the value of Kp after the decomposition of sulfur trioxide (S03). 2SO3 (g)--2SO2 (g) +O2 (g) is 1.79 x10^-5 at 350 degree celcius. if SO3 initially at 0.200 atm pressure comes to equilibrium in an otherwise empty tank at 350 degree celcius, determine the partial pressure of O2 at equilibrium.
To determine the equilibrium partial pressure of O2 (g) at 350 degrees Celsius, we can use the expression for the equilibrium constant (Kp):
Kp = (PO2)^n / (PSO3)^m
Where n and m are the coefficients of O2 and SO3 in the balanced equation, respectively.
From the balanced equation:
2SO3 (g) -> 2SO2 (g) + O2 (g)
We can see that n = 1 (coefficient of O2) and m = 2 (coefficient of SO3).
Given that Kp = 1.79 x 10^-5, we can substitute the known values into the equation:
1.79 x 10^-5 = (PO2)^1 / (0.200 atm)^2
To solve for PO2, let's rearrange the equation:
PO2 = (1.79 x 10^-5) * (0.200 atm)^2
PO2 = 7.16 x 10^-8 atm
Therefore, the partial pressure of O2 (g) at equilibrium is approximately 7.16 x 10^-8 atm.
To determine the partial pressure of O2 at equilibrium, we need to use the equilibrium constant expression, Kp, for the given reaction: SO3 (g) ⇌ SO2 (g) + O2 (g).
The equilibrium constant expression for this reaction is given by:
Kp = (pSO2^2 * pO2) / pSO3^2
where pSO2, pO2, and pSO3 represent the partial pressures of SO2, O2, and SO3, respectively.
Given that the value of Kp is 1.79 x 10^-5 and that the partial pressure of SO3 is 0.200 atm, we can use the following equation to determine the partial pressure of O2 at equilibrium:
1.79 x 10^-5 = (pSO2^2 * pO2) / (0.200^2)
Now, we need to solve this equation to find the partial pressure of O2.
Rearranging the equation, we have:
pO2 = (1.79 x 10^-5 * 0.200^2) / pSO2^2
However, we don't have the partial pressure of SO2, so we need additional information to solve the problem.