PCl5→ PCl3 + Cl2; Kc = 1.844. 3.0 moles of PCl5 is introduced in a 1 L closed reaction vessel at 380K. The number of moles of PCl5 at equilibrium is × 10–3. (Round off to the nearest integer)

PCl5→ PCl3 + Cl2 KC =PCl3Cl2PCl5=x23-x=1.604Moles of PCl5= 3 – 1.604 = 1.396 = 1400Hence, the correct option 1400.

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$P C l_{5} \to P C l_{3} + C l_{2}$ ; Kc = 1.844. 3.0 moles of PCl₅ is introduced in a 1 L closed reaction vessel at 38⁰K. The number of moles of PCl₅ at equilibrium is × 10^–3. (Round off to the nearest integer)

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answer is 1400.

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$P C l_{5} \to P C l_{3} + C l_{2} K_{C} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = \frac{x^{2}}{3 - x} = 1.604$

Moles of $P C l_{5}$ = 3 – 1.604 = 1.396 = 1400

Hence, the correct option 1400.

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