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Question

One mole of N2(g) is mixed with 2 moles of H2(g) in a 4 litre vessel. If 50% of N2(g) is converted to NH3(g) by the following reaction:
N2(g)+3H2(g)2NH3(g)
What will be the value of Kc for the following equilibrium ?
NH3(g)12N2(g)+32H2(g)

A
256
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B
16
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C
116
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D
None of these
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Solution

The correct option is B 116
N2(g)+3H(g)2NH3(g)
initial moles120
at eqm1x23x2x
where x=0.5
Kc=(1/4)2(0.54)2(0.54)3=256
equilibrium constant for the reaction
NH3(g)12N2(g)+32H2(g)
K1C=1KC=116

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