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Hydrogen and iodine gas react to produce hydrogen iodide gas. What is the value of Keq for this reaction if at equilibrium the concentrations of H2, I2, and HI were 0.25 M, 0.035 M, and 0.55 M, respectively?

Respuesta :

joseaaronlara

Answer:

The answer to your question is Keq = 34.57

Explanation:

Data

[H₂] = 0.25 M

[I₂] = 0.035 M

[HI] = 0.55 M

Balanced chemical reaction

                      H₂  +  I₂   ⇒  2 HI

Formula

The coefficient of HI will be the exponent.

Keq = [HI]² / [H₂][I₂]

Substitution

Keq = [0.55]² / [0.25][0.035]

Simplification

Keq = 0.3025 / 0.00875

Result

Keq = 34.57

Eduard22sly

The value of the equilibrium constant Keq for the reaction having 0.25 M H₂, 0.035 M I₂ and 0.55 M HI at equilibrium is 34.57

Data obtained from the question

  • Concentration of Hydrogen [H₂] = 0.25 M
  • Concentration of iodine [I₂] = 0.035 M
  • Concentration of Hydrogen iodide [HI] = 0.55 M
  • Equilibrium constant (Keq) =?

Balanced equation

H₂ + I₂   <=> 2HI

How to determine the equilibrium constant

The equilibrium constant for the reaction can be obtained as follow:

Keq = [HI]² / [H₂][I₂]

Keq = [0.55]² / [0.25][0.035]

Keq = 0.3025 / 0.00875

Keq = 34.57

Thus, the equilibrium constant for the reaction is 34.57

Learn more about equilibrium constant:

https://brainly.com/question/17960050

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