Question

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2SO3(g) -> 2SO2(g)+O2(g)
Calculate Keq for this reaction if the equalibrium concentrations are: |SO2|=0.42M; |O2|=0.21M; |SO3|=0.072M
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Answer 1

Answer:

The equilibrium constant for the reaction comes out to be 7.145

Explanation:

Given concentration of all the species at equilibrium are shown below

$\left [ SO_{2} \right ] = 0.42 \textrm{ M}, \left [ O_{2} \right ] = 0.21 \textrm{ M}, \left [ SO_{3} \right ] = 0.072 \textrm{ M}$ \\

Given reaction is shown below

$2\textrm{SO}_{3}\left ( g \right )\rightleftharpoons 2\textrm{SO}_{2}\left ( g \right )+\textrm{O}_{2}\left ( g \right )$

$K_{eq} = \displaystyle \frac{\left [ SO_{2} \right ]^{2}\times \left [ O_{2} \right ]}{\left [ SO_{3} \right ]^{2}} \\K_{eq} = \displaystyle \frac{\left ( 0.42 \right )^{2}\times 0.21}{\left ( 0.072 \right )^{2}} \\K_{eq} = 7.145$

$K_{eq}$ for the reaction is 7.145