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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
$K_1 , K_2$ and $K_3$ are the equilibrium constants of the following reactions $(I), (II)$ and $(III)$ respectively

$(I)\,\,\,\,{N_2} + 2{O_2} \rightleftharpoons 2N{O_2}$

$(II)\,\,\,\,2N{O_2} \rightleftharpoons {N_2} + 2{O_2}$

$(III)\,\,\,\,N{O_2} \rightleftharpoons \frac{1}{2}{N_2} + 2{O_2}$

The correct relation from the following is

  • A
    ${K_1} = \frac{1}{{{K_2}}} = \frac{1}{{{K_3}}}$
  • ✓
    ${K_1} = \frac{1}{{{K_2}}} = \frac{1}{{{(K_3)^2}}}$
  • C
    ${K_1} = \sqrt {{K_2}}  = {K_3}$
  • D
    ${K_1} = \frac{1}{{{K_2}}} = {K_3}$
✓

Answer

Correct option: B.
${K_1} = \frac{1}{{{K_2}}} = \frac{1}{{{(K_3)^2}}}$
b
$(I)\,{N_2} + 2{O_2}\overset {{K_1}} \longleftrightarrow 2N{O_2}$

${K_1} = \frac{{{{[N{O_2}]}^2}}}{{[{N_2}]{{[{O_2}]}^2}}}...(i)$

$(II)\,2N{O_2}\overset {{K_2}} \longleftrightarrow {N_2} + 2{O_2}$

${K_2} = \frac{{[{N_2}]{{[{O_2}]}^2}}}{{{{[{N_2}O]}^2}}}...(ii)$

$(III) \,N{O_2}\overset {{K_3}} \longleftrightarrow \frac{1}{2}{N_2} + {O_2}$

${K_3} = \frac{{{{[{N_2}]}^{1/2}}[{O_2}]}}{{[N{O_2}]}}$

$\therefore \,{({K_3})^2} = \frac{{[{N_2}]{{[{O_2}]}^2}}}{{{{[N{O_2}]}^2}}}...(iii)$

$\therefore $ from equation $(i),(ii) and (iii)$

${K_1} = \frac{1}{{{K_2}}} = \frac{1}{{{{({K_3})}^2}}}$

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