Consider the reaction: NO2(g) ® NO(g) + ½ O2(g). The data below were collected for the concentration of NO2 as a function of time.
a)Determined the average rate of the reaction between 10 and 20 s.
b)Determine the rate of appearance of O2(g) between 10.0 and 20.0 sec
Time (s) |
0.0 |
10.0 |
20.0 |
30.0 |
40.0 |
50.0 |
[NO2] (M) |
1.000 |
0.951 |
0.904 |
0.860 |
0.818 |
0.778 |
(a) Average rate = - [A]/t = [A]final - [A]initial / tfinal - tinitial
(Here A is reactant and negative sign is to make quantity positive)
For the given reaction average rate = - [NO2]/t = -(0.904 - 0.951 / 20 - 10)= -(-0.047)/10 = 0.0047 M/s
(b) 1 mole of NO2 reacts to give 1/2 mole of oxygen
For the given reaction we can also write - 1/2[NO2]/t = [O2]/t
Thus average rate of formation of O2 = 0.0047/2 = 0.00235 M/s
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