The equilibrium constant for the reaction PCl5g- PCl3g - Cl2g is 2-4- 10-3- Determine the equilibrium constant for the reverse reaction at same temperature-

PC I_ 3 ⇌ PC I_ 3 +C l_ 2 K_ C =2.4× 10^ 3 for Reverse Reaction K^1c = 1/Kc⇒1/2.4 ×10^ 3⇒1000/2.4 = 416.67 ...

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Introduction

The equilibri...

Question

The equilibrium constant for the reaction
$P C l_{5 (g)} ⇌ P C l_{3 (g)} + C l_{2 (g)}$ is $2.4 \times 10^{- 3}$ . Determine the equilibrium constant for the reverse reaction at same temperature.

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Solution

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P C l_{5} (g) ⇋ P C l_{3} (g) + C l_{2} (g)

2.4 \times 10^{- 3}

P C l_{3} (g) + C l_{2} (g) ⇋ P C l_{5} (g)

At a given temperature, the equilibrium constant for reaction

$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ is $2.4 \times 10^{- 3}$ . At the same temperature, the

equilibrium constant for reaction $P C l_{3} (g) + C l_{2} (g ⇌ (P C l_{5} (g)$ is

P C l_{5}, P C l_{3}

C l_{2}

0.5 \times 10^{- 3} M P C L_{5}

1.5 \times 10^{- 2} M P C L_{3}

3 \times 10^{- 2} M C l_{2}

K_{c}

P C L_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

P {C l}_{5} (g) ⇌ P {C l}_{3} (g) + {C l}_{2} (g)

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