For the equation, SO2(g) + 1/2 O2(g)-> SO3(g) Kp=9.5
EQUAL partial pressures of SO2 and O2 are placed in an empty reaction vessel and heated to 600 degrees celsius. Assuming that 62% of the SO2 partial pressure has reacting to give S03, calculate the partial pressure of S03 at equilibrium.
So far, I did:
R SO2 + 1/2 O2 -> SO3
I X X 0
C -.62X -.31X. +.62X
E X-.62X X-.31X .62X
And 9.5=(.62X)/(X-.62X)(X-.31X)
But I'm not sure if I set this up right or how to solve for X. I am not getting the correct answer. Please help!! Thank you!!!
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Sorry- I forgot to add that the last value, (X-.31X) is raised to the 1/2 power
1 answer