Ask Question
4 February, 01:57

The decomposition of hydrogen peroxide in dilute sodium hydroxide at 20 °C is first order in H2O2. H2O2 (aq) H2O (l) + ½ O2 (g) During one experiment it was found that the H2O2 concentration dropped from 5.16*10-2 M at the beginning of the experiment to 1.95*10-2 M in 832 min. What is the value of the rate constant for the reaction at this temperature? min-1

+5
Answers (1)
  1. 4 February, 04:20
    0
    The rate constant is 1.98*10^-3 min^-1

    Explanation:

    The reaction follows a first order

    Rate = k[H2O2] = change in concentration of H2O2/time

    Initial concentration of H2O2 = 5.16*10^-2 M

    Concentration of H2O2 in 832 min = 1.95*10^-2 M

    Change in concentration = 5.16*10^-2 - 1.95*10^-2 = 3.21*10^-2 M

    Time = 832 min

    1.95*10^-2k = 3.21*10^-2/832

    k = 3.86*10^-5/1.95*10^-2 = 1.98*10^-3 min^-1
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “The decomposition of hydrogen peroxide in dilute sodium hydroxide at 20 °C is first order in H2O2. H2O2 (aq) H2O (l) + ½ O2 (g) During one ...” 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