Respuesta :
S₂O₈²⁻
(aq) + 2I⁻
(aq) → I₂(aq) + 2SO₄
²⁻(aq)
2S₂O₃²⁻
(aq) + I₂(aq) → S₄O₆²⁻
(aq) + 2I⁻
(aq)
Explanation:
S₂O₈²⁻
(aq) + 2I⁻
(aq) → I₂(aq) + 2SO₄
²⁻(aq)
To measure the rate of this reaction we must measure the rate of concentration change of one of the reactants or products. To do this, we will include (to the reacting S₂O₈
²⁻ and I⁻
i) a small amount of sodium thiosulfate, Na₂S₂O₃,
ii) some starch indicator.
The added Na₂S₂O₃ does not interfere with the rate of above reaction, but it does consume the I₂ as soon as it is formed.
2S₂O₃²⁻
(aq) + I₂(aq) → S₄O₆²⁻
(aq) + 2I⁻
(aq)
This reaction is much faster than the previous, so the conversion of I2 back to I⁻ is essentially instantaneous.
[tex]rate = \frac{dI2}{dt} = \frac{1/2 [S2O3^2^-]}{t}[/tex]
The rate of appearance of blue black color has been proportional to the persulfate ion, as the rate of reaction has been dependent on the reactant concentration.
The reaction for the consumption of iodine has been given as:
[tex]\rm H_2O_2\;+\;2\;I^-\;+\;2\;H^+\;\rightarrow\;I_2\;+\;2\;H_2O\\2\;S_2O_3^2^-\;+\;I_2\;\rightarrow\;S_4O_6^2^-\;+\;2\;I^-[/tex]
The reaction has been mediated by the addition of persulfate in the reaction that results in the formation of iodine. The iodine reacting with the starch has been forming the blue-black color complex.
The iodine formation for the reaction has been based on the presence of persulfate.
Thus, the higher the concentration of persulfate in the reaction, the higher will be the iodine reactant resulting in the lesser blue-black color complex. Therefore with decrease in persulfate, more iodine will be available resulting in the blue black complex.
For more information about rate of reaction, refer to the link:
https://brainly.com/question/8592296
