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Question

Two moles of ammonia is introduced in a evacuated 500 ml vessel at high temperature. The decomposition reaction is:

2NH3(g)N2(g)+3H2(g)
At the equilibrium NH3 becomes 1 mole then the Kc would be:

A
0.42
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B
108
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C
1.7
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D
1.5
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Solution

The correct option is A 108
Given that,
Volume =500 ml=0.5L

2NH3N2+3H2
2 0 0 --- Initial mole
1 1 3 ---- At equilibrium mole
2 2 6 ---- At equilibrium conc.

Now,
Kc=[N2][H2]3[NH3]2=2×(6)3(2)2=108

Kc=108

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