4C.7 vaporization enthalpy

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Hannah_1G
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Joined: Wed Sep 18, 2019 12:22 am

4C.7 vaporization enthalpy

Postby Hannah_1G » Mon Feb 10, 2020 10:11 pm

How do you go about solving this? The moles of liquid methane are given and the heat required. The only equation I know for enthalpy of vaporization is Hvap= Hm(vapor) - Hm(liquid).

BNgo_2L
Posts: 95
Joined: Wed Sep 11, 2019 12:17 am

Re: 4C.7 vaporization enthalpy

Postby BNgo_2L » Mon Feb 10, 2020 10:15 pm

To find delta H(vap), you need the energy required for vaporization per mole, so all you do is divide 4.76kJ by 0.579 moles.

Paul Hage 2G
Posts: 105
Joined: Thu Jul 25, 2019 12:17 am

Re: 4C.7 vaporization enthalpy

Postby Paul Hage 2G » Mon Feb 10, 2020 10:17 pm

For part a), we know that the enthalpy of vaporization can be written in units of kJ/mol. And because CH4 is at its boiling point, we can divide the heat value given in the problem by the amount in moles given in the problem. This will give the enthalpy of vaporization.
For part b), you would do the same thing, but you would first have to convert 22.45g to moles using the molar mass of ethanol. After you have the amount in moles, you can divide the heat value given in the problem by the amount in moles to give the enthalpy of vaporization.


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