A certain substance has a heat of vaporization of 59.88 kJ/mol. At what Kelvin temperature will the vapor pressure be 3.50 times higher than it was at 325 K?

Question

Ive tried to solve this many times but still couldn't come up with the right answer, please help me and show the steps to get to the right answer! THANK YOU!

Answer 1

Hi Keithlyn M.,

What you need is the equation relating ΔH(v) to K(eq). That would be the van't Hoff equation:

ln (K2/K1) = (-ΔH/R)*(1/T₂ - 1/T₁) . In short, the natural logarithm of K scales linearly with -ΔH and inversely with T.

So let's see what tha means to your problem. We need a 3.5 x boost on the vapor pressure, that's a 3.5 x boost in K(eq) =K₂/K₁ .

K₂/K₁ = 3.5

ln(3.5) = (-59.88 kJ/mol / 8.314 J/ mol K)*(1/T₂ - 1/325)

T₂ = 344.47 K, rounds to 344 K (note that you cannot assume that 325 K was an exact temperature, therefore round to its precision).

Note: Lots of opportunity to miswrite equations and drop minus signs!

-- Cheers, -- Mr. d.