Question

Calculate the heat (in kJ) required to completely vaporize 11.42 g of

n-octane, C8H18, if the sample is initially at –20 ºC. Pertinent data are: m.p. = -57 ºC, b.p. = 126 ºC,

ΔHfus = 20.65 kJ/mol, ΔHvap = 38.6 kJ/mol, Cp(liq.) = 254 J/K-mol.

Answer 1

Initial temperature of octane is -20oC = 273-20 = 253K
As the melting point is -57oC, we can consider given a sample as a liquid. So initially sample will obtain heat to raise the temperature from 253 K its boiling point 126oC i.e. (273+126=399K). The heat will be needed to convert all the liquid octane at 399K to vaporize.

Thus,
Let's first calculate heat required to raise the temperature from 253 to 399 K.
           Q1   = m*Cp*(T)
               = (11.42g/114g/mol)*254(J/K*mol) * (399-253)K
               = 3.715*10^3 J
               = 3.715 kJ
Now, heat required to convert liquid octane to gaseous octane,
           Q2   = moles*Hvap
               = (11.42g/114g/mol)*38.6 kJ/mol
               = 3.867 kJ
Thus total heat required    = Q1+Q2
               = 3.867 kJ