Question

The equilibrium of 2H 2 O(g) 2H 2 (g) + O 2 (g) at 2,000 K has a Keq value of 5.31 x 10-10. What is the Keq expression for this system? (Enter subscripts after the letters: for example, H2O = Hs2O. Also, dont forget to use proper chemical shorthand for chemical symbols. For instance, chlorine = Cl but not cl or cL.)

Answer 1

`5.31*10^(-10) = ([]H_(2)]^(2)[O_(2)])/([H_(2)O]^(2))`

Step-by-step explanation:

For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.

For a hypothetical reaction:

xA + yB ⇄ zC

The equilibrium constant is :

`Keq = ([A]^(x)[B]^(y))/([C]^(z) )`

The given reaction involves the decomposition of H2O into H2 and O2

`2H_(2)O\rightleftharpoons 2H_(2) + O_(2)`

The equilibrium constant is expressed as :

`Keq = ([]H_(2)]^(2)[O_(2)])/([H_(2)O]^(2))`

Since Keq = 5.31*10^-10

`5.31*10^(-10) = ([]H_(2)]^(2)[O_(2)])/([H_(2)O]^(2))`

Answer 2

Answer : The equilibrium constant expression will be,

`k_(eq)=([H_2]^2[O_2])/([H_2O]^2)`

Explanation :

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given balanced chemical reaction is,

`2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)`

So, the equilibrium constant expression will be,

`k_(eq)=([H_2]^2[O_2])/([H_2O]^2)`