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.

Answer 2

From data provided and the calculations, the total heat 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 is -30,540 J.

What is the quantity of heat involved in 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?

The quantity of Heat involved in 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 involves three heat changes:

• Q1: the heat involved in converting ethanol gas at 301°C to ethanol gas at 78.5°C
• Q2: the heat involved in converting ethanol gas at 78.5°C to liquid ethanol at 78.5°C
• Q3: the heat involved in converting liquid ethanol at 78.5°C to liquid ethanol at 25.0°C.

Calculating Q1:

• Q₁ = m × c × ΔT

Where,

• m = mass of ethanol
• c =  The specific heat of the gas = 1.43 J/g°C
• ΔT = Final temperature - Initial temperature

Mass of ethanol = moles × molar mass

molar mass of ethanol = 46 g/mol

Moles of ethanol = 0.499 moles

mass of ethanol = 0.499 moles × 46 g/mol

Mass of ethanol = 22.995 g

ΔT = 78.5 - 301°C = - 222.5 °C

Thus:

Q₁ = 22.995 × 1.43 × -222.5

Q₁ = -7316.43 J

Calculating Q₂

• Q₂ = moles × ΔHcondensation

ΔH vaporization = 40.5 kJ/mol = 40500 J/mol

• Since a condensation occurs, ΔHcondensation = - ΔHvaporization

ΔHcondensation = -40500 J/mol

Q₂ = 0.499 moles × (- 40500 J/mol)

Q₂ = -20209.5 J

Calculating Q3

Q₃ = m × C ×ΔT

where,  

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

ΔT = Final temperature - Initial temperature

ΔT = 25.0 - 78.5 °C = - 53.5 °C

Q₃ = 22.995 × 2.45 × -53.5

Q₃ = 3008.69

Therefore,

• Total heat, Q = Q1 + Q2 + Q3

Q = -7316.43 J - 20209.50 J - 3014.07 J

Q = -30,540 J

Therefore, from calculated values, the total heat 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 is -30,540 J.

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