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What is the correct equilibrium constant expression for the following heterogeneous equilibrium? 4HCl (g) + O2 (g) ⇄ 2H2O (l) + 2Cl2 (g) View Available Hint(s) What is the correct equilibrium constant expression for the following heterogeneous equilibrium?
4HCl (g) + O2 (g) ⇄ 2H2O (l) + 2Cl2 (g)
Kc=[Cl2]/[HCl][O2] Kc=[HCl]4[O2]/[Cl2]2
Kc=[Cl2]2[H2O]2/[HCl]4[O2]
Kc=[Cl2]2/[HCl]4[O2]

Answer :

dfdoradod

Answer:

Kc=[Cl2]2/[HCl]4[O2]

Explanation:

For determinate the equillibrium constant, water and pure susbtances are not taken account on the calculus. In this case, water are present in the equation, so we have to remove it in the Kc determination.

sebassandin

Answer:

[tex]Kc=[Cl2]2/[HCl]4[O2][/tex]

Explanation:

Hello,

In this case, the law of mass action, allows us to study the mathematical expression regarding an equilibrium chemical reaction allowing us to see a relationship between the equilibrium constant and the concentration of both the products and reactants at equilibrium for either gaseous or aqueous substances only. Such relationship is assembled as the quotient between the concentration of products at equilibrium over the concentration of reactants at equilibrium equaling the equilibrium constant. Thus, for the given chemical reaction, such expression will have the concentration of chlorine at the numerator and both the concentrations of hydrogen chloride and oxygen at the denominator since water is liquid so it is not included in the shown below equation:

[tex]Kc=\frac{[Cl]^2}{[HCl]^4[O_2]}[/tex]

Therefore the answer is: Kc=[Cl2]2/[HCl]4[O2].

Best regards.

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