Question

From data below, calculate the total heat (in J) associated with the conversion of 0.499 mol ethanol gas (C2H6O) at 301°C and 1 atm to liquid ethanol at 25.0°C and 1 atm.
(Pay attention to the sign of the heat.) Answer should be in J.

Boiling point at 1 atm 78.5°C
cgas 1.43 J/g°C
cliquid 2.45 J/g°C
H°vap 40.5 kJ/mol

Answer 1

Answer:

The total heat associated is -30,520.3 J.

Explanation:

Moles of ethanol = 0.499 moles

Molar mass of ethanol = 46 g/mol  

Mass = Moles × Molar mass = 0.499 moles × 46 g/mol = 22.954 g

m is the mass of ethanol = 22.954 g

Q₁ is heat involved in the conversion of 22.954 g ethanol gas at 301°C to ethanol gas at 78.5°C

Thus, Q₁ = m × c × ΔT

Where,  

c =  The specific heat of the gas = 1.43 J/g°C

ΔT = Final temperature - Initial temperature = 78.5 - 301°C  

= - 222.5 °C

Applying the values in the above equation as:-

$Q_1 = 22.954 g\times 1.43 J/g^0C\times (-222.5^oC) = -7303.3J$

Q₂ is the enthalpy of condensation from gas to liquid for the given mass of ethanol .

Thus, Q₂ = moles×ΔH condensation  

Given that:- ΔH vaporization = 40.5 kJ/mol

Enthalpy of condensation of gaseous ethanol to liquid ethanol = - 40.5 kJ/mol

Considering, 1 kJ = 1000 J

So,  

ΔH condensation = - 40.5 ×1000 J/mol = - 40500 J/mol

Thus, Q₂ = 0.499 moles × (- 40500 J/mol) = -20209.5 J

Q₃ is heat involved in the conversion of 22.954 g gaseous ethanol at 78.5°C to ethanol liquid at 25.0°C.

Thus, Q₃ = m × C ×ΔT

Where,  

C = The specific heat of the liquid = 2.45 J/g°C

ΔT = Final temperature - Initial temperature = 25.0 - 78.5 °C  

= - 53.5 °C

Applying the values in the above equation as:-

$Q_3 = 22.954 g\times 2.45 J/g^0C\times (-53.5 ^0C)=-3007.5 J$

Applying the values as:

Total heat = $Q_1+Q_2+Q_3$

= -7303.3 J - 20209.5 J - 3007.5 J

= -30,520.3 J

The total heat associated is -30,520.3 J.