Self-Test 8.9B

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Tram Ha 3I
Posts: 11
Joined: Fri Sep 25, 2015 3:00 am

Self-Test 8.9B

Postby Tram Ha 3I » Sun Jan 10, 2016 7:52 pm

The same heater was used to heat a 23-g sample of ethanol, C2H5OH,
to its boiling point. It was found that 22 kJ was required to vaporize all the ethanol.
What is the enthalpy of vaporization of ethanol at its boiling point?
I don't know how to apply delta Hvap= H(vapor)- H(liquid) to solve this problem, or if this formula is the right one to use. How do I solve for the enthalpy of vaporization?

Sandeep Gurram 2E
Posts: 42
Joined: Fri Sep 25, 2015 3:00 am

Re: Self-Test 8.9B

Postby Sandeep Gurram 2E » Sun Jan 10, 2016 8:08 pm

You wouldn't use the Hvap= H(vapor)- H(liquid) formula for this question. All you have to do is find the enthalpy of formation in terms of kJ/mol. This involves converting the amount of methanol they gave in grams to mols and then dividing the amount of heat it took to vaporize to this amount by your calculated answer. Shown below:

Given: Takes 21.2 kJ of energy to vaporize 22.45 grams of ethanol.

Step 1: Convert grams of ethanol to mols of ethanol. 22.45 grams ethanol / 46.069 grams/mol = .4873 mols ethanol.
Step 2: Divide energy needed to vaporize the given amount by previous answer. 21.2 kJ / .4873 mols = 43.5 kJ/mol.

43.5 kJ/mol is your enthalpy of vaporization.


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