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Standard IX

Chemistry

Ways of Representing the Concentration of a Solution

An equilibriu...

Question

An equilibrium mixture ${S O}_{2} (g) + {N O}_{2} (g) ⇌ {S O}_{3} (g) + N O (g)$ was found to contain $0.6$ mole of ${S O}_{2}$ , $0.2$ mole of ${N O}_{2}$ , $0.8$ mole of ${S O}_{2}$ and $0.3$ mole of $N O$ in a liter. vessel. How many moles of ${N O}_{2}$ per it. will have to be added to the vessel in order to increase the $N O$ concentration to $0.5$ mole/liter.

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Solution

$S O_{2} + N O_{2} ⇌ S O_{3} + N O_{(g)}$
At equation $0.6 0.25 0.8 0.3$
suppose volume of vessel well be $1 l t$
$K_{C} = [S O_{3}] [N O]$
$[S O_{2}] [N O_{2}]$
$K_{C} = \frac{0.8 \times 0.3}{0.6 \times 0.2}$
when coil increased to $0.5$ of $N O$
$S O_{2} + N O_{2 (g)} ⇌ S O_{3} + N O_{(g)}$
$0.6 x 0.8 0.5$
$K_{C} = [S O_{3}] [N O]$
$[S O_{2}] [N O_{2}]$
$2 = \frac{0.8 \times 0.5}{0.6 \times x}$
$1.2 x = 0.4$
$x = \frac{0.4}{1.2} = 0.33$
Amount added = Amount required -Amount present
$= 0.33 - 0.2$
$= 0.13, o l e s 1 l t s h o u l d b e a d d e d$

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An equilibrium mixture SO2(g)+NO2(g)⇌SO3(g)+NO(g) was found to contain 0.6 mole of SO2, 0.2 mole of NO2, 0.8 mole of SO2 and 0.3 mole of NO in a liter. vessel. How many moles of NO2 per it. will have to be added to the vessel in order to increase the NO concentration to 0.5 mole/liter.