Question

The rate law for the reactionC₂H₄Br₂ + 3I⁻ → C₂H₄ + Br⁻ + I⁻ is rate = k[C₂H₄Br₂][I⁻]. The rate of the reaction is found to be 1.1 x 10⁻⁴ M/s when the concentrations of C₂H₄Br₂ and I⁻ are 0.12 Mand 0.18 M respectively. Calculate the rate constant of the reaction.

Answer 1

Answer:

Rate constant $k = 50.92\times 10^{-4}$ (M sec)^-1

Explanation:

Since we have given the reaction (C₂H₄Br₂ + 3I⁻ → C₂H₄ + Br⁻ + I⁻)

Rate law for the reaction is (rate = k[C₂H₄Br₂][I⁻]).......................(1)

where k = Rate constant.

rate = 1.1 x 10⁻⁴ M/s

Concentrations of C₂H₄Br₂ and I⁻ are 0.12 M and 0.18 M

using equation (1) we get

1.1 x 10⁻⁴ = k x 0.12 x 0.18

So, Rate constant $k = 50.92\times 10^{-4}$ (M sec)^-1