Question

8. An instructor asked the class, "Can you substitute hydrogen in nitrous oxide booster in your car engine, and if so what would be a representative reaction occurring in your car engine?" A half alert student thought he heard the instructor ask for the reaction between hydrogen and nitric oxide (instead of nitrous oxide NO2) and took off to perform the calculations. He assumed the reaction would be: 2H2 (g) + 2NO (g) → 2H,0 (g) + N2 (g) The student collected the following initial rate data for this reaction. EXPERIMENT HIM 0.010 0.0050 0.010 [NOJ M 10.025 0.025 0.0125 Initial Ratel Ms 2.4 x 10 1.2 x 10 0.60 x 10° 2 3 a) What is the rate law for the reaction? b) Calculate the rate constant for the reaction

Answer 1

Answer rate law rate = k H27 No] Initial rate a (ro) is given with initial concentration o = k (tab. wol from exp- I I . 2.4810-6= k (0-01039 (0.025) -0 from exp- 1-2x10-6 = (0.0050)" (0.026) -O From sexp00 0.60 x10-6 = k (0-010) (0.0125) 8 - © Dividin ey O2 !! 2:48 10- k (0.010) (0.020) 8 T2x10-6. klo.opsoja (0.025)" 2= (2)9 d=1] Dividing em? 0) 2.4x10-6 0.60 x10-6 4 = = 6 (.0lo ) p (0.02538 k 10.0109 (0.012578 (218

(2)2 = 120 B=2 @ so rate law becomes rate rate = .k [Hz] ² [NO]² = .k - Now putting as B in ea?-o 2.4x10-6 mst = k 10.0 10 mg)'(0.025mj2 I k= 0.384 M-2 5-1) rate constant