Wednesday
May 22, 2013

# Homework Help: Chem help

Posted by Duffy on Sunday, June 7, 2009 at 10:10pm.

An equilibrium reaction which has K1 = 6.5 x 10¯2 at 100oC and has ∆Horxn = -49.5 kJmol¯1. What is the value of K at 250 oC?

I can't get the right answer, which is supposed to be 6.68 x 10^-4.

How does one do this? Thanks

• Chem help - Duffy, Sunday, June 7, 2009 at 10:10pm

Should be 100 degrees C and 250 degrees C

• Chem help - smiley, Monday, June 8, 2009 at 5:09am

ln(K1/K2)=(∆Horxn/R)((1/T2)-(1/T1))
ln((6.5x10^-2)/K2)=((-49.5 kJmol^-1)/R)((1/(250+273.15))-(1/(100+273.15)))

That's it! Just solve for K2. Don't forget that your R must have the units of kJ/molK.

Hope this helps. Cheers!

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