PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). In the above reaction, the partial pressure of PCl₃, Cl₂ and PCl₅ are 0.3, 0.2 and 0.6 atm, respectively. If partial pressure of PCl₃ and Cl₂ was increased twice at the new

Question

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). In the above reaction, the partial pressure of PCl₃, Cl₂ and PCl₅ are 0.3, 0.2 and 0.6 atm, respectively. If partial pressure of PCl₃ and Cl₂ was increased twice at the new equilibrium, what will be the new partial pressure of PCl₅ (in atm)?

Prashant Kotian · Accepted Answer

Correct answer: C. The equilibrium constant Kp = (P_PCl₃ × P_Cl₂) / P_PCl₅ = (0.3 × 0.2) / 0.6 = 0.1 atm. If at the new equilibrium P_PCl₃ is 0.6 atm and P_Cl₂ is 0.4 atm (increased twice), we use Kp to find P_PCl₅: 0.1 = (0.6 × 0.4) / P_PCl₅ => P_PCl₅ = 0.24 / 0.1 = 2.4 atm. Therefore, correct answer is C.

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