For the formation of $NH_3$ from $N_2$ and $H_2$ at $500 \ K$,the concentrations of $N_2, H_2$ and $NH_3$ at equilibrium are $1.5 \times 10^{-2} \ M, 3.0 \times 10^{-2} \ M$ and $1.2 \times 10^{-2} \ M$,respectively. The equilibrium constant for the reverse reaction is

  • A
    $3.56 \times 10^2$
  • B
    $2.81 \times 10^{-3}$
  • C
    $3.56 \times 10^{-2}$
  • D
    $2.81 \times 10^3$

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