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The "iodine clock reaction" is a popular chemical demonstration. As part of that demonstration, the I−3 ion is generated in the reaction

S2O2−8(aq)+3I−(aq)⟶2SO2−4(aq)+I−3(aq)

In one trial, the unique rate of reaction was 6.93 μmol⋅L−1⋅s−1. What was the rate of reaction of iodide ions?
1. Initial Rate = _______ μmol⋅L−1⋅s−1 ??????

What was the rate of formation of sulfate ions?
2. Rate of formation = _______ μmol⋅L−1⋅s−1 ??????

Respuesta :

In one trial, the unique rate of reaction was 6.93 μmol⋅L−1⋅s−1, the initial rate is 17.4 μmol⋅[tex]l^-^1[/tex]⋅[tex]s^-^1[/tex] and rate of formation of sulfate ions was 11.6 μmol⋅[tex]l^-^1[/tex]⋅[tex]s^-^1[/tex]

What is the rate of recation?

Reaction rate, in chemistry, is the speed at which a chemical reaction proceeds.

[tex](S_2O_8)^-^2 (aq) + 3 I^-(aq)[/tex] → [tex]2 S0_4^-^2 (aq) + 3I^- (aq)[/tex]

Rate of the reaction = -Δ [[tex]S_20_8^-^2[/tex]] /Δ t = -1/3 Δ [[tex]I^-[/tex]] /Δ t

5.8  μmol⋅[tex]l^-^1[/tex]⋅[tex]s^-^1[/tex] = -1/3 Δ [[tex]I^-[/tex]] /Δ t

-Δ [[tex]I^-[/tex]] /Δ t =17.4 μmol⋅[tex]l^-^1[/tex]⋅[tex]s^-^1[/tex]

So the rate of reaction of iodide ion =17.4 μmol⋅[tex]l^-^1[/tex]⋅[tex]s^-^1[/tex]

+1/2Δ [[tex]S0_4^-^2[/tex]] /Δ t  = rate of reaction

+1/2Δ [[tex]S0_4^-^2[/tex]] /Δ t  = 5.8 μmol⋅[tex]l^-^1[/tex]⋅[tex]s^-^1[/tex]

+ Δ [[tex]S0_4^-^2[/tex]] /Δ t  = 5.8 X 2 = 11.6 μmol⋅[tex]l^-^1[/tex]⋅[tex]s^-^1[/tex]

Rate of formation of sulfate ions=11.6 μmol⋅[tex]l^-^1[/tex]⋅[tex]s^-^1[/tex]

Hence, the initial rate is 17.4 μmol⋅[tex]l^-^1[/tex]⋅[tex]s^-^1[/tex] and rate of formation of sulfate ions was 11.6 μmol⋅[tex]l^-^1[/tex]⋅[tex]s^-^1[/tex]

Learn more about the rate of reaction here:

https://brainly.com/question/13228022

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