# If equilibrium constant for the reaction ${N}_{2}+3{H}_{2}⇋2N{H}_{3}$ at 298K is 2.54, the value of equilibrium constant for the reaction will be $d\frac{1}{2}{N}_{2}+\frac{3}{2}{H}_{2}⇋N{H}_{3}$

0.395

5.08

3.18

1.59

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