Given the following reaction at equilibrium,if K_p = 1.10 at 250.0 °C,K_c = ________. PCl₅ (g) PCl₃ (g)+ Cl₂ (g)

Question

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The equilibrium constant expression for the given reaction is: K_p = [PCl₃][Cl₂]/[PCl₅] We are given K_p = 1.10 at 250.0 °C. Now, if x is the concentration of PCl₅ that reacts to form PCl₃ and Cl₂ at equilibrium, then the equilibrium concentrations of PCl₅, PCl₃, and Cl₂ can be expressed in terms of x as follows: [PCl₅] = [PCl₅]0 - x [PCl₃] = x [Cl₂] = x where [PCl₅]0 is the initial concentration of PCl₅. Substituting these concentrations in the equilibrium constant expression, we get: K_p = x^2/([PCl₅]0 - x) Solving for x using the quadratic formula, we get: x = ([PCl₅]0K_p)/(1 + K_p) = [PCl₅]eq Substituting the given values, we get: [PCl₅]eq = (1.10 × 1.00)/(1 + 1.10) = 0.556 Therefore, the equilibrium concentrations of PCl₅, PCl₃, and Cl₂ are: [PCl₅]eq = 0.556 mol/L [PCl₃]eq = [Cl₂]eq = 0.556 mol/L The equilibrium constant expression for the reverse reaction can be written as: K_c = [PCl₅]/[PCl₃][Cl₂] Substituting the equilibrium concentrations, we get: K_c = (0.556)/(0.556 × 0.556) = 2.56 × 10^-12 Therefore, the answer is (B).

Given the Following Reaction at Equilibrium,if Kp = 1

Given the Following Reaction at Equilibrium,if K_p = 1