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Concentration Terms - An Introduction (w/w etc)

28 g of N2 ...

Question

$28 g$ of $N_{2}$ and $6 g$ of $H_{2}$ were kept at $400^{\circ} C$ in $1$ litre vessel, the equilibrium mixture contained $27.54 g$ of $N H_{3}$ . The approximate value of $K_{c}$ for the above reaction can be: (in $m o l e^{- 2} l i t r e^{2})$ .

A

75

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B

50

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C

25

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D

100

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Solution

The correct option is A $75$
$N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}$
$(g)$ $(g)$ $(g)$
At $28 g$ ${6 g}^{1}$ $0 g$
$t = 0$
$⇓$ $⇓$ $⇓$
1 mol 3 mol 0 mol
$A$ $1 - α$ $3 - 3 α$ $2 α$
$t = t_{e q}$
Given ( $2 α$ moles) weight $= 27.54$
$2 α = 1.62$
$α = 0.81$
$0.19$ $0.57$ $1.62$
$K_{C} = \frac{{(1.62)}^{2}}{{(0.57)}^{3} \times (0.19)} \approx 75$

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