For which of the following equilibria does `"k"_("eq") = ["o"_2]`? a. o2(l) o2(g) b. 2o3(g) 3o2(g) c. 2h2o(l) 2h2(g) + o2(g) d. 2hg(s) +o2(g) 2hgo(s)

For plato users

the answer is a. O2(l) O2(g)

hope this helps!

For a hypothetical reaction:

aA ↔ bB

the equilibrium constant is given as:

Keq = [B]^b/[A]^a

activity of pure solids and liquids = 1

For the given reactions:

A) O2(l) ↔ O2(g)

Keq = [O₂]

B) 2O3(g) ↔ 3O2(g)

Keq = [O₂]³/[O₃]²

C) 2H2O(l) ↔2H2(g) + O2(g)

Keq = [H₂]²[O₂]

D) 2Hg(s) + O2(g) ↔ 2HgO(s)

Keq = [Hg]²[O₂]

Ans: A

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer