For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the equilibrium constant is $64$. If the volume of the container is reduced to $1/4$ of its original value,the value of the equilibrium constant will be .......
- A$16$
- B$32$
- C$64$
- D$128$
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