Question

The following information is given for benzene, CH at 1 atm: boiling point So.1 °C AHap(80.1 °C)=30.7 kJ/mol melting point AHfus(5.50 °C)9.95 kJ/mol 5.50 °C specific heat liquid 1.74 J/g°C At a pressure of 1 atm, what is AH in kJ for the process of condensing a 49.4 g sample of gaseous benzene at its normal boiling point of S0.1 °C kJ

Answer 1

Molar mass of benzene = 78.11 gram

so, 49.4 gram of benzene = (49.4/ 78.11) mol = 0.632 mol

during condensation no temperature change occurs as mentioned in the question so,

At constant pressure, enthalpy change ∆H ( negative change as heat will be released ) = -(∆H_vap x 0.632 ) KJ = -(30.7 x 0.632) KJ = -19.4024 KJ

=> ∆H  = -19.4024 KJ

So, heat released will be -19.4024 KJ