Question

Problem 11.46

The fluorocarbon compound C2Cl3F3 has a normal boiling point of 47.6 ∘C. The specific heats of C2Cl3F3(l) and C2Cl3F3(g) are 0.91 J/g⋅K and 0.67 J/g⋅K, respectively. The heat of vaporization for the compound is 27.49 kJ/mol.

Part A

Calculate the heat required to convert 72.0 g of C2Cl3F3 from a liquid at 10.30 ∘C to a gas at 76.05 ∘C.

Express your answer using two significant figures.

Q =

14.83

  kJ  

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Answer 1

Molar mass of C2Cl3F3 = 2 * 12 + 3 * 35.5 + 3 * 19 = 187.5 gm/mol

Number of moles of C2Cl3F3 = 72/187.5 = 0.384 moles

Heat of vaporization for 0.384 moles = 0.384 moles * 27.49 KJ/mol = 10.55616 KJ

Heat required to convert 72g of C2F3Cl3 from the liquid temperature of 10.30C to 47.6C

=> 72 * 0.91 * (47.6-10.3)

=> 2443.896 J

Heat required to convert 72g of C2F3Cl3 from the gaseous temperature of 47.6C to 76.05C

=> 72 * 0.67 * (76.05-47.6)

=> 1372.428 J

Total heat required = 10556.16 J + 1372.428 J + 2443.896 J

=> 14372.424 J = 14.372 KJ