Calculate Heat of Vaporization and normal boing point (in degree C).?

[laughingnahga]

Using line equation y = -4058.7x + 16.10 with the experiment pressure measured in kPa instead of atm.

I already solved for heat of vaporization thus:

(-4058.7K)(-8.314 J/mol \astK)=heat of vaporization = 33444 J/mol = 33.44 kJ/mol.

Also, the vapor pressure for normal boiling point:
(1atm)(101325 Pa/1atm)(1 kPa/1000 Pa)=101.32 kPa

I attempted to solve for T by rearranging the linear form of the Clausius-Clapeyron equation, ln Pvap = (-\DeltaH/R)(1/T) + ln \beta into something like this (ln101.32 kPa)(-33444 J/mol \div 8.413 J/mol * K)\div1 - ln16.10 = T but I got something insane like 21857.7 K which even with subtracting 273 to get C is no where near close to the answer choices provided.

Any help with the last part would be greatly appreciated.

 

[Borek]

Telling us what is x and what is y should slightly increase chances that someone will try to understand the problem and what you did.

 

[laughingnahga]

I had to turn the assignment in this afternoon so it really doesn't matter at this point.

Thanks anyway...