Consider the balanced chemical equation. H2O2(aq) + 3 I^-(aq) + 2 H+(aq) S I3^-(aq) + 2 H2O(l) In the first 10.0 seconds of the reaction, the concentration of I^- drops from 1.000 M to 0.868 M. (a) Calculate the average rate of this reaction in this time interval.

(b) Predict the rate of change in the concentration of H (that is, ∆[H ]>∆t) during this time interval.

Based on the data provided,  the rate of the given reactions are:

a. reaction rate =  0.0132 M/sec

b. rate of change of [H⁺] = 8.8 * 10⁻³ M/sec

Reaction rate = change in concentration of reactant or product/ time taken

Change in concentration of I⁻ = 1.000 M - 0.868 M= 0.132

time taken = 10 s

reaction rate = 0.132/10

Rate of reaction = 0.0132 M/sec

From the equation of the reaction 3 moles of I⁻ are used up for every 2 moles of H⁺

Hence, the rate of change of [H⁺] = 0.0132 M/sec * 2/3

rate of change of [H⁺] = 8.8 * 10⁻³ M/sec

Therefore, the rate of the given reactions are:

a. reaction rate =  0.0132 M/sec

b. rate of change of [H⁺] = 8.8 * 10⁻³ M/sec

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the answer is symmetrical : )