Ask Question
1 November, 10:25

The data below were determined for the reaction shown below. s2o82 - + 3i - (aq) → 2so42 - + i3 - expt. # [s2o82-] (m) [i - ] (m) initial rate 1 0.038 0.060 1.4 * 10 - 5 m/s 2 0.076 0.060 2.8 * 10 - 5 m/s 3 0.076 0.030 1.4 * 10 - 5 m/s the rate law for this reaction must be:

+4
Answers (1)
  1. 1 November, 13:40
    0
    The rate law for this reaction must be = k[S₂O₈⁻][I⁻].

    Explanation:

    To solve this problem, I will rewrite the data for clarification:

    Exp. [S₂O₈⁻], M [I⁻], M initial rate

    1 0.038 0.060 1.4 x 10⁻⁵ M/s

    2 0.076 0.060 2.8 x 10⁻⁵ M/s

    3 0.076 0.030 1.4 x 10⁻⁵ M/s

    The initial rate method to determine the order of the reaction is one of the most accurate methods to determine the order. The rate law for this reaction = k[S₂O₈⁻]ᵃ[I⁻]ᵇ,

    where, k is the rate constant of the reaction,

    a is the order of the reaction with respect to [S₂O₈⁻].

    b is the order of the reaction with respect to [I⁻].

    From Exp 1 and 2,

    The concentration of [S₂O₈⁻] changes while [I⁻] is constant, the initial rate of the reaction changes. So, the rate of the reaction depends on [S₂O₈⁻].

    (initial rate) ₁ = k[S₂O₈⁻]₁ᵃ[I⁻]₁ᵇ (1)

    (initial rate) ₂ = k[S₂O₈⁻]₂ᵃ[I⁻]₂ᵇ (2)

    By dividing (1) over (2)

    ∴ (initial rate) ₁ / (initial rate) ₂ = [k[S₂O₈⁻]₁ᵃ[I⁻]₁ᵇ] / [k[S₂O₈⁻]₂ᵃ[I⁻]₂ᵇ]

    ∴ (1.4 x 10⁻⁵ M/s) / (2.8 x 10⁻⁵ M/s) = [k[0.038]ᵃ[0.06]ᵇ] / [k[0.076]ᵃ[0.06]ᵇ]

    ∴ (0.5) = [0.038]ᵃ / [0.076]ᵃ = [0.5]ᵃ

    Taking log for the both sides,

    log (0.5) = a log (0.5)

    ∴ a = 1.

    ∴ the reaction is first order reaction with respect to [S₂O₈⁻].

    From Exp. 2 and 3,

    The concentration of [S₂O₈⁻] is constant while [I⁻] changes, the initial rate of the reaction changes. So, the rate of the reaction depends on [I⁻].

    (initial rate) ₂ = k[S₂O₈⁻]₂ᵃ[I⁻]₂ᵇ (3)

    (initial rate) ₃ = k[S₂O₈⁻]₃ᵃ[I⁻]₃ᵇ (4)

    By dividing (3) over (4)

    ∴ (initial rate) ₂ / (initial rate) ₃ = [k[S₂O₈⁻]₂ᵃ[I⁻]₂ᵇ] / [k[S₂O₈⁻]₃ᵃ[I⁻]₃ᵇ]

    ∴ (2.8 x 10⁻⁵ M/s) / (1.4 x 10⁻⁵ M/s) = [k[0.076]ᵃ[0.06]ᵇ] / [k[0.076]ᵃ[0.03]ᵇ]

    ∴ (2.0) = [0.06]ᵇ / [0.03]ᵇ = [2.0]ᵇ

    Taking log for the both sides,

    log (2.0) = a log (2.0)

    ∴ b = 1.

    ∴ the reaction is first order reaction with respect to [I⁻].

    ∴ the rate law for this reaction must be = k[S₂O₈⁻][I⁻].
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “The data below were determined for the reaction shown below. s2o82 - + 3i - (aq) → 2so42 - + i3 - expt. # [s2o82-] (m) [i - ] (m) initial ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers