Ask Question
2 December, 04:11

Initial Concentration mol/L[A] Initial Concentration mol/L[B] Initial Rate mol/Ls 0.20 0.10 20 0.20 0.20 40 0.40 0.20 160 Given the data in the accompanying table, what is the rate law for the reaction?

+3
Answers (1)
  1. 2 December, 05:53
    0
    The rate equation is as follows;

    Rate = k [A]ˣ[B]ⁿ

    where k - rate constant

    x and n are order of the reaction with respect to [A] and [B] respectively

    the data is tabulated as follows

    experiment [A] [B] rate mol/Ls

    1 0.20 0.10 20

    2 0.20 0.20 40

    3 0.40 0.20 160

    to find x lets take the data for experiment 2 and 3 where [B] is constant for both

    40 mol/Ls = k [0.20 mol/L] ˣ[0.20 mol/L]ⁿ - - - 1)

    160 mol/Ls = k [0.40 mol/L]ˣ[0.20 mol/L]ⁿ - - 2)

    divide 2) by 1)

    160/40 = (0.4/0.2) ˣ

    4 = 2ˣ

    x = 2

    to find n we have to take data from experiment 1 and 2 where [A] is constant

    40 mol/Ls = k [0.20 mol/L]ˣ[0.20 mol/L]ⁿ - - - 1)

    20 mol/Ls = k [0.20 mol/L]ˣ[0.10 mol/L]ⁿ - - - 2)

    divide 1) by 2)

    40/20 = (0.2/0.1) ⁿ

    2 = 2ⁿ

    n = 1

    order of the reaction with respect to [A] is 2

    and order of the reaction with respect to [B] is 1

    rate law for the equation is

    Rate = k [A]²[B]¹

    This a third order reaction as the sum of the orders of the reactants is 3
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Initial Concentration mol/L[A] Initial Concentration mol/L[B] Initial Rate mol/Ls 0.20 0.10 20 0.20 0.20 40 0.40 0.20 160 Given the data in ...” 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