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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
At $450K, K_p= 2.0 \times 10^{10}/bar$ for the given reaction at equilibrium.
$2\text{SO}_2\text{(g) + O}_2\text{(g)}\rightleftharpoons\text{2SO}_3\text{(g)}$
What is $K_c$ at this temperature?
✓

Answer

For the given reaction,
$\Delta\text{n}=2-3=-1$
$\text{T}=450\text{K}$
$\text{R}=0.0831 \text{ bar L bar K}^{-1}\text{mol}^{-1}$
$\text{K}_{\text{p}}=2.0\times10^{10}\text{ bar}^{-1}$
We know that,
$\text{K}_{\text{p}}=\text{K}_{\text{c}}(\text{RT})\Delta\text{n}$
$\Rightarrow2.0\times10^{10}\text{ bar}^{-1}=\text{K}_{\text{c}}(0.0831\text{ L bar K}^{-1}\text{mol}^{-1}\times450\text{K})^{-1}$
$\Rightarrow\text{K}_{\text{c}}=\frac{2.0\times10^{-10}\text{ bar}^{-1}}{(0.0831\text{ L bar}^{-1}\text{ K}^{-1}\text{mol}^{-1}\times450\text{K})^{-1}}$
$=(2.0\times10^{10}\text{ bar}^{-1})(0.0831\text{ L bar K}^{-1}\text{mol}^{-1}\times450\text{K})$
$=74.79\times10^{10}\text{L mol}^{-1}$
$=7.48\times10^{11}\text{L mol}^{-1}$
$=7.48\times10^{11}\text{M}^{-1}$

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