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:

The rate law for this reaction must be = k[S₂O₈⁻][I⁻].

Explanation:

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

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,

(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,

(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⁻].

these nuts

im pretty sure its b