1 / 7

Review Problems

Review Problems. Water can be made from hydrogen and oxygen: H 2 (g) + O 2 (g) → H 2 O (g) Some experiments are run in a sealed 1 L flask at 350 K to determine the initial rate of reaction for different concentrations of reactants: [H2 ] M [O2] M Rate of formation of H2O M/sec

bradley-humphrey
Download Presentation

Review Problems

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Review Problems

  2. Water can be made from hydrogen and oxygen: H2 (g) + O2 (g) →H2O (g) Some experiments are run in a sealed 1 L flask at 350 K to determine the initial rate of reaction for different concentrations of reactants: [H2 ] M [O2] M Rate of formation of H2O M/sec 0.050 0.100 0.0048 0.050 0.050 0.0025 0.100 0.025 0.00120 If I began with 0.050 moles of both reactants, how long would it take to make 0.01 moles of water?

  3. Orders of reaction [H2 ] M [O2] M Rate of formation of H2O M/sec 0.050 0.100 0.0048 0.050 0.050 0.0025 0.100 0.025 0.00120 Rate = k[H2]x[O2]y Rate 1 = k[0.050]x[0.100]y Rate 2 = k[0.050]x[0.050]y 0.0048 = [0.100]y 0.0025 = [0.050]y 1.92 =2y Y =1

  4. Orders of reaction [H2 ] M [O2] M Rate of formation of H2O M/sec 0.050 0.100 0.0048 0.050 0.050 0.0025 0.100 0.025 0.00120 Rate = k[H2]x[O2]y Rate 2 = k[0.050]x[0.050] Rate 3 = k[0.100]x[0.025] 0.0025 = [0.100]x 2 0.0012 = [0.050]x 2.08 =2*2x 1.04 =2x X=0

  5. Rate = k[O2] [H2 ] M [O2] M Rate of H2O M/sec 0.050 0.100 0.0048 0.050 0.050 0.0025 0.100 0.025 0.00120 0.0048 M/s = k [0.100 M] K =0.048 s-1 0.0025 = k [0.050] K =0.05 s-1 0.0012 =k[0.025] K =0.048 s-1 Kavg = 0.0487 s-1

  6. 1st order kinetics ln[O2]t= - kt [O2]0 2H2 (g) + O2 (g) →2H2O (g) If I began with 0.050 moles of both reactants, how long would it take to make 0.01 moles of water? 0.010 mol H2O* 1 mol O2 = 0.005 mol O2 2 mol H2O 0.050 mol O2 started – 0.005 mol O2 reacted = 0.045 mol left

  7. 1st order kinetics ln [O2]t= - kt [O2]0 Ln (0.045/0.05) = - 0.0487s-1 t t = 2.16 seconds!

More Related