1. How much energy (in kilojoules) is released when 33.0 g of ethanol vapor at 9

Question

1. How much energy (in kilojoules) is released when 33.0 g of ethanol vapor at 92.5 C is cooled to -12.0 C? Ethanol has mp = -114.5 C, bp = 78.4 C, Hvap = 38.56 kJ/mol and Hfusion = 4.60 kJ/mol. The molar heat capacity is 113 J/(Kmol) for the liquid and 65.7 J/(Kmol) for the vapor.

2. How much heat (kJ) is required to convert 3.12 moles of liquid benzene at 75.1°C to gaseous benzene at 115.1°C? The following information may be useful.
b.p = 80.1C
Cm (liquid benzene) = 136.0 J/ (mol * °C)
Hvap = 30.72 kJ/mol
Cm (gaseous benzene) = 36.6 J/ (mol * °C)





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Explanation / Answer

1) Solution-

(33.0 g CHO) * (1 mole CHO / 46.068 g CHO)

= 0.716 moles CHO


The temperature changes (T) in °C are the same number in Kelvin.

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Step 1: Heat released to cool from 88.0 °C to bp = 78.4 °C :

q = ncT

Here n is number of moles

c= molar heat capacity

= (0.716 moles) * [65.7 J/(Kmol)] * (-14.1 K)

= -663.28 J

Step 2: Heat released converting 78.4 °C vapor to 78.4 °C liquid:

q = n Hvap

= (0.6925 moles) * (38.56 kJ/mol)

= -26.7 kJ

= -26,700 J

---
Step 3: Heat released to cool from 78.4 °C to -12.0 °C :

q = ncT

= (0.6925 moles) * [113 J/(Kmol)] * (66.4 K)

= -5195.97J

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Total heat released

q = -663.28 J - 26,700 J - 5195.97 J

= -32559.25 J

= -32.6 kJ

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