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Standard XII

Chemistry

Law of Mass Action

Two moles of ...

Question

Two moles of $N_{2}$ and two moles of $H_{2}$ are taken in a closed vessel of $5 L$ capacity and suitable conditions are provided. When the equilibrium is reached, it is found that $0.6 m o l e$ of $N_{2}$ is used up. The equilibrium concentration of ${N H}_{3}$ is________.

0.4

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0.08

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0.8

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0.24

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Solution

The correct option is D $0.24$
$N_{2} + {3 H}_{2} ⇌ 2 {N H}_{3} V = S L$
At $2$ $2$ $0$
$t = 0$
At $2 - x$ $2 - 3 x$ $2 x$
$t = t_{e q}$
Given $0.6$ mole of $N_{2}$ is used $O P$
Hence $n = 0.6$
So $N H g = 1.2 m o l e s$
concentration $= \frac{m o l e s}{v o l u m e i n l i t} = \frac{1.2}{5} = 0.24$

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Two moles of N2 and two moles of H2 are taken in a closed vessel of 5 L capacity and suitable conditions are provided. When the equilibrium is reached, it is found that 0.6 mole of N2 is used up. The equilibrium concentration of NH3 is________.

The correct option is D 0.24 N2+3H2⇌2NH3V=SLAt 2 2 0t=0At 2−x 2−3x 2xt=teqGiven 0.6 mole of N2 is used OPHence n=0.6So NHg=1.2molesconcentration =molesvolumeinlit=1.25=0.24

Two moles of $N_{2}$ and two moles of $H_{2}$ are taken in a closed vessel of $5 L$ capacity and suitable conditions are provided. When the equilibrium is reached, it is found that $0.6 m o l e$ of $N_{2}$ is used up. The equilibrium concentration of ${N H}_{3}$ is________.