In the reaction,$A + B \rightleftharpoons 2C$,at equilibrium,the concentration of $A$ and $B$ is $0.20 \ mol \ L^{-1}$ each and that of $C$ was found to be $0.60 \ mol \ L^{-1}$. The equilibrium constant of the reaction is

  • A
    $2.4$
  • B
    $18$
  • C
    $4.8$
  • D
    $9$

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If for ${H_2(g)} + \frac{1}{2}{S_2(s)} \rightleftharpoons {H_2S(g)}$ and ${H_2(g)} + {Br_2(g)} \rightleftharpoons 2{HBr(g)}$ the equilibrium constants are $K_1$ and $K_2$ respectively,the reaction ${Br_2(g)} + {H_2S(g)} \rightleftharpoons 2{HBr(g)} + \frac{1}{2}{S_2(s)}$ would have equilibrium constant

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If $K_1$ and $K_2$ are respective equilibrium constants for the two reactions:
$XeF_{6(g)} + H_2O_{(g)} \rightleftharpoons XeOF_{4(g)} + 2HF_{(g)}$
$XeO_{4(g)} + XeF_{6(g)} \rightleftharpoons XeOF_{4(g)} + XeO_3F_{2(g)}$
The equilibrium constant for the reaction $XeO_{4(g)} + 2HF_{(g)} \rightleftharpoons XeO_3F_{2(g)} + H_2O_{(g)}$ will be:

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