Question

Ethanol (C2H5OH) boils at 78∘C. Its density is 0.789 g/mL. The enthalpy of vaporization is 38.56 kJ/mol. The specific heat of liquid ethanol is 2.3 J/g−K.

1. How much heat is required to convert 25.0 g of ethanol at 30 ∘C to the vapor phase at 78 ∘C?

2. Place the following substances in order of decreasing volatility: CH4, CBr4, CH2Cl2, CH3Cl, CHBr3, and CH2Br2. Rank from most volatile to least volatile. To rank items as equivalent, overlap them.

3. How do the boiling points vary through this series? Rank from highest to lowest boiling point. To rank items as equivalent, overlap them. CH4, CBr4, CH2Cl2, CH3Cl, CHBr3, and CH2Br2.

Answer 1

Q1) heat of enthalpy vaporization = 38.56 KJ/mol = 38560 Joules/mol

Molar mass of Ethanol (C2H5OH) = 12 * 2 + 1 * 6 + 16 = 46 gm/mol

Heat of enthalpy vaporization per gm = 38560/46 = 838.26 J/gm

Heat Required = 25 * 2.3 * (78-30) + 838.26 * 25

=> 25 * 2.3 * 48 + 25 * 838.26

=> 23716.6

= 23.716 KJ

Q2) The order of decreasing voltality will be

assuming that voltality is same per gm of compound, the least weighted compound will be the most volatile and heavier compounds will be less volatile, so the final order will be

CH4 > CH3Cl > CH2Cl2 > CH2Br2 > CHBr3 > CBr4

Q3) Lower the voltality, higher will be the boiling point, hence in Q3, the reverse answer of Q2 will be used

So the final order is

CH4 < CH3Cl < CH2Cl2 < CH2Br2 < CHBr3 < CBr4