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!!!
Sorry- I forgot to add that the last value, (X-.31X) is raised to the 1/2 power
You have correctly set up the initial and change tables for this equilibrium calculation. However, there is an error in your expression for the equilibrium constant. The correct expression for the equilibrium constant Kp is:
Kp = (P_SO3) / (P_SO2)^0.62 * (P_O2)^0.31
To solve for X, you need to substitute the given information into the equilibrium constant expression and then solve for X. Let's go step by step:
1. The given equilibrium constant is Kp = 9.5.
2. You are told that 62% of the initial partial pressure of SO2 has reacted, so the partial pressure of SO2 at equilibrium is 0.38X. Similarly, the partial pressure of O2 at equilibrium is 0.69X (1/2 of 0.31X).
3. Now, substitute these values into the equilibrium constant expression:
9.5 = (P_SO3) / (P_SO2)^0.62 * (P_O2)^0.31
9.5 = (0.62X) / (0.38X)^0.62 * (0.69X)^0.31
4. Simplify the equation by raising each term to the appropriate exponent:
9.5 = (0.62X) / (0.38^0.62 * X^0.62) * (0.69^0.31 * X^0.31)
9.5 = 0.62 / (0.38^0.62 * 0.69^0.31) * X^0.31
5. Rearrange the equation to solve for X:
X^0.31 = (0.62 * (0.38^0.62 * 0.69^0.31)) / 9.5
X = [(0.62 * (0.38^0.62 * 0.69^0.31)) / 9.5]^(1/0.31)
Now you can use a calculator to evaluate the right-hand side of the equation. The resulting value of X will give you the partial pressure of SO3 at equilibrium.