If the equilibrium constant for the reaction,$2 SO_2 + O_2 \rightleftharpoons 2 SO_3$ is $64$ at $500 \ K$,then the equilibrium constant for the reaction $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$ at the same temperature is

  • A
    $8$
  • B
    $\frac{1}{8}$
  • C
    $32$
  • D
    $\frac{1}{64}$

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The value of the equilibrium constant for the reaction:
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At $T \ K$,the equilibrium constant for the reaction $a A_{(g)} \rightleftharpoons b B_{(g)}$ is $K_c$. If the reaction takes place in the following form $2a A_{(g)} \rightleftharpoons 2b B_{(g)}$,its equilibrium constant is $K_c^{\prime}$. The correct relationship between $K_c$ and $K_c^{\prime}$ is

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