The hydrogenation of nitric oxide at a certain temperature is 2 NO + 2 H2 N2 + 2 H20 From the following data, derive the rate law expression expt [NO] in M [H2] in M initial rate (M/s) 0.005 0.002 1.3 x 105 0.01 0.002 5.0 x 10-5 0.01 0.004 10.0 x 10-5 Unone of the above Orate = k [NO]²[H2] %3D Orate = k [NO][H2] Orate = k [NO]²[H2]²

Chemistry
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Chapter1: Chemical Foundations
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The hydrogenation of nitric oxide at a certain temperature is
2 NO + 2 H2 N2 + 2 H20
From the following data, derive the rate law expression
expt [NO] in M [H2] in M initial rate (M/s)
+-----+
1
0.005
0.002
1.3 x 10-5
+--
0.01
0.002
5.0 x 10-5
3
0.01
0.004
10.0 x 10-5
Unone of the above
Drate = k [NO]²[H>]
%3D
Orate
k [NO][H2]
Orate = k [NO]²[H2]²
2.
Transcribed Image Text:The hydrogenation of nitric oxide at a certain temperature is 2 NO + 2 H2 N2 + 2 H20 From the following data, derive the rate law expression expt [NO] in M [H2] in M initial rate (M/s) +-----+ 1 0.005 0.002 1.3 x 10-5 +-- 0.01 0.002 5.0 x 10-5 3 0.01 0.004 10.0 x 10-5 Unone of the above Drate = k [NO]²[H>] %3D Orate k [NO][H2] Orate = k [NO]²[H2]² 2.
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