The following data are for the decomposition of hydrogen
peroxide in dilute sodium hydroxide at 20 °C.
H2O2(aq) -->
H2O(l) + ½ O2(g)
| [ H2O2], M | 2.58×10-2 | 1.29×10-2 | 6.45×10-3 | 3.23×10-3 |
| time, min | 0 | 8.68 | 17.4 | 26.0 |
Hint: It is not necessary to graph these
data.
(1)
The half life observed for this reaction is ____
min .
(2)
Based on these data, the rate constant for this
(zero/first/second?) order reaction is
______min -1.
Half-life is the time required for a quantity to reduce to half of its initial value.
H2O2 takes 8.68min to be half ( i.e.1.29×10-2 M) of its initial concentration (i.e. 2.58×10-2 M), so its half age will be 8.68min
(1) = 8.68min
(2) first of all we calculate order of reaction

The following data are for the decomposition of hydrogen peroxide in dilute sodium hydroxide at 20...
The decomposition of hydrogen peroxide in dilute sodium hydroxide at 20 °C H2O2(aq) ------> H2O(l) + ½ O2(g) is first order in H2O2 with a rate constant of 1.10×10-3min-1. If the initial concentration of H2O2 is 5.52×10-2 M, the concentration of H2O2 will be 1.56×10-2 M after _____ min have passed.
The decomposition of hydrogen peroxide in dilute sodium hydroxide at 20 °C H2O2(aq)H2O(l) + ½ O2(g) is first order in H2O2 with a rate constant of 1.10×10-3 min-1. If the initial concentration of H2O2 is 5.66×10-2 M, the concentration of H2O2 will be 9.34×10-3 M after _______ min have passed.
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