Consider the equilibrium,$H_2 + I_2 \rightleftharpoons 2 HI$. Calculate the equilibrium constant of the reverse reaction when the equilibrium concentrations of $H_2$,$I_2$,and $HI$ are $1.14 \times 10^{-2} \ mol \ L^{-1}$,$0.12 \times 10^{-2} \ mol \ L^{-1}$,and $2.52 \times 10^{-2} \ mol \ L^{-1}$,respectively.

  • A
    $46.4$
  • B
    $0.021$
  • C
    $18.42$
  • D
    $0.054$

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