Hydrogen gas and gaseous iodine will react to form hydrogen iodide, as described by the following chemical equation. H2 (g) + I2 (g) ↽--⇀HI (g) Kc (400∘C) = 50.0Assume that all of the HI (g) is removed from a vessel containing this reaction, and equilibrium is re-established. What will be the new equilibrium concentration of HI if the equilibrium concentrations of H2 and I2 are both equal to 0.450M?

Question

Izayah

Chemistry

25 October, 01:03

Hydrogen gas and gaseous iodine will react to form hydrogen iodide, as described by the following chemical equation. H2 (g) + I2 (g) ↽--⇀HI (g) Kc (400∘C) = 50.0Assume that all of the HI (g) is removed from a vessel containing this reaction, and equilibrium is re-established. What will be the new equilibrium concentration of HI if the equilibrium concentrations of H2 and I2 are both equal to 0.450M?

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Answers (1)

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Leia Aguirre · Answer 1

3.18 M

Explanation:

Given data

[H₂]eq = [I₂]eq = 0.450 M

[HI]eq = ?

Let's consider the following reaction at equilibrium.

H₂ (g) + I₂ (g) ⇄ 2 HI (g) Kc (400°C) = 50.0

The concentration equilibrium constant (Kc) is:

Kc = [HI]²eq / [H₂]eq * [I₂]eq

[HI]²eq = Kc * [H₂]eq * [I₂]eq

[HI]eq = √ (Kc * [H₂]eq * [I₂]eq)

[HI]eq = √ (50.0 * 0.450 * 0.450)

[HI]eq = 3.18 M