4I.3

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Camille Marangi 2E
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Joined: Fri Sep 28, 2018 12:26 am

4I.3

Postby Camille Marangi 2E » Tue Feb 05, 2019 2:27 pm

The standard entropy of vaporization of benzene is approximately 85 J/Kmol at its boiling point. (a) Estimate the standard enthalpy of vaporization of benzene at its boiling point of 80 C. (b) What is the standard entropy change of the surroundings when 10. g of benzene, C6H6, vaporizes at its boiling point? Could someone just walk me through this problem?

Jessica Chen 1F
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Joined: Fri Sep 28, 2018 12:17 am

Re: 4I.3

Postby Jessica Chen 1F » Tue Feb 05, 2019 2:53 pm

In part (a), the equation you use to calculate the standard enthalpy of vaporization is delta S (vap) = delta H (vap) / T(boiling point in K).
So you get delta H (vap) = (85 J/K)(80 + 273 K) = 30,005 J or 30.0 kJ.

For part (b), the standard entropy change of the system is calculated by multiplying the # of moles by the standard entropy of vaporization (you convert the 10.g of benzene to moles and then multiply by the standard entropy of vaporization. The standard entropy change of the surroundings will be the opposite sign of this number, so if the standard entropy change of the system is positive, the standard entropy change of the surroundings will be negative, and vice versa.

Sarah Kiamanesh 1D
Posts: 30
Joined: Fri Sep 28, 2018 12:22 am

Re: 4I.3

Postby Sarah Kiamanesh 1D » Tue Feb 05, 2019 2:55 pm

Entropy (S) of vaporization = delta H (vaporization) / Boiling Point in Kelvin
Using this equation and plugging in the values we already know:
85 = delta H (vap) / (80 + 273)
which results in:
Delta H (vap) = 30.0KJ/mol

Now for part B: we know that for 1 mole of benzene, delta S (system) = 85 J/mole
We can convert 10g C6H6 to 0.128 moles C6H6
Using a proportion and solving for x:
1/85 = 0.128/x
x = 10.88J/K
delta S (system) + delta S (surroundings) = 0
delta S (surroundings) = -10.88J/K

Alyssa Bryan 3F
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Re: 4I.3

Postby Alyssa Bryan 3F » Sun Feb 17, 2019 10:45 pm

Jessica Chen 1F wrote:In part (a), the equation you use to calculate the standard enthalpy of vaporization is delta S (vap) = delta H (vap) / T(boiling point in K).
So you get delta H (vap) = (85 J/K)(80 + 273 K) = 30,005 J or 30.0 kJ.

For part (b), the standard entropy change of the system is calculated by multiplying the # of moles by the standard entropy of vaporization (you convert the 10.g of benzene to moles and then multiply by the standard entropy of vaporization. The standard entropy change of the surroundings will be the opposite sign of this number, so if the standard entropy change of the system is positive, the standard entropy change of the surroundings will be negative, and vice versa.


Where did you get the equation delta S (vap) = delta H (vap) / T for part a?


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