4 mole of PCl5 is heated in a 2L vessel to establish an equilibrium [PCl5(g)⇌ PCl3(g)+Cl2(g)]. At equilibrium molar concentration of Cl2 is one. The value of Kc for the above equilibrium is

PCl54(g)⇌PCl30+Cl204−x2 x2x2As x2 = 1 ∴ x =2 1 1 1Kc=1×11= 1 mole/L

4 mole of PCl5 is heated in a 2L vessel to establish an equilibrium [PCl5(g)⇌ PCl3(g)+Cl2(g)]. At equilibrium molar concentration of Cl2 is one. The value of Kc for the above equilibrium is

PCl54(g)⇌PCl30+Cl204−x2 x2x2As x2 = 1 ∴ x =2 1 1 1Kc=1×11= 1 mole/L

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4 mole of PCl₅ is heated in a 2L vessel to establish an equilibrium $[\underset{(g)}{P C l_{5}} \overset{}{⇌} \underset{(g)}{P C l_{3}} + \underset{(g)}{C l_{2}}] .$ At equilibrium molar concentration of Cl₂ is one. The value of K_c for the above equilibrium is

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$\frac{1}{6} m o l e / L$

$\frac{1}{3} m o l e / L$

$1 m o l e / L$

$2 M$

answer is C.

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Detailed Solution

$\underset{4}{P C l_{5}} (g) ⇌ \underset{0}{P C l_{3}} + \underset{0}{C l_{2}}$
$\frac{4 - x}{2} \frac{x}{2} \frac{x}{2}$
As $\frac{x}{2} = 1 ∴ x = 2$
1 1 1
$K_{c} = \frac{1 \times 1}{1} = 1 m o l e / L$

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