Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, PCl_5(g) leftharpoons PCl_3(g) + Cl_2(g) . If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl_5 is x , the partial pressure of PCl_3 will be:

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, PCl_5(g) leftharpoons PCl_3(g) + Cl_2(g) . If total pressure at equilibr - Chemistry MCQ AIEEE 2006 2026 | ExamTest.live

mediumMCQAIEEE 20062026Chemistry

1 attempts0% success rate1 mark

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g) . If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl_{5} is x, the partial pressure of PCl_{3} will be:

Reaction: PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g) Initial moles: 1, 0, 0 .At equilibrium: 1 - x, x, x .Total moles = (1 - x) + x + x = 1 + x .Partial pressure of PCl_{3} = Mole fraction of PCl_{3} \times Total Pressure $ p_{PCl_{3}} = \frac{x}{1 + x} P $

Try it yourself before checking the explanation above.

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g) . If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl_{5} is x, the partial pressure of PCl_{3} will be:

(\frac{x}{x + 1}) P

(\frac{2 x}{1 - x}) P

(\frac{x}{x - 1}) P

(\frac{x}{1 - x}) P

Practice with timed mock tests and track your performance across Chemistry.