Ask Question
8 April, 22:16

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:

+3
Answers (1)
  1. 9 April, 01:08
    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?
Get 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 there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers