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Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

One mole of ...

Question

One mole of ${S O}_{3}$ 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 $0.2$ mole $K M n O_{4}$ in acidic medium. Hence, $K_{c}$ is :

0.50

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0.25

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0.125

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None of these

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Solution

The correct option is D $0.125$
2 moles of $K M n O_{4}$ in acidic medium oxidize 5 moles of $S O_{2}$
0.2 moles of $K M n O_{4}$ in acidic medium oxidize 0.5 moles of $S O_{2}$
The initial number of moles of $S O_{3}, S O_{2}$ and $O_{2}$ are 1, 0 and 0 respectively.
The equilibrium number of moles of $S O_{3}, S O_{2}$ and $O_{2}$ are $1 - 2 x = 0.5, 2 x = 0.5$ and $x = 0.25$ respectively.
The equilibrium concentrations of $S O_{3}, S O_{2}$ and $O_{2}$ are $\frac{0.5}{2} = 0.25 M, \frac{0.5}{2} = 0.25 M$ and $\frac{0.25}{2} = 0.125 M$
The equilibrium constant expression is $K_{c} = \frac{[S O_{2}]^{2} [O_{2}]}{[S O_{3}]^{2}} = \frac{(0.25)^{2} \times 0.125}{(0.25)^{2}} = 0.125$

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One mole of SO3 was placed in a two litre vessel at a certain temperature. The following equilibrium was established in the vessel : 2SO3(g)⇌2SO2(g)+O2(g)The equilibrium mixture reacted with 0.2 mole KMnO4 in acidic medium. Hence, Kc is :

One mole of ${S O}_{3}$ 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 $0.2$ mole $K M n O_{4}$ in acidic medium. Hence, $K_{c}$ is :