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Characteristics of Equilibrium Constant

28 gms of N...

Question

28 gms of $N_{2}$ , and 6 gms of $N_{2}$ , were heated in a closed 1 litre vessel. At equilibrium, 25.5 gms of $N H_{3}$ , is present. The approximate value of $K_{c}$ . is:

5.55

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21.33

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2.54

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3.16

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Solution

The correct option is A 21.33
$28 g$ of $N_{2}$ contains= $\frac{28}{28}$ mole of $N_{2} = 1$ mole of $N_{2}$
$6 g$ of $H_{2}$ contains= $\frac{6}{2}$ mole of $H_{2} = 3$ mole of $H_{2}$
$25.5 g$ of $N H_{3}$ contains= $\frac{25.5}{17} = 1.5$ mole of $N H_{3}$
moles $N_{2} + 3 H_{2} ⟶ 2 N H_{3}$
at $t = 0$ $1$ $3$ -
at $E q b^{m}$ $1 - x$ $3 - 3 x$ $2 x$
as mole of $N H_{3}$ produced= $1.5 = 2 x, [N H_{3}] = \frac{1.5}{1} M$
$x = \frac{1.5}{2}$
Thus mole of $N_{2}$ at $E q b^{m} = 1 - \frac{1.5}{2} = 0.25, [N_{2}] = \frac{0.25}{1} M$
Mole of $H_{2}$ at $E q b^{m} = 3 - 3 \times \frac{1.5}{2} = 0.75, [H_{2}] = \frac{0.75}{1} M$
$\Rightarrow K_{C} = \frac{[N H_{3}]^{2}}{[N_{2}] [H_{2}]^{3}} = \frac{[1.5]^{2}}{[0.25] [0.75]^{3}} = 21.33 M^{- 2}$
$\Rightarrow K_{C} = 21.33 M^{- 2}$

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28 gms of N2 , and 6 gms of N2, were heated in a closed 1 litre vessel. At equilibrium, 25.5 gms of NH3, is present. The approximate value of Kc. is:

28 gms of $N_{2}$ , and 6 gms of $N_{2}$ , were heated in a closed 1 litre vessel. At equilibrium, 25.5 gms of $N H_{3}$ , is present. The approximate value of $K_{c}$ . is: