Question

QUESTION 3 How much heat (in kJ) is required to raise the temperature of 122 g of ethanol (mw 46.07) from 11.29 °C to its boiling point of 78.37 °C and then vaporize it completely creating ethanol gas at the boiling temperature? (specific heat - 2.46 J/g°C, AHyap = 48.6 kJ/mol) QUESTION 4 What is the change in enthalpy (in kJ) when 40.1 g of ethanol (mw-46.07) is condensed at its boiling temperature? (specific heat 2.46 J/g°C, AHvap 48.6 kJ/mol) (Hint: be sure to think about the sign)

Answer 1

3)

Ti = 11.29 oC

Tf = 78.37 oC

here

Cl = 2.46 J/g.oC

Heat required to convert liquid from 11.29 oC to 78.37 oC

Q1 = m*Cl*(Tf-Ti)

= 122 g * 2.46 J/g.oC *(78.37-11.29) oC

= 20132.0496 J

Hvap = 48.6KJ/mol =

48600J/mol

Lets convert mass to mol

Molar mass of C2H5OH = 46.07 g/mol

number of mol

n= mass/molar mass

= 122.0/46.07

= 2.6483 mol

Heat required to convert liquid to gas at 78.37 oC

Q2 = n*Hvap

= 2.6483 mol *48600 J/mol

= 128705.392 J

Total heat required = Q1 + Q2

= 20132.0496 J + 128705.392 J

= 148837 J

= 148.8 KJ

Answer: 148.8 KJ

4)

Molar mass of C2H5OH = 46.07 g/mol

mass(C2H5OH)= 40.1 g

use:

number of mol of C2H5OH,

n = mass of C2H5OH/molar mass of C2H5OH

=(40.1 g)/(46.07 g/mol)

= 0.8705 mol

Given:

ΔH = -48.6 KJ/mol [negative because condensation is opposite of vapourisation]

use:

Q = ΔH * number of mol

= -48.6 KJ/mol * 0.8705 mol

= -42.3 KJ

Answer: -42.3 KJ