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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium1 Mark
MCQ
The equibrium constant $\ce{K_c}$ for the reaction : $\text{H}_2(\text{g})+\text{I}_2\text{(g)}\rightleftharpoons2\text{HI}(\text{g})\text{ at }700\text{K}\text{ is }49$ The $'K\ '$ for reaction $\text{HI(g)}\rightleftharpoons\frac{1}{2}\text{H}_2(\text{g})+\ \frac{1}{2}\text{I}_2(\text{g}).$
  • A
    $49$
  • B
    $0.02$
  • ✓
    $\frac{1}{7}$
  • D
    $1.43$
✓

Answer

Correct option: C.
$\frac{1}{7}$
$\text{K}=\frac{[\text{KI}^2]}{[\text{H}_2][\text{I}_2]}=49$
$\Rightarrow\text{K}'=\frac{[\text{H}_2]^{\frac{1}{2}}[\text{I}^2]^{\frac{1}{2}}}{[\text{HI}]}$
$=\frac{1}{\sqrt{\text{K}}}=\frac{1}{\sqrt{49}}=\frac{1}{7}$

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