From the data below, calculate the total heat (in J) needed to convert 0.707 mol of gaseous ethanol at 300.0°C and 1 atm to liquid ethanol at 25.0°C and 1 atm

Question

From the data below, calculate the total heat (in J) needed to convert 0.707 mol of gaseous ethanol at 300.0 °C and 1 atm to liquid ethanol at 25.0°C and 1 atm:

b.p. at 1 atm: 78.5 °C. ΔH^0 vap : 40.5 kJ/mol

c_gas: 1.43 J/g °C. c_liquid: .45 J/g·°C

Whats the J ?

Answer 1

First, notice that ∆Hvap is given in kj/mol while c_gas and c_liquid are in J/gºC. It is useful to use either mole or g in all cases. Here I will choose to use g and so will convert ∆Hvap to kJ/g using the molar mass of ethanol of 46.07 g/mol.

∆Hvap = 40.5 kJ/mol x 1 mol/46.07 g = 0.879 kJ/g = 879 J/g

mass of ethanol = 0.707 mol x 46.07 g/mol = 32.6 g of ethanol

Step 1: convert 32.6 g ethanol gas @ 300º to gas @ 78.5º

q = mc_gas∆T = (32.6 g)(1.43 J/g/deg)(221.5 degrees) = 10,326 J

Step 2: convert 32.6 g ethanol gas @ 78.5º to liquid @ 78.5º (phase change)

q = m∆Hvap = 32.6 g x 879 J/g = 28,655 J

Step 3: convert 32.6 g of liquid ethanol @ 78.5º to liquid @ 25º:

q = mc_liquid∆T = (32.6 g)(0.45 J/g/degree)(53.5 degrees) = 785 J

Add joules from all steps:

10,326 + 28,655 + 785 = 39766 J = 39,800 J (to 3 significant figures)

(note: the question asks for heat needed, but this is actually the heat generated in the process)