A container of capacity V L contains an equilibrium mixture that consists of 2 moles each of PCl₅, P — Chemical Equilibrium Chemistry Question
Question
A container of capacity V L contains an equilibrium mixture that consists of 2 moles each of PCl₅, PCl₃ and Cl₂ (all as gases). The pressure is 30.3975 kPa and temperature is T K. A certain amount ('x' mole) of Cl₂(g) is now introduced keeping the pressure and temperature constant, until the equilibrium volume becomes 2V L. The value of '90x' is
💡 Solution & Explanation
Initial eq: PCl₅ ⇌ PCl₃ + Cl₂. Initial moles = 2 + 2 + 2 = 6. K_p = P_PCl₃ P_Cl₂ / P_PCl₅ = (1/3 P)(1/3 P) / (1/3 P) = P/3. Since pressure and temperature are kept constant, volume is proportional to total moles. The volume doubles to 2V, so the final total moles must double to 12. Let y moles of Cl₂ react backward. New eq moles: PCl₅ = 2+y, PCl₃ = 2-y, Cl₂ = 2+x-y. Total moles = (2+y) + (2-y) + (2+x-y) = 6+x-y = 12, so x - y = 6. Using K_p = P/3: [((2-y)/12)P × ((2+x-y)/12)P] / [((2+y)/12)P] = P/3. This simplifies to (2-y)(2+x-y) / (12(2+y)) = 1/3 => (2-y)(2+x-y) = 4(2+y). Substituting x-y = 6 (so 2+x-y = 8): (2-y)(8) = 8 + 4y => 16 - 8y = 8 + 4y => 12y = 8 => y = 2/3. Then x = y + 6 = 20/3. The value of 90x = 90(20/3) = 600. Therefore, correct answer is 0600.