Question

Consider this equilibrium H2(g) + O2(g) H2O(g). The equilibriumlaw expression for the backward balanced chemical equation would be a) [H2][02]^2 / [H20]^2 b) [H2] / [02][H20] c)[H2][02] / [H20]^2 d)[H2]^2[02] / [H20]^2 e)[H2][02] / [H20]

Answer 1

1) First, let's balance the equation. The same number atom of each elements must be the same on both sides of the reaction:

2 H₂(g) + O₂(g) --> 2 H₂O(g)

Reactants side:

H - 4

O - 2

Products side:

H - 4

O - 2

2) Now let's write the equilibrium law expression. The question asks us the backward balanced equation. So it is the reverse equation, the one that goes to the left. So, in this case, the products are 2 H₂(g) + O₂(g) and the reactant is 2 H₂O.

To write the equilibrium law expression ou need to:

Kc = [Products]^stoichiometric coefficient/[Reactant]^stoichiometric coefficient

So in this case the backward balanced chemical equation would be:

Kc = [H2]^2[O2]/[H2O]^2

Answer: Alternative "d"