Asked by kellie
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!!!
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!!!
Answers
Answered by
kellie
Sorry- I forgot to add that the last value, (X-.31X) is raised to the 1/2 power
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