Question

How much heat is required to convert 22.8 g of liquid benzene (C6H6) at 58 ∘C to gaseous benzene at 100 ∘C? The boiling point of benzene is 80.1 ∘C and Cm [C6H6(l)] = 136.0 J/(mol⋅∘C), and ΔHvap = 30.72 kJ/mol, Cm [C6H6(g)] = 82.4 J/(mol⋅∘C).

Answer 1

Moles of benzene = mass / molar mass = 22.8 / 78.11

Heat required to convert liquid benzene from 58 C to 80.1 ( Boiling point)

       = Molar heat capacity x Moles of benzene x temperature change

            = ( 136 ) x ( 22.8/78.11) x ( 80.1-58) = 877.3 J

Heat required to boil bezene at 80.1 C = moles of benzene x Heat of vaporisation

       = ( 22.8/78.11) x ( 30.72) = 8.967 KJ = 8967J

Heat required to rise temp of gas benzene at 80.1 to 100 C

    = molar heat capacity of gas benzene x moles x temp change

       = ( 82.4) x ( 22.8/78.11) x ( 100-80.1) = 478.64 J

Combining all we get = ( 877.3+8967 + 478.64) = 10322.6 J   = 10.32 KJ