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Solubility Product

2 moles each ...

Question

$2$ moles each of $S O_{3}, C O, S O_{2}$ and $C O_{2}$ is taken in a one litre vessel. If $K_{c}$ for
$S O_{3} (g) + C O (g) ⇌ S O_{2} (g) + C O_{2} (g)$ is $\frac{1}{9}$ then:

Total number of moles of gases at equilibrium is less than

8

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At equilibrium, the sum of moles of

S O_{3}

and

C O_{2}

4

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At equilibrium

[\frac{n (S O_{2})}{n (C O)}] < 1

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Both (B) and (C)

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Solution

The correct option is D Both (B) and (C)
Given reaction is, $S O_{3} (g) + C O (g) ⇌ S O_{2} (g) + C O_{2} (g)$
We know, $K_{c} = \frac{[C O_{2}] [S O_{2}]}{[S O_{3}] [C O]}$
Given $K_{c} = \frac{1}{9}$
Since $K_{c}$ has a value which is less than one, thus backward reaction will be favoured.

$\begin{matrix} S O_{3} (g) & + & C O (g) & ⇌ & S O_{2} (g) & + & C O_{2} (g) Initially & 2 & 2 & 2 & 2 At equilibrium & (2 + x) & (2 + x) & 2 - x & 2 - x \end{matrix}$
$K_{c} = \frac{[C O_{2}] [S O_{2}]}{[S O_{3}] [C O]}$
$\Rightarrow \frac{1}{9} = (\frac{2 - x}{2 + x})^{2}$
$\Rightarrow 9 = (\frac{2 + x}{2 - x})^{2} \Rightarrow x = 1$
Therefore, total number of moles of gases at equilibrium
$= 3 + 3 + 1 + 1 = 8$
At equilibrium, the sum of moles of $S O_{2}$ and $C O_{2}$ is
$= n (S O_{3}) + n (C O_{2}) = 4$

$\frac{n (S O_{2})}{n (C O)} = \frac{1}{3} < 1$
Therefore, (D) option is correct.

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Similar questions

2

mole each of

S O_{3}, C O, S O_{2}

and

C O_{2}

is taken in a one lit. vessel. If

K_{c}

for

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

1 / 9

then:

Q. 2 moles each of

S O_{3} (g), C O (g), S O_{2} (g) a n d C O_{2} (g)

is taken in a one lit. vessel. If

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

is 1/9 then :

Q. A mixture of

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

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2 moles each of SO3, CO, SO2 and CO2 is taken in a one litre vessel. If Kc for SO3(g)+CO(g)⇌SO2(g)+CO2(g) is 19 then:

$2$ moles each of $S O_{3}, C O, S O_{2}$ and $C O_{2}$ is taken in a one litre vessel. If $K_{c}$ for
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