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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?

Respuesta :

dsdrajlin

Answer:

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

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