MCQ
${A_{(g)}} + 3{B_{(g)}}$ $\rightleftharpoons$ $4{C_{(g)}}$. Starting concentration of $A$ is equal to $B$, equilibrium concentration of $A$ and $C$ are same. ${K_c} = $
  • A
    $0.08$
  • B
    $0.8$
  • $8$
  • D
    $80$

Answer

Correct option: C.
$8$
(c)   $ A $       $+$       $3B $     $ \rightleftharpoons $       $4C$

        $a$                  $b$                      $0$

      $(a -x)$          $(b -3x)$             $4x$

${K_C} =  \frac{{{{[C]}^4}}}{{[A]{{[B]}^3}}} = \frac{{4x.4x.4x.4x}}{{(a - x)(b - 3x)}}$

$a=b,\,a-x=4x\Rightarrow a=5x=b$

${{K}_{C}}=\frac{4x.4x.4x.4x}{(5x-x)(5x-3x)}=\frac{4x.4x.4x.4x}{4x.2x.2x.2x}=8$

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