Consider the balanced chemical equation. H2O2(aq)+3I−(aq)+2H+(aq)→I3−(aq)+2H 2O(l) In the first 15.0 s of the reaction, the concentration of I− drops from 1.000 M to 0.871 M .Part A Predict the rate of change in the concentration of H2O2 (Δ[H2O2]/Δt). Express the rate to three significant figures and include the appropriate units. Δ[H2O2]/Δt Δ [ H 2 O 2 ] / Δ t =

Part B Predict the rate of change in the concentration of I3− (Δ[I3−]/Δt). Express the rate to three significant figures and include the appropriate units. Δ[I3−]/Δt Δ [ I 3 − ] / Δ t =

Part A. 8.60 × 10⁻³ M/s

Part B. 8.60 × 10⁻³ M/s

Explanation:

Let's consider the following balanced chemical equation.

H₂O₂(aq) + 3 I⁻(aq) + 2 H⁺(aq) → I₃⁻(aq) + 2 H₂O(l)

Part A

The rate of change in the concentration of H₂O₂ is:

-Δ[H₂O₂]/Δt = -(0.871 M - 1.000 M)/15.0 s = 8.60 × 10⁻³ M/s

Part B

The molar ratio of H₂O₂(aq) to I₃⁻(aq) is 1:1. The rate of change in the concentration of I₃⁻ is:

8.60 × 10⁻³ mol H₂O₂/L.s × (1 mol I₃⁻/1 mol H₂O₂) = 8.60 × 10⁻³ mol I₃⁻/L.s = 8.60 × 10⁻³ M/s

what are the choices?

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years