The equilibrium constant Kp of following reaction is 900 atm at 800K. ...
Equilibrium Reaction
The reaction under consideration is:
2SO2 + O2 ⇌ 2SO3
Given:
- Kp = 900 atm at 800 K
- Initial partial pressures: SO3 = 1 atm, O2 = 2 atm
Setting Up the ICE Table
To find the equilibrium partial pressures, we set up an ICE (Initial, Change, Equilibrium) table:
- Initial:
- P(SO3) = 1 atm
- P(O2) = 2 atm
- P(SO2) = 0 atm
- Change:
Let x be the change in pressure of SO2 formed at equilibrium.
- 2SO2: +2x
- O2: -x
- 2SO3: -2x
- Equilibrium:
- P(SO3) = 1 - 2x
- P(O2) = 2 - x
- P(SO2) = 2x
Applying the Equilibrium Constant
Using the expression for Kp:
Kp = (P(SO3)^2) / (P(SO2)^2 * P(O2))
Substituting the equilibrium pressures:
900 = ((1 - 2x)^2) / ((2x)^2 * (2 - x))
Solving for x
This equation can be solved to find the value of x. After simplifying and solving the quadratic equation, you will find a suitable value for x.
Final Partial Pressures
Once x is determined, substitute back to find the equilibrium partial pressures:
- P(SO3) = 1 - 2x
- P(O2) = 2 - x
- P(SO2) = 2x
Calculate these values to obtain the final equilibrium partial pressures of SO3, O2, and SO2 at 800 K.
This approach ensures the system reaches equilibrium while adhering to the given conditions.