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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3/4 mol2 liter-2

27/64 mol2 liter-2

27/32 mol2 liter-2

27/16 mol2 liter-2

answer is D.

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Detailed Solution

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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