Question:
The equilibrium constant for the reaction
is $\mathrm{K}_{\mathrm{p}}=4$. At equilibrium, the partial pressure of $\mathrm{O}_{2}$ is______ atm. (Round off to the nearest integer)
Solution:
$\mathrm{k}_{\mathrm{p}}=\mathrm{Po}_{2}^{1 / 2}=4$
$\therefore \mathrm{Po}_{2}=16 \mathrm{bar}=16 \mathrm{~atm}$
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