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Derivation of Kp and Kc

In a study of...

Question

In a study of equilibrium
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ .
Starting with 2 mole $S O_{2}$ and 1.5 mole $O_{2}$ in 5 litre flask.Equilibrium mixture required 0.4 mole $K M n O_{4}$ in acidic medium.Hence $K_{C}$ is:

A

0.2

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B

5

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C

675

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D

None of these

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Solution

The correct option is B 5
Initially, number of moles of $S O_{2}, O_{2} a n d S O_{3}$ are 2, 1.5 and 0 respectively.
2 moles of $K M n O_{4}$ reacts with 5 moles of $S O_{2}$ .
0.4 moles of $K M n O_{4}$ will react with 1 moles of $S O_{2}$ .
Thus at equilibrium number of moles of $S O_{2}, O_{2} a n d S O_{3}$ are 1, 1 and 1 respectively.
The volume is 5 L.
Thus the equilibrium concentrations of $S O_{2}, O_{2} a n d S O_{3}$ are 0.2, 0.2 and 0.2 respectively.
The expression for the equilibrium constant is $K_{p} \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]} = \frac{{0.2}^{2}}{{0.2}^{2} \times 0.2} = 5$ .

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Similar questions

{S O}_{3}

was placed in a two litre vessel at a certain temperature. The following equilibrium was established in the vessel :

{2 S O}_{3} (g) ⇌ {2 S O}_{2} (g) + O_{2} (g)

The equilibrium mixture reacted with

0.2

K M n O_{4}

in acidic medium. Hence,

K_{c}

Q. For a reaction,

2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)}, 1.5

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0.35 m o l L^{- 1}

K_{c}

for the reaction would be ?

Q. A mixture of

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for the reaction,

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

m o l^{- 1} l i t r e

At equilibrium.
a) If no. of mole of

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in flask are same, how many mole of

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b) If no. of mole of

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in flask are twice the no.of mole of

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Q. Homogeneous equilibrium :

K_{c}

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A mixture of

S O_{3}, S O_{2}

O_{2}

gases is maintained at equilibrium in

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for the reaction,

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{2} (g)

100 m o l^{- 1}

litre. At equilibrium.
(a) If no. of mole of

S O_{3}

S O_{2}

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O_{2}

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(b) If no. of mole of

S O_{3}

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K

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