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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium2 Marks
Question
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g});$
$\text{K}=0.50\text{ at }673\text{K.}$
Write the equilibrium expression and equilibrium constant for reverse reaction.
✓

Answer

$2\text{NH}_3(\text{g})\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2(\text{g})$
$\text{K}'_{\text{c}}=\frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2}$ for reverse reaction
$\text{K}_{\text{c}}=0.50$
$\Rightarrow\text{K}'_{\text{c}}=\frac{1}{\text{K}_{\text{c}}}=\frac{1}{0.50}=2$

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