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From the data below, calculate the total heat (in J) associated with the conversion of 0.310 mol ethanol gas (C2H6O) at 201°C and 1 atm to liquid ethanol at 25.0°C and 1 atm. (Pay attention to the sign of the heat.)

Boiling point at 1 atm 78.5 °C
Cgas1.43 J/g°C
Cliquid 2.45 J/g°C
H° vap 40.5 kJ/mol

User Dan Doe
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1 Answer

3 votes

Answer:

-16,873.25 J

Step-by-step explanation:

This process will occur in steps. First, the temperature will decrease until the boiling point, then the ethanol will suffer a phase change from gas to liquid, and then, the temperature will decrease to 25.0°C. Because the temperature is decreasing, it is losing heat, and so, for all process the heat is negative, also H°vap is negative.

For the first and last step, the heat occur with a temperature change, and it can be calculated by:

Q = m*c*ΔT

Where m is the mass, c is the specif heat, and ΔT is the temperature variation. The mass is the number of moles multiplied by the molar mass (46.07 g/mol for ethanol), so m = 0.310*46.07 = 14.28 g.

For the second step, the heat lost is:

Q = - n*H°vap

At first step the substance is a gas, so c = cgas, and at the third step, the substance is a liquid, and c =cliquid.

Q1 = 14.28*1.43*(78.5 - 201) = -2501.5 J

Q2 = -0.310*40.5 = -12.5 kJ = -12500 J

Q3 = 14.28*2.45*(25.0 - 78.5) = -1871.75 J

Q = Q1 + Q2 + Q3

Q = -16873.25 J

User Fenster
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