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
2 answers
Should be 100 degrees C and 250 degrees C
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!
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!