Answer:
a.
[tex]K=\frac{[BaO]^2}{[Ba]^2[O_2]}[/tex]
b.
[tex]K=\frac{[MgO]^2}{[Mg]^2[O_2]}[/tex]
c.
[tex]K=\frac{[P_4O_{10}]^2}{[P_4][O_2]^5}[/tex]
Explanation:
Hello!
In this case, since the equilibrium expression is set up by dividing the products over the reactants and powering to the stoichiometric coefficient, we can proceed as follows:
a. 2 Ba + O2 ⇌ 2 BaO.
[tex]K=\frac{[BaO]^2}{[Ba]^2[O_2]}[/tex]
b. 2 Mg + O2 ⇌ 2MgO
[tex]K=\frac{[MgO]^2}{[Mg]^2[O_2]}[/tex]
c. P4 + 5 O2 ⇌ P4O10
[tex]K=\frac{[P_4O_{10}]^2}{[P_4][O_2]^5}[/tex]
Best regards.
