Posted by **Anonymous** on Tuesday, July 19, 2011 at 2:38pm.

Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling point of 80.1 degrees celsius. At what temperature does benzene boil when the external pressure is 480 torr? Can someone show me an equation and how to solve this?

- Chemistry -
**DrBob222**, Tuesday, July 19, 2011 at 2:52pm
Use the Clausius-Clapeyron equation. Remember the definition of boiling point; i.e., the temperature at which the vapor pressure of the liquid equals atmospheric pressure. So you know p1 and p2 and T1. T2 is the only unknown.

- Chemistry -
**Anonymous**, Tuesday, July 19, 2011 at 4:38pm
I know im using that equation but would you set it up like this

1/t2= (760mm hg/480torr)/(30720j/8.314)+(1/353.25degrees celsius) when i do it out i get a number that does not make sense.

- Chemistry -
**DrBob222**, Tuesday, July 19, 2011 at 6:39pm
I didn't work it out but I suspect the problem is that you are trying to solve for 1/t2 from the get go. It is far simpler if you plug the numbers into the equation and solve that way.

ln(P2/p1) = DHvap/R (1/T1 - 1/T2)

ln(480/760) = (30720/8.314)(1/353.25 - 1/T2)

First term = ln(480/760) = -0.4595

30720/8.314 = 3694.97

1/353.25 = 0.002831

1/t2 = 1/t2

-0.4595 = 3694.97(0.002831-1/T2)

-0.4595 = (10.4599 - 3694.97/T2)

Check my work. I'll let you finish.

- Chemistry -
**Anonymous**, Tuesday, July 19, 2011 at 9:17pm
I have -.4595= -3684.51/T2. Would you then just divide -3684.51/-.4595? and how did you het 10.4599 in the last equation?

- Chemistry -
**DrBob222**, Tuesday, July 19, 2011 at 10:20pm
No. I think your algebra is cockeyed.

Assuming my work is ok to where I stopped, then multiply through by T2 to get rid of the T2 term in the denominator.

-0.4595 = (10.4599 - 3694.97/T2)

-0.4595T2 = 10.4599T2 - 3694.97

-10.919T2 = -3694.97

T2 = 338.4 or thereabouts which is approximately 65 C. Check my work but that sounds reasonable to me. Lower pressure means lower boiling point. I can see why 8,000 K didn't sound right to you.

- Chemistry -
**1994**, Wednesday, November 28, 2012 at 5:53am
Isn't the formula suppose to be ln(p2/p1)=Dvap/R(1/T2 -1/T1) ? That is what my textbook says.

## Answer This Question

## Related Questions

- problem - Benzene has a heat of vaporization of 30.72 kj/mol and a normal ...
- chemistry - Benzene has a heat of vaporization of 30.72kJ/mol and a normal ...
- Chem. - Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling...
- science - benzene has a heat of vaporization of 30.72kj/mol and a normal boiling...
- chemistry - Benzene has a heat of vaporization of 30.72kJ/mol and a normal ...
- Chemistry - Benzene has a heat of vaporization of 30.72kJ/mol and a normal ...
- chemistry - Benzene has a heat of vaporization of 30.72kJ/mol and a normal ...
- chemistry - Benzene has a heat of vaporization of 30.72kj/mol and a normal ...
- Chem. - Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling ...
- chemistry - Benzene has a heat of vaporization of 30.72kj/mol and a normal ...

More Related Questions