The equilibrium $P_{4} (s) + 6 C l_{2} (g) ⇋ 4 P C l_{3} (g)$ Is attained by mixing equal moles of $P_{4} a n d C l_{2}$ inana evacuated vessel. Then at equilibrium

[Cl₂ ]> [PCl₃]

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[Cl₂]>[P₄]

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[P₄] > [Cl₂]

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[PCl₃]> [P₄]

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Solution

The correct option is C [P₄] > [Cl₂]
Solution:Correct option is c).

$\begin{matrix} P_{4} (s) + & 6 C l_{2} (g) ⇋ & 4 P C l_{3} (g) I n i t i a l m o l e s & 1 & 1 & 0 A t e q u i l i b r i u m & 1 - x & 1 - 6 x & 4 x \end{matrix}$

$A s (1 - x) > (1 - 6 x), H e n c e [P_{4}] > [C l_{2}] .$

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Similar questions

Q. The equilibrium

P_{4} (s) + 6 C l_{2} (g) ⇋ 4 P C l_{3} (g)

Is attained by mixing equal moles of

P_{4} a n d C l_{2}

inana evacuated vessel. Then at equilibrium

Q. The equilibrium:

P_{4} (g) + 6 {C l}_{2} (g) ⇌ 4 P {C l}_{3} (g)

is attained by mixing equal moles of

P_{4}

and

{C l}_{2}

in an evacuated vessel. Then at equilibrium:

P_{4} + C l_{2} (e x c e s s) ⟶

P_{4} (e x c e s s) + C l_{2} \to X

P_{4} + C l_{2} (e x c e s s) \to Y

Then the correct statement is:

$P C l_{5} ⇌ P C l_{3} + C l_{2}$ in the reversible reaction, the moles of $P C l_{5}$ , $P C l_{3}$ and $C l_{2}$ are a, b and c respectively at equilibrium and total pressure is P, then value of $K_{p}$ is:

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The equilibrium P4 (s)+6Cl2(g)⇋ 4 PCl3 (g) Is attained by mixing equal moles of P4 and Cl2 inana evacuated vessel. Then at equilibrium

The correct option is C [P4] > [Cl2]Solution:Correct option is c). P4(s) +6 Cl2(g) ⇋4 PCl3(g)Initial moles110At equilibrium1−x1−6x4x As(1−x)>(1−6x),Hence[P4]>[Cl2].

The equilibrium $P_{4} (s) + 6 C l_{2} (g) ⇋ 4 P C l_{3} (g)$ Is attained by mixing equal moles of $P_{4} a n d C l_{2}$ inana evacuated vessel. Then at equilibrium

The correct option is C [P₄] > [Cl₂]
Solution:Correct option is c).

$\begin{matrix} P_{4} (s) + & 6 C l_{2} (g) ⇋ & 4 P C l_{3} (g) I n i t i a l m o l e s & 1 & 1 & 0 A t e q u i l i b r i u m & 1 - x & 1 - 6 x & 4 x \end{matrix}$

$A s (1 - x) > (1 - 6 x), H e n c e [P_{4}] > [C l_{2}] .$