Kinetics Test-9

SCH4U Kinetics Test March 2011 Look at experiments 4 and 5. The H, concentration DOUBLES and the rate DOUBLES. 2" =2 therefore n = 1 since 2'=2 So the rate law expression can be rewritten as rate = k [NOJ?[H,]' Now to determine the value of 'k'. Choose any one of the experiments. Using experiment 1. Using the rate law, above fill in the values from the data table. 0.002 mol/L sec = k (0.001 mol/L)** (0.004 mol/L) 0.002 mol/L sec = k * (0.000001 mol*/L?) * (0.004 mol/L) 0.002 mol/L sec = k * 0.000 000 009 mol*/L’ k =0.002 mol/L sec 0.000 000 004 mol*/L* = 500,000 sec/mol*L?or sec mol?L> Therefore the rate law equation for this reaction is rate = 500,000 sec mol>L*[NO mole/L]*[H, mol/L] REF: 1 OBJ: 6.3 STO: EC2.06 Tutoring Pages Page 9