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Standard XII

Chemistry

Characteristics of Equilibrium Constant

Two moles of ...

Question

Two moles of ${N H}_{3}$ gas are introduced into a previously evacuated one litre vessel in which it partially dissociates at high temperature as ${2 N H}_{3} (g) ⇋ . N_{2} (g) + {3 H}_{2} (g)$ . At equilibrium, one mole of ${N H}_{3} (g)$ remain. The value of $K_{C}$ is:

3

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27 / 16

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27 / 64

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Solution

The correct option is C $27 / 16$
2 moles of $N H 3$ gives 1 mole of $N_{2}$ and 3 moles of $H_{2}$ .

$2 {N H_{3}}_{(g)} ⇋ {N_{2}}_{(g)} + 3 {H_{2}}_{(g)}$

Given that, at equillibrium, 1 mole of ${N H_{3}}_{(g)}$ remain.

$Initial amount : At equillibrium : 2 {N H_{3}}_{(g)} 2 1 ⇋ {N_{2}}_{(g)} 0 1 / 2 + 3 {H_{2}}_{(g)} 0 3 / 2$

$∴ K_{c} = \frac{[(\frac{1}{2}) {(\frac{3}{2})}^{3}]}{[{(1)}^{2}]} = \frac{27}{16}$

Hence, $K_{c}$ is $\frac{27}{16}$ .

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Two moles of NH3 gas are introduced into a previously evacuated one litre vessel in which it partially dissociates at high temperature as 2NH3(g)⇋.N2(g)+ 3H2(g). At equilibrium, one mole of NH3(g) remain. The value of KC is:

The correct option is C 27/162 moles of NH3 gives 1 mole of N2 and 3 moles of H2.2NH3(g)⇋N2(g)+3H2(g)Given that, at equillibrium, 1 mole of NH3(g) remain.Initial amount:At equillibrium:2NH3(g)21⇋N2(g)01/2+3H2(g)03/2∴Kc=[(12)(32)3][(1)2]=2716Hence, Kc is 2716.

Two moles of ${N H}_{3}$ gas are introduced into a previously evacuated one litre vessel in which it partially dissociates at high temperature as ${2 N H}_{3} (g) ⇋ . N_{2} (g) + {3 H}_{2} (g)$ . At equilibrium, one mole of ${N H}_{3} (g)$ remain. The value of $K_{C}$ is: