Equilibrium involving SO2(g), O2(g) & SO3(g) is important in sulfuric acid production. When a 0.0200 mol sample of SO3 is introduced into an evacuated 1.52 L vessel at 900 K, 0.0142 mol, SO3 is found to be present at equilibrium. What is the value of Kp for the dissociation of SO3(g) at 900 K?
Use PV = nRT and solve for pSO3 at the beginning and at equilibrium. I obtained about 1 for pSO3 initially but you need to get a more accurate answer.
Then
..........2SO3 ==> 2SO2 + O2
I.........1.0.......0......0
C.........-2x.......2x.....x
E........1.0-2x.....2x.....s
The problem tells you that 1.0-2x (or whatever more accurate answer you get for that 1.0 number) = pSO3 at equilibrium. You get that number from the 0.142 mol and PV = nRT.
Evaluate each equilibrium value from above and substitute into the Kp expression. Solve for Kp.
0.236
To find the value of Kp for the dissociation of SO3(g) at 900 K, we need to use the given information about the initial and equilibrium amounts of the reactant.
The balanced chemical equation for the dissociation of SO3(g) is:
2 SO3(g) ⇌ 2 SO2(g) + O2(g)
From the given information, we know that initially, there were 0.0200 mol of SO3(g) in a 1.52 L vessel. At equilibrium, there is 0.0142 mol of SO3(g) present.
Using the ideal gas law, we can calculate the initial and equilibrium partial pressures of SO3(g). At the given temperature, the ideal gas law can be expressed as:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in K)
For the initial conditions:
P_initial × V = n_initial × R × T
Solving for P_initial:
P_initial = (n_initial × R × T) / V
Substituting the given values:
P_initial = (0.0200 mol × 0.0821 L·atm/(mol·K) × 900 K) / 1.52 L
Calculating P_initial:
P_initial ≈ 0.484 atm
Similarly, for the equilibrium conditions:
P_equilibrium × V = n_equilibrium × R × T
Solving for P_equilibrium:
P_equilibrium = (n_equilibrium × R × T) / V
Substituting the given values:
P_equilibrium = (0.0142 mol × 0.0821 L·atm/(mol·K) × 900 K) / 1.52 L
Calculating P_equilibrium:
P_equilibrium ≈ 0.311 atm
Now, we can use the equilibrium partial pressures of the reactants to calculate the value of Kp. For the given balanced equation:
2 SO3(g) ⇌ 2 SO2(g) + O2(g)
The expression for Kp is given by:
Kp = (P_SO2^2 × P_O2) / P_SO3^2
Substituting the calculated values:
Kp = (0.311^2 × 0.484) / (0.0142^2)
Calculating Kp:
Kp ≈ 238.539
Therefore, the value of Kp for the dissociation of SO3(g) at 900 K is approximately 238.539.