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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
For the reactions
$2NO\, + {O_2} \rightleftharpoons 2N{O_2}\,;\,{K_1}$
$4NO\, + 2C{l_2} \rightleftharpoons 4NOCl\,;\,{K_2}$
$N{O_2}\, + \frac{1}{2}C{l_2} \rightleftharpoons NOCl + \frac{1}{2}{O_2}\,;\,{K_3}$
where $K_1, K_2, K_3$ are equilibrium constants then $K_3^2$ equal to
  • A
    $\sqrt {{K_2}/{K_1}} $
  • B
    $\sqrt {{K_1}{K_2}} $
  • ✓
    $\sqrt {{K_2}} /{K_1}$
  • D
    $\frac{1}{{{K_1}{K_2}}}$
✓

Answer

Correct option: C.
$\sqrt {{K_2}} /{K_1}$
c
$2 \mathrm{NO}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}_{2}$

$\mathrm{K}_{1}=\frac{\left[\mathrm{N} \mathrm{O}_{2}\right]^{2}}{[\mathrm{N} \mathrm{O}]^{2}\left[\mathrm{O}_{2}\right]} \ldots$ (ii)

$4 \mathrm{NO}+2 \mathrm{Cl}_{2} \rightleftharpoons 4 \mathrm{NOCl}$

$\mathrm{K}_{2}=\frac{[\mathrm{NOCl}]^{4}}{[\mathrm{NO}]^{4}\left[\mathrm{Cl}_{2}\right]^{2}} \ldots$ (ii)

$\mathrm{NO}_{2}+\frac{1}{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{NOCl}+\frac{1}{2} \mathrm{O}_{2}$

$\mathrm{K}_{3}=\frac{[\mathrm{NOCl}]\left[\mathrm{O}_{2}\right]^{1 / 2}}{\left[\mathrm{NO}_{2}\right]\left[\mathrm{Cl}_{2}\right]^{1 / 2}} \ldots$ (iii)

divide $\mathrm{K}_{2}$ by square of $\mathrm{K}_{1}^{2}$ $\frac{[\mathrm{NOCl}]^{4}}{[\mathrm{NO}]^{4}\left[\mathrm{Cl}_{2}\right]^{2}} \times \frac{[\mathrm{NO}]^{4}\left[\mathrm{O}_{2}\right]^{2}}{\left[\mathrm{NO}_{2}\right]^{4}}=\frac{[\mathrm{NOCl}]^{4}\left[\mathrm{O}_{2}\right]^{2}}{\left[\mathrm{NO}_{2}\right]^{4}\left[\mathrm{Cl}_{2}\right]^{2}}$

$\mathrm{K}_{2} \times \frac{1}{\mathrm{K}_{1}^{2}}=\mathrm{K}_{3}^{4} \Rightarrow \frac{\mathrm{K}_{2}}{\mathrm{K}_{1}^{2}}=\mathrm{K}_{3}^{4}$

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