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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

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Review Problems

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  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!

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