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Standard XII

Chemistry

Introduction

A vessel cont...

Question

A vessel contains 1 mole $P C l_{5 (g)}$ at 4 atm and 0.5 mole $P C l_{3}$ formed at equilibrium. Now, equilibrium pressure of mixture is (assume ideal behavior)

16 atm

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6 atm

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2 atm

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4.5 atm

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Solution

The correct option is B 16 atm
${P C l}_{5} ⇋ {P C l}_{3} + {C l}_{2} i n i t i a l l y 100 a t e q u l i b r i u m 1 - 0.5 0.5 0.5$
Total pressure = g
Initial pressure = 4 atm
$(\frac{0.5}{1.5}) g = 4 ⟹ g = 4 \times 3 = 12 K_{P} = \frac{P_{{P C l}_{3}} \cdot P_{{C l}_{2}}}{P_{{P C l}_{5}}} = \frac{(\frac{0.5}{1.5}) g \cdot (\frac{0.5}{1.5}) g}{(\frac{0.5}{1.5}) g} = \frac{1}{3} g = \frac{1}{3} (12) = 4$
Equilibrium Pressure = 4 $\times$ 4 = 16 atm

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Similar questions

Q. In an evacuated closed isolated chamber at

250^{o} C

.02

mole

P {C l}_{5}

and

.01

mole

{C l}_{2}

are mixed

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

. At equilibrium density of mixture was

2.49 g / L

and pressure was

1

atm. The number of total moles at equilibrium will be approximately:

a

moles of

{P C l}_{5}

are heated in a closed container to equilibrate

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

at a pressure of

P

atm. If

x

moles of

{P C l}_{5}

dissociate at equilibrium, then:

Q. For the given reaction,

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

Calculate

K_{p}

, if at equilibrium

2

moles each of

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

and

C l_{2}

is maintained at

250^{o} C

and a total pressure of

3 atm

Q. A quantity of

P C l_{5}

was heated in a 10 litre vessel at

250^{\circ} C

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

.
At equilibrium, the vessel contains 0.1 mole of

P C l_{5}

, 0.2 mole of

P C l_{3}

and 0.2 mole of

C l_{2}

. The equilibrium constant is :

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

. If 1 mole of

P C l_{5}

be put in a container of volume V litre and at equilibrium, x moles of it was decomposed, find its

K_{p}

and

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A vessel contains 1 mole PCl5(g) at 4 atm and 0.5 mole PCl3 formed at equilibrium. Now, equilibrium pressure of mixture is (assume ideal behavior)

The correct option is B 16 atmPCl5⇋PCl3+Cl2initially100atequlibrium1−0.50.50.5Total pressure = gInitial pressure = 4 atm(0.51.5)g=4⟹g=4×3=12KP=PPCl3⋅PCl2PPCl5=(0.51.5)g⋅(0.51.5)g(0.51.5)g=13g=13(12)=4Equilibrium Pressure = 4 × 4 = 16 atm

A vessel contains 1 mole $P C l_{5 (g)}$ at 4 atm and 0.5 mole $P C l_{3}$ formed at equilibrium. Now, equilibrium pressure of mixture is (assume ideal behavior)