Benzene (C6H6) has a normal boiling point of 80.1 C and normal melting point 5.4

Question

Benzene (C6H6) has a normal boiling point of 80.1 C and normal melting point 5.42 C. The molar heat capacity of solid benzene is 118.4 J/K mol; of liquid benzene is 134.8 J/K mol; and the molar heat capacity of benzene vapor is 82. 4 J/K mol. the heat of evaporation for benzene is 30.8 kj/mol and heat of fusion 9.9 kj/mol.

a) how much heat is needed to heat 154 g of benzene from 2.7 C to 98.5 C?

b) sketch a heating curve for benzene showing the temperature changes and phase transitions that occur when heat is added to the above sample from 2.7 C to 98.5 C.

Explanation / Answer

a) heat required(q) = nsolid*molarheatcapacity of benzene solid* DT1 + n*DHfus + nliquid*Cliquid*DT2 + n*DHvap + nvapor*Cvapor*DT3

    = (154/78)*118.4*(5.42-2.7) + (154/78)*9.9*10^3 + (154/78)*134.8*(80.1-5.42) + (154/78)* 30.8*10^3 + (154/78)*82.4*(98.5 - 80.1)

    = 103861.3 joule

    = 103.86 kj

solid benzene at 2.7 C ------ > solid benzene at 5.42 C ( n*C*DT)

   solid benzene at 5.42 C -------> liquid benzene at 5.42 C (DHfus)

liquid benzene at 5.42 C -----> liquid benzene at 80.1 C (n*C*DT)

    liquid benzene at 80.1 C ----> vapor benzene at 80.1 C (DHvap)

   vapor benzene at 80.1 C ----- > vapor benzene at 98.5 C

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