The equilibrium constant for the equilibrium PCl 5(g) leftharpoons PCl 3(g)+ Cl 2(g) at a particular temperature is 2 × 10-2 mol L -1. The number of moles of PCl 5 that must be taken in a one litre flask at the same temperature to obtain a concentration of 0.20 mole of chlorine at equilibrium is

Question

The equilibrium constant for the equilibrium PCl 5(g) leftharpoons PCl 3(g)+ Cl 2(g) at a particular temperature is 2 × 10-2 mol L -1. The number of moles of PCl 5 that must be taken in a one litre flask at the same temperature to obtain a concentration of 0.20 mole of chlorine at equilibrium is (A) 2.2 (B) 2.0 (C) 1.8 (D) 0.2

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/193096f2da3ce937421b94197bd44e10-.png" /> So, equilibrium constant, K=([ PCl 3][ Cl 2]/[ PCl 5]) [given, K=2 × 10-2 mol / L] ∴ 2 × 10-2= (0.2 × 0.2/x-0.2) ⇒ x=2.2 moles

2.2

2.0

1.8

0.2

0.1

So, equilibrium constant,

$K = \frac{[ PC l _{3} ] [ C l _{2} ]}{[ PC l _{5} ]}$

[given, $K = 2 \times 1 0^{- 2} m o l / L$ ]

$∴ 2 \times 1 0^{- 2} = \frac{0.2 \times 0.2}{x - 0.2}$

$\Rightarrow x = 2.2$ moles

Q. The equilibrium constant for the equilibrium PCl5​(g)⇌PCl3​(g)+Cl2​(g) at a particular temperature is 2×10−2molL−1. The number of moles of PCl5​ that must be taken in a one litre flask at the same temperature to obtain a concentration of 0.20 mole of chlorine at equilibrium is

2.2

2.0

1.8

0.2

0.1

So, equilibrium constant, K=[PCl5​][PCl3​][Cl2​]​ [given, K=2×10−2mol/L] ∴2×10−2=x−0.20.2×0.2​ ⇒x=2.2 moles

So, equilibrium constant,

$K = \frac{[ PC l _{3} ] [ C l _{2} ]}{[ PC l _{5} ]}$

[given, $K = 2 \times 1 0^{- 2} m o l / L$ ]

$∴ 2 \times 1 0^{- 2} = \frac{0.2 \times 0.2}{x - 0.2}$

$\Rightarrow x = 2.2$ moles