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Standard XII

Chemistry

Introduction to Le Chatelier's Principle

When sulphur ...

Question

When sulphur in the form of $S_{8}$ is heated at $900 K$ , the initial pressure of one $a t m$ of $S_{(8)}$ falls by 29% at equilibrium. This is because of conversion of some $S_{8}$ to $S_{2} .$ Find the value of equilibrium constant for this reaction.

1.16

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0.71

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2.55

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5.1

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Solution

The correct option is C $2.55$
According to the given Reaction,
$\begin{matrix} S_{8} (g) & ⇌ & 4 S_{2} (g) A t s t a r t & 1 & 0 A t e q b m & 1 - 0.29 & 4 \times 0.29 = 0.71 a t m & = 1.16 a t m \end{matrix}$

$K_{p} = \frac{[p S_{2}]^{4}}{[p S_{8}]}$

putting the values,
$K_{p} = \frac{(1.16)^{4}}{(0.71)} = 2.55$

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Introduction to Le Chatelier's Principle

Standard XII Chemistry

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When sulphur in the form of S8 is heated at 900 K, the initial pressure of one atm of S(8) falls by 29% at equilibrium. This is because of conversion of some S8 to S2. Find the value of equilibrium constant for this reaction.

The correct option is C 2.55According to the given Reaction, S8 (g)⇌4S2 (g)At start10At eqbm1−0.294×0.29=0.71 atm=1.16 atm Kp=[pS2]4[pS8] putting the values, Kp=(1.16)4(0.71)=2.55

When sulphur in the form of $S_{8}$ is heated at $900 K$ , the initial pressure of one $a t m$ of $S_{(8)}$ falls by 29% at equilibrium. This is because of conversion of some $S_{8}$ to $S_{2} .$ Find the value of equilibrium constant for this reaction.