what is the equilibrium partial pressure of C? Kp for this reaction is 11.2. A (g) ⇌ B (g) + C (g)

Question

An empty steel vessel is charged with 0.280 atm of C and 0.280 atm of A. Once the system reaches equilibrium according to the reaction below, what is the equilibrium partial pressure of C? Kp for this reaction is 11.2. A (g) ⇌ B (g) + C (g)

Answer 1

For the reaction A (g) ⇌ B (g) + C (g), the equilibrium constant expression, Kp, is:

Kp = PC/(PA × PB)

where PA, PB, and PC are the partial pressures of A, B, and C at equilibrium.

We are given the value of Kp as 11.2, and the initial partial pressures of A and C as 0.280 atm each. Let the equilibrium partial pressure of C be represented by PC atm. At equilibrium, the partial pressure of A and B will each decrease by PC atm (since 1 mole of C is produced for every 1 mole of A consumed, and 1 mole of C is produced for every 1 mole of B consumed). Therefore, the partial pressures of A and B at equilibrium will be 0.280 - PC atm each.

Substituting these equilibrium partial pressures into the Kp expression, we get:

11.2 = PC/[(0.280 - PC) × (0.280 - PC)]

Simplifying and solving for PC, we get:

PC = 0.0916 atm

Therefore, the equilibrium partial pressure of C is 0.0916 atm.