Two moles of NH3 when put into a previously evacuated vessel (one liter), partially dissociated into N2 and H2. If at equilibrium one mole of NH3, is Present, the equilibrium constant is :

2NH3(g)⇌N2(g)+3H2(g)t=0 ; 2 moles 0 0teq. ; 2-2x x 3xGiven : 2-2x=1 ⇒x = 0.5 Now, kc=N2H23NH32=0.51.531=2716mol2liter-2

Two moles of NH3 when put into a previously evacuated vessel (one liter), partially dissociated into N2 and H2. If at equilibrium one mole of NH3, is Present, the equilibrium constant is :

2NH3(g)⇌N2(g)+3H2(g)t=0 ; 2 moles 0 0teq. ; 2-2x x 3xGiven : 2-2x=1 ⇒x = 0.5 Now, kc=N2H23NH32=0.51.531=2716mol2liter-2

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Two moles of NH₃ when put into a previously evacuated vessel (one liter), partially dissociated into N₂and H_2. If at equilibrium one mole of NH₃, is Present, the equilibrium constant is :

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$3 / 4 {mol}^{2} {liter}^{- 2}$

$27 / 32 {mol}^{2} {liter}^{- 2}$

$27 / 64 {mol}^{2} {liter}^{- 2}$

$27 / 16 {mol}^{2} {liter}^{- 2}$

answer is D.

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Detailed Solution

$2 {NH}_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$

t=0 ; 2 moles 0 0

t_eq.; 2-2x x 3x

Given : 2-2x=1 $\Rightarrow$ x = 0.5

$Now, k_{c} = \frac{[N_{2}] {[H_{2}]}^{3}}{{[{NH}_{3}]}^{2}} = \frac{(0.5) {(1.5)}^{3}}{1} = \frac{27}{16} {mol}^{2} {liter}^{- 2}$

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Two moles of NH3 when put into a previously evacuated vessel (one liter), partially dissociated into N2 and H2. If at equilibrium one mole of NH3, is Present, the equilibrium constant is :

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Two moles of NH₃ when put into a previously evacuated vessel (one liter), partially dissociated into N₂and H_2. If at equilibrium one mole of NH₃, is Present, the equilibrium constant is :