Difficult
On a given condition,the equilibrium concentrations of $HI$,$H_2$,and $I_2$ are $0.80 \ mol/L$,$0.10 \ mol/L$,and $0.10 \ mol/L$ respectively. The equilibrium constant for the reaction $H_2 + I_2 \rightleftharpoons 2HI$ will be:
- A$64$
- B$12$
- C$8$
- D$0.8$
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