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Derivation of Kp and Kc

1 mole of P...

Question

1 mole of $P C l_{3} (g)$ with 2 moles of $C l_{2} (g)$ were placed in a 3 lit vessel and heated to 400K. When equilibrium was established, only $0.7$ moles of $P C l_{3}$ remained. What is the value of equilibrium constant $K_{c}$ for the reaction $P C l_{3} + C l_{2} \Leftrightarrow P C l_{5}$ ?

A

0.25

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B

1.31

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C

0.76

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D

2.6

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Solution

The correct option is B 0.76
The following table lists the number of moles of different species.

$P C l_{3}$
$C l_{2}$
$P C l_{5}$
Initial
1
2
0
Change
-x
-x
x
Equilibrium
1-x
2-x
x

But at equilibrium, $[P C l_{3}] = 0.7 = 1 - x ⟹ x = 0.3$ .
Thus, the number of moles of $P C l_{3}, C l_{2}$ and $P C l_{5}$ are $0.7, 1.7$ and $0.3$ respectively.
The molar concentrations of $P C l_{3}, C l_{2}$ and $P C l_{5}$ are $\frac{0.7}{3} M, \frac{1.7}{3} M$ and $\frac{0.3}{3} M$ respectively.

So, According to definition of equilibrium constant,
So, $K_{c} = \frac{(0.3) (3)}{(1.7) (0.7)} = 0.76$ .

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Similar questions

Q. A quantity of

P C l_{5}

was heated in a 10 litre vessel as

250^{o} 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}

P C l_{3}

and 0.2 mole of

C l_{2}

. The equilibrium constant of the reaction is:

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}

P C l_{3}

and 0.2 mole of

C l_{2}

. The equilibrium constant is :

Q. A quantity of

P {C l}_{5}

was heated in a

10

litre vessel at

250^{o} C

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

At equilibrium the vessel contains

0.25

P {C l}_{5}

0.3

P {C l}_{3}

0.30

{C l}_{2}

(a) Compute the equilibrium constant for the formation of

P {C l}_{5}

250^{o} C

Δ G^{o}

for the dissociation of

P {C l}_{5}

log 0.036 = - 1.443

Q. Two moles of

P C l_{5}

327^{\circ} C

in a closed two litre vessel and when equilibrium was achieved,

P C l_{5}

was found to be 40% disociated into

P C l_{3}

C l_{2}

. Calcutta the equilibrium constant

(K_{c})

for the reaction.

0.6

P C l_{5}

0.3

P C l_{3}

0.5

C l_{2}

1

L flask to obtain the following equilibrium:

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

If the equilibrium constant

K_{c}

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0.2

. Predict the direction of the reaction.

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