When $1 \ mol$ of $H_2$ and $1 \ mol$ of $N_2$ are enclosed in a $5 \ L$ vessel and the reaction is allowed to attain equilibrium,it is found that at equilibrium there is $x \ mol$ of $H_2$. The number of moles of $NH_3$ would be

  • A
    $\frac{2x}{3}$
  • B
    $\frac{2(1+x)}{3}$
  • C
    $\frac{2(1-x)}{3}$
  • D
    $\frac{1-x}{2}$

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