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:
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
PPCl5 = we need to calculate
PCl3 = 1.40 atm ,
PCl2 = 4.07 atm
Kp =321
Step 3: Calculation of partial pressure of PPCl5
PCl5 (g) <-----> PCl3 (g) + Cl2 (g)
we know, the Kp expression can be written as
Kp = ( PCl3 × PCl2 ) / PPCl5
on substituting the values of partial pressure and Kp
=> 321 = (1.43 atm ) × 4.07 atm ) / PPCl5
=> PPCl5 = (1.43 atm ) × 4.07 atm ) / 321
=> PPCl5 = ( 5.8201 atm ) / 321
=> PPCl5 = 0.01813 ≈ 1.81E-2
Hence, the equilibrium partial pressure of PCl5 is 1.81E-2 atm ( i.e 1.81×10-2 atm )
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