3Fe(s) + 4H2O(g) ⇄ Fe3O4(s) + 4H2(g). The equilibrium constant, K, of the reaction above is represented as
\(\dfrac{[\mathrm{Fe_3O_4}][\mathrm{H_2}]}{([\mathrm{Fe}][\mathrm{H_2O}])}\)
\(\dfrac{[\mathrm{H_2O}]^{4}}{[\mathrm{H_2}]^{4}}\)
\(\dfrac{[\mathrm{H_2}]^{4}}{[\mathrm{H_2O}]^{4}}\)
\(\dfrac{[\mathrm{Fe}]^{3}[\mathrm{H_2O}]^{2}}{([\mathrm{Fe_3O_4}][\mathrm{H_2}]^{4})}\)
0
Correct answer: [H2]4 / [H2O]4
Explanation: For a heterogeneous equilibrium, pure solids have activity ≈ 1 and do not appear in the equilibrium constant. In
3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g), only the gases enter K:
K = ([H2]4) / ([H2O]4)
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