Question

For the reaction PCl_5(g) ⇌ PCl_3(g) + Cl_2(g) at equilibrium, K_p = 321. The equilibrium partial pressures of PCl_3 and Cl_2 are 1.43E0 atm and 4.07E0 atm, respectively. What is the equilibrium partial pressure of PCl_5?

Answer 1

Answer:

Step 1: Explanation

Kp is the equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures

Step 2: Write the balanced the chemical equation

PCl₅ (g) <-----> PCl₃ (g) + Cl₂ (g)

All value of partial pressure at equilibrium is

P_PCl5 =  we need to calculate

P_Cl3 = 1.40 atm ,

P_Cl2 = 4.07  atm

K_p =321

Step 3: Calculation of partial pressure of P_PCl5

PCl₅ (g) <-----> PCl₃ (g) + Cl₂ (g)

we know, the K_p expression can be written as

K_p = ( P_Cl3 × P_Cl2 ) / P_PCl5

on substituting the values of partial pressure and K_p

=> 321 = (1.43 atm ) × 4.07 atm ) / P_PCl5

=> P_PCl5 = (1.43 atm ) × 4.07 atm ) / 321

=>   P_PCl5 = ( 5.8201 atm ) / 321

=> P_PCl5 = 0.01813 ≈ 1.81E-2

Hence, the equilibrium partial pressure of PCl₅ is 1.81E-2 atm ( i.e 1.81×10^-2 atm )