MCQ
The equilibrium constant for equilibrium is  $2H{X_{(g)}}\,\, \rightleftharpoons \,{H_{2(g)}}\, + \,{X_{2(g)}}$ is  $1\, \times \,10^{-5}$. What will be  the equilibrium concentration of $HX$  if the equilibrium concentrations for $H_2$  and $X_2$  are $1.2\, \times \,10^{-3}\, M$  and  $1.2 \times \, 10^{-4}\, M$  respectively
  • A
    $12\,\times \,10^{-4}\,M$
  • B
    $12\,\times \,10^{-3}\,M$
  • $12\,\times \,10^{-2}\,M$
  • D
    $12\,\times \,10^{-1}\,M$

Answer

Correct option: C.
$12\,\times \,10^{-2}\,M$
c
At equilibrium $2 \mathrm{HX} \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+\mathrm{X}_{2}(\mathrm{g})$

$\mathrm{K}=\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{X}_{2}\right]}{[\mathrm{HX}]^{2}}$

$10^{-5}=\frac{1.2 \times 10^{-3} \times 1.2 \times 10^{-4}}{[\mathrm{HX}]^{2}}$

$[\mathrm{HX}]=\sqrt{\frac{1.2 \times 1.2 \times 10^{-7}}{10^{-5}}}$

$=1.2 \times 10^{-1}$

$=12 \times 10^{-2} \mathrm{M}$

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