The oxidation of iodide ion by arsenic acid is described by: 3I^- (aq) - H_3AsO_

Question

The oxidation of iodide ion by arsenic acid is described by: 3I^- (aq) - H_3AsO_4 (aq) - 2H^+ (aq) rightarrow I_3^- (aq) - H_3 AsO_3 (aq) + H_2O (l) If - Delta [I^-]/Delta t = 4.8 Times 10^-4 M/s, what is the value of Delta [I_3^-]/Delta t during the same rime interval? What is the average rate of consumption of H^+ during that time interval? Draw a reaction energy diagram and label each axis as well as the reactants, products, transition state, activation energy and tell whether you drew an exo or endo thermic reaction.

Explanation / Answer

Given equation for the oxidation of Iodide ion by arsenic acid is describe as,

3I– (aq.) + H3AsO4 (aq.) + 2H+ (aq.) ---------> I3–(aq.) + H3AsO3 (aq.) + H2O (l)

Then we write,

–(1/3)[I–]/T = [H3AsO4]/T = –(1/2)[H+]/T = [I3–]/T = [H3AsO3]/T = [H2O]/T

Hence for rate of formation of I3–.

–(1/3)[I–]/T = [I3–]/T

i.e. [I3–]/T = (1/3) (–[I–]/T)

[I3–]/T = (1/3) x (4.8 x 10–4) …………….(Given value)

[I3–]/T =1.6 x 10–4) M/s.

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2) And for rate of formation of H+,

–(1/3)[I–]/T = –(1/2)[H+]/T

i.e. –[H+]/T = – 2 x (1/3)[I–]/T

–[H+]/T = – 2 x (1/3) x (4.8 x 10–4)

–[H+]/T = 3.2 x x 10–4 M/s.

Average rate of consumption of H+ ions is 3.2 x x 10–4 M/s.

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