For the reaction 2 A(g) ⇌ B(g) + 2 C(g), a reaction vessel initially contains only A at a pressure of PA = 255 mmHg. At equilibrium, PA = 55 mmHg. Calculate the value of Kp. (Assume no changes in volume or temperature.)

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1. Identify the initial and equilibrium pressures of A. Initially, the pressure of A is 255 mmHg, and at equilibrium, it is 55 mmHg.
2. Determine the change in pressure of A. The change in pressure, \( \Delta P_A \), is the initial pressure minus the equilibrium pressure: \( \Delta P_A = 255 \text{ mmHg} - 55 \text{ mmHg} \).
3. Use the stoichiometry of the reaction to find the changes in pressure for B and C. Since 2 moles of A produce 1 mole of B and 2 moles of C, the change in pressure for B is \( \Delta P_B = \frac{1}{2} \Delta P_A \) and for C is \( \Delta P_C = \Delta P_A \).
4. Calculate the equilibrium pressures of B and C. Since B and C are initially absent, their equilibrium pressures are equal to their changes in pressure: \( P_B = \Delta P_B \) and \( P_C = \Delta P_C \).
5. Write the expression for \( K_p \) using the equilibrium pressures: \( K_p = \frac{P_B \cdot P_C^2}{P_A^2} \), and substitute the equilibrium pressures to find \( K_p \).