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Equilibrium — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium3 Marks
Question
$\begin{matrix}\text{A }\rightleftharpoons\text{ B }&\text{K}_1=1&\text{B }\rightleftharpoons\text{ C}&\text{K}_2=2\\\text{C }\rightleftharpoons\text{ D}&\text{K}_3=3&\text{D }\rightleftharpoons\text{ E}&\text{K}_4=4\end{matrix}$
What is value of K for $\text{A }\rightleftharpoons\text{ E}?$
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Answer

$\text{A }\rightleftharpoons\text{ E K}=\frac{[\text{E}]}{[\text{A}]}$

$\text{K}_1=\frac{[\text{B}]}{[\text{A}]},\text{K}_2=\frac{[\text{C}]}{[\text{B}]},$

$\text{K}_3=\frac{[\text{D}]}{[\text{C}]}\text{ and }\text{K}_4=\frac{[\text{E}]}{[\text{D}]}$

$\because\text{K}=\text{K}_1\times\text{K}_2\times\text{K}_3\times\text{K}_4=\frac{[\text{E}]}{[\text{A}]}$

$=1\times2\times3\times4=24$

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