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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
For the reaction: $\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$at 400K, $K_p = 41$. Find the value of $K_p$ for each of the following reactions at the same temperature.
  1. $\text{2NH}_3(\text{g})\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2(\text{g})$
  2. $\frac{1}{2}\text{Na}(\text{g})+\frac{3}{2}\text{H}_2(\text{g})\rightleftharpoons\text{NH}_3(\text{g})$
  3. $2\text{N}_2(\text{g})+6\text{H}_2(\text{g})\rightleftharpoons4\text{NH}_3(\text{g})$
✓

Answer

  1. $\text{K}_{\text{p}}=\frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}=41$
$2\text{NH}_3\text{(g)}\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2(\text{g})$

$\text{K}'_{\text{p}}=\frac{\text{[N}_2][\text{H}_2]^3}{[\text{NH}_3]^2}$

$=\frac{1}{\text{K}_{\text{p}}}=\frac{1}{41}=0.024$
  1. $\frac{1}{2}\text{N}_2(\text{g})+\frac{3}{2}\text{H}_2\rightleftharpoons\text{NH}_3(\text{g})$
$\text{K}''_{\text{p}}=\frac{[\text{NH}_3]}{[\text{NH}_2]^{\frac{1}{2}}[\text{H}_2]^{\frac{3}{4}}}$

$=\sqrt{\text{K}_{\text{p}}}=\sqrt{41}=\sqrt{6.4}$
  1. $2\text{N}_2\text{(g)}+6\text{H}_2(\text{g})\rightleftharpoons4\text{NH}_3(\text{g})$
$\text{K}'''_{\text{p}}=\frac{[\text{NH}_3]^4}{[\text{N}_2]^2[\text{H}_2]^6}$

$=(\text{K}_{\text{p}})^2=(41)^2=1681$

$=1.68\times10^3$

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