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How much heat energy is required to convert 49.4 g of solid ethanol at − 114.5 ° C to gasesous ethanol at 140.7 ° C ? The molar heat of fusion of ethanol is 4.60 kJ/mol , and its molar heat of vaporization is 38.56 kJ/mol . Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C . The specific heat capacity of liquid ethanol is 2.45 J / g ⋅ ° C , and that of gaseous ethanol is 1.43 J / g ⋅ ° C .

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Answer:

Step-by-step explanation:

mass of ethanol (s)
m=49.4\ gm

molar specific heat of fusion
\Delta H_f=4.6\ KJ/mol

molar specific heat of vaporization
\Delta H_v=38.56\ KJ/mol

specific heat of liquid
c_l=2.45\ J/g-^(\circ)C

specific heat of gas
c_v=1.43\ J/g-^(\circ)C

no of moles of ethanol
n=(49.4)/(46)=1.074\ mol


Q_1=n* \Delta H_f=1.074* 4.6=4.94\ KJ


Q_2=mc_l\Delta T=49.4* 2.45* (192.9)


Q_2=23.346\ KJ


Q_3=n* \Delta H_v=1.074* 38.56=41.41\ KJ


Q_4=mc_v\Delta T=49.4* 1.43* (192.9)=13.626\ KJ


Q_(net)=Q_1+Q_2+Q_3+Q_4


Q_(net)=83.322\ KJ

How much heat energy is required to convert 49.4 g of solid ethanol at − 114.5 ° C-example-1
User Dean MacGregor
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