Calculate the value of equilibrium constant when 2 moles of PCl5 is heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of PCl5 is dissociated into PCl3 and Cl2 .

Question

Calculate the value of equilibrium constant when 2 moles of PCl5 is heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of PCl5 is dissociated into PCl3 and Cl2 . (A) 0.266 (B) 0.53 (C) 2.66 (D) 5.3

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-dxlt8efemso7.png" /> Volume of container = 2 litre . Kc=((2 × 40/100 × 2) × (2 × 40/100 × 2)/(2 × 60)100 × 2)=0.266

Q. Calculate the value of equilibrium constant when 2 moles of $PC l_{5}$ is heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of $PC l_{5}$ is dissociated into $PC l_{3}$ and $C l_{2}$ .

0.266

0.53

2.66

5.3

Volume of container = $2 l i t re$ .

$K_{c} = \frac{\frac{2 \times 40}{100 \times 2} \times \frac{2 \times 40}{100 \times 2}}{\frac{2 \times 60}{100 \times 2}} = 0.266$

Q. Calculate the value of equilibrium constant when 2 moles of PCl5​ is heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of PCl5​ is dissociated into PCl3​ and Cl2​ .

0.266

0.53

2.66

5.3

Volume of container = 2litre . Kc​=100×22×60​100×22×40​×100×22×40​​=0.266

Q. Calculate the value of equilibrium constant when 2 moles of $PC l_{5}$ is heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of $PC l_{5}$ is dissociated into $PC l_{3}$ and $C l_{2}$ .

Volume of container = $2 l i t re$ .

$K_{c} = \frac{\frac{2 \times 40}{100 \times 2} \times \frac{2 \times 40}{100 \times 2}}{\frac{2 \times 60}{100 \times 2}} = 0.266$