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6.1. Equilibrium - 1 (chemical Equilibrium) — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry6.1. Equilibrium - 1 (chemical Equilibrium)1 Mark
MCQ
$0.1$ $mole$ of ${N_2}{O_{4(g)}}$ was sealed in a tube under one atmospheric conditions at $25\,^oC$. Calculate the number of moles of $N{O_{2(g)}}$ present, if the equilibrium ${N_2}{O_{4(g)}}$ $\rightleftharpoons$ $2N{O_{2(g)}}$ $({K_p} = 0.14)$ is reached after some time
  • A
    $1.8\; \times \;{10^2}$
  • B
    $2.8\; \times \;{10^2}$
  • ✓
    $0.034$
  • D
    $2.8\; \times \;{10^{ - 2}}$
✓

Answer

Correct option: C.
$0.034$
(c) $\mathop {{N_2}{O_4}}\limits_{0.1} $ $ \rightleftharpoons $ $\mathop {2N{O_2}}\limits_0 $

     $(1-$$\alpha$$)$      $2\alpha$

$\therefore P \propto 0.1$

If $V$ and $T$ are constant $(P \propto 0.1 + \alpha)$

${\rm{P}} = {\rm{(0}}{\rm{.1}} + \alpha )/0.1$

${K_p} = \frac{{{{[2\alpha ]}^2}}}{{[0.1 - \alpha ]}} \times \left[ {\frac{P}{{0.1 + \alpha }}} \right]$ or ${K_p} = \frac{{40{\alpha ^2}}}{{[0.1 - \alpha ]}} = 0.14$

$\alpha = 0.017$

$N{O_2} = 0.017 \times 2 = 0.034$ $mole$

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