Based on the following thermodynamic data, calculate the boiling pressure of ethanol in degrees celcius

Substance Delta Hf ( Kj/mol) Delta Sf ( J/K*mol)

C2H5OH (L) --- -277.7 160.6

C2H5OH (G) --- -235.1 282.6

76.03

Explanation:

We have the equation

T= Delta H/ Delta S

For this we need to find the Delta H of vaporization for the boiling point. We can obtain this by taking delta H(products) - delta H(reactants). In this case we will take (-235.1 kj/mol) - (-277.1 kj/mol) = 42.6 kj/mol.

We will then need the delta S of vaporization, which can be obtained by taking delta S(products) - delta S(reactants). Here we will take (282.6 j/k*mol) - (160.6 j/k*mol) = 122 j/k*mol. We still need to convert this to kj/mol so we divide 122 by 1000 to get 0.122 kj/mol as delta S of vaporization.

Then we plug in these values into the equation mentioned above as follows.

T= 42.6/0.122

T= 349.18 K

All we need to do now is to convert this into degrees Celsius by taking 349.18 - 273.15 to end up with an answer of 76.03 degrees Celsius.

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