to vapour 

The heat required to convert 0.333 mol liquid to vapour is

螖H2=40500J/mol=40500*0.333J螖H2=13486.5J

03

Heat needed to convert 0.333 mol liquid at 78.5 oCto 300 oC

The heat required to convert 0.333 mol (15.34 g) liquid at 78.5oCto300oC is calculated as:

螖H3=Csolid(J/goC)*Weight(g)*螖T(oC)

Hence, the total heat energy required in the process is

螖H=螖H1+螖H2+螖H3=2010.69+13486.5+4858.87螖H=20356.06J

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Most popular questions from this chapter

Use these data to draw a qualitative phase diagram for ethylene (C2H4 ). Is C2H4(s) more or less dense thanC2H4 (l)?

bp at 1 atm: -103.7oC

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