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Q. The equilibrium constant, $K_p$ for the reaction
$2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)$
is $4.0 atm^{-1}$ at $1000 K.$ What would be the partial pressure of $O_{2}$ if at equilibrium the amount of $SO_{2}$ and $SO_{3}$ is the same?

Equilibrium

Solution:

$2SO_{2} (g) + O_{2} \rightleftharpoons 2SO_{3}(g)$
$K_{p} = 4.0\,\,atm^{-2}$
$K_{p} =\frac{(SO_{3})^{2}}{(SO_{2})^{2} (O_{2})}$
Given that at equilibrium the amount of $SO_{2}$ and $SO_{3}$ is the same so
$\frac {(SO_{3})^{2}}{(SO_{2})^{2} (O_{2})} =4 \Rightarrow [O_{2}] =\frac {1}{2} =0.25$ atm