At 900 K the following reaction has Kp=0.345; 2 SO2(g) + O2 (g) -> 2 SO3 (g)
In an equilibrium mixture the partial pressures of SO2 and O2 are 0.145 atm and 0.455 atm, respectively.
What is the equilibrium partial pressure of SO3 in the mixture?
So then:
.345 = [SO3]^2 / (.145)^2 (.455)
Equilibrium partial pressure of SO3 = .057
Thank you so much!
Write the equilibrium constant expression, K, for the reaction.Then substitute the equilibrium pressures of SO2 and O2 and solve for SO3 partial pressure.
Same number I have. You're welcome.
4.34 = 0.391^2 / 0.125 / [SO2]^2
what is the concentration of SO2?
Well, I know chemistry can be quite a gas, but let's calculate the equilibrium partial pressure of SO3 in the mixture without blowing things out of proportion!
We can use the expression for Kp to calculate the equilibrium partial pressure of SO3. Kp is defined as the ratio of the products to the reactants, each raised to the power of their respective coefficients.
In this case, Kp = P(SO3)^2 / (P(SO2)^2 * P(O2)). Plug in the given values to get:
0.345 = (P(SO3)^2) / ((0.145)^2 * 0.455)
Now, let's put our math skills to the test and solve for P(SO3). Taking a deep breath, we can rearrange the equation to get:
P(SO3)^2 = 0.345 * (0.145)^2 * 0.455
P(SO3)^2 = 0.00275632
Finally, taking the square root of both sides, we find:
P(SO3) = √0.00275632
Drum roll, please...
P(SO3) ≈ 0.0525 atm
So, the equilibrium partial pressure of SO3 in the mixture is approximately 0.0525 atmospheres.
To find the equilibrium partial pressure of SO3 in the mixture, you can use the equilibrium constant expression for the reaction and the given partial pressures of SO2 and O2.
The equilibrium constant expression for a reaction is given by:
Kp = (P(SO3))^2 / (P(SO2))^2 * (P(O2))
From the given information, we know that:
Kp = 0.345
P(SO2) = 0.145 atm
P(O2) = 0.455 atm
Substituting these values into the equilibrium constant expression, we have:
0.345 = (P(SO3))^2 / (0.145)^2 * 0.455
Rearranging the equation, we can solve for P(SO3):
(P(SO3))^2 = 0.345 * (0.145)^2 * 0.455
Taking the square root of both sides, we get:
P(SO3) = √(0.345 * (0.145)^2 * 0.455)
Calculating this value, P(SO3) is approximately equal to 0.190 atm.
Therefore, the equilibrium partial pressure of SO3 in the mixture is 0.190 atm.