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Standard XII

Chemistry

Charles' Law

2.0 mole of ...

Question

2.0 mole of $P C l_{5}$ were introduced in a vessel of 5.0 L capacity of a particular temperature. At equilibrium, $P C l_{5}$ was found to be 35% dissociated into $P C l_{3}$ and $C l_{2}$ . The value of $K_{c}$ for the reaction:-
$P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$

0.075

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0.377

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1.33

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13.3

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Solution

The correct option is A 0.075
$P C l_{5} ⇌ P C l_{3} + C l_{2}$
$2$ $0$ $0$
$(- 2 α)$ $2 α$ $2 α$
$E$ $2 (1 - α)$ $2 α$ $2 α$
Degree of dissociation= $35$ %= $0.35$
Volume of equilibrium mixture= $5 L$
$[P C l_{5}] = \frac{2 (1 - α)}{V} = \frac{2 (1 - 0.35)}{5} = 0.26$ mol/lit
$[P C l_{3}] = \frac{2 α}{V} = \frac{2 (0.35)}{5} = 0.14$ mol/lit
$[C l_{2}] = \frac{2 α}{V} = \frac{2 \times 0.35}{5} = 0.14$ mol/lit
$K_{C} = \frac{0.14 \times 0.14}{0.26} = 0.075$ mol/lit

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Similar questions

Q. 2.0 mol of

P C l_{5}

were introduced in a vessel of 5.0 L capacity at a particular temperature. At equilibrium,

P C l_{5}

was found to be 35% dissociated into

P C l_{3}

and

C l_{2}

. The value of

K_{C}

for the reaction is-

A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 × 10^–1 mol L^–1. If value of K_c is 8.3 × 10^–3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?

Q. Two moles of

P {C l}_{5}

were introduced in a

2

litre flask and heated at

600 K

to attain equilibrium.

P {C l}_{5}

was found to be

40

% dissociated into

P {C l}_{3}

{C l}_{2}

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K_{c}

Q. A sample of pure

{P C l}_{5}

was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of

{P C l}_{5}

was found to be

0.5 \times 10^{- 1} m o l L^{- 1}

. If value of

K_{c}

8.3 \times 10^{- 3}

, what are the concentrations of

{P C l}_{3}

and

{C l}_{2}

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

A sample of pure $P C l_{5}$ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl_5 was found to be $0.5 \times 10^{- 1} m o l L^{- 1} . I f v a l u e o f K_{c} i s 8.3 \times 10^{- 3}$ , what are the concentrations of $P C l_{3}$ and $C l_{2}$ at equilibrium?\\
$P C l_{5} (g) \leftrightarrow P C l_{3} (g) + C l_{2} (g)$

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2.0 mole of PCl5 were introduced in a vessel of 5.0 L capacity of a particular temperature. At equilibrium, PCl5 was found to be 35% dissociated into PCl3 and Cl2. The value of Kc for the reaction:-PCl3(g)+Cl2(g)⇌PCl5(g)

The correct option is A 0.075 PCl5⇌PCl3+Cl2 2 0 0 (−2α) 2α 2αE 2(1−α) 2α 2αDegree of dissociation= 35%=0.35Volume of equilibrium mixture= 5L[PCl5]=2(1−α)V=2(1−0.35)5=0.26 mol/lit[PCl3]=2αV=2(0.35)5=0.14 mol/lit[Cl2]=2αV=2×0.355=0.14 mol/lit KC=0.14×0.140.26=0.075 mol/lit

2.0 mole of $P C l_{5}$ were introduced in a vessel of 5.0 L capacity of a particular temperature. At equilibrium, $P C l_{5}$ was found to be 35% dissociated into $P C l_{3}$ and $C l_{2}$ . The value of $K_{c}$ for the reaction:-
$P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$