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Review Problems

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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 0.0500.1000.0048 0.0500.0500.0025 0.1000.0250.00120 If I began with 0.050 moles of both reactants, how long would it take to make 0.01 moles of water?

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Orders of reaction [H2 ] M[O2] M Rate of formation of H2O M/sec 0.0500.1000.0048 0.0500.0500.0025 0.1000.0250.00120 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 =2 y Y =1

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Orders of reaction [H2 ] M[O2] M Rate of formation of H2O M/sec 0.0500.1000.0048 0.0500.0500.0025 0.1000.0250.00120 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*2 x 1.04 =2 x X=0

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Rate = k[O 2 ] [H2 ] M[O2] M Rate of H2O M/sec 0.0500.1000.0048 0.0500.0500.0025 0.1000.0250.00120 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 K avg = 0.0487 s -1

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1 st order kinetics ln [O 2 ] t = - kt [O 2 ] 0 2H 2 (g) + O 2 (g) → 2H 2 O (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 H 2 O * 1 mol O 2 = 0.005 mol O 2 2 mol H 2 O 0.050 mol O2 started – 0.005 mol O2 reacted = 0.045 mol left

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1 st order kinetics ln [O 2 ] t = - kt [O 2 ] 0 Ln (0.045/0.05) = - 0.0487s -1 t t = 2.16 seconds!

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Stoichiometry Notes. Equations to MEMORIZE 1 mol = (molar mass) g – Get it from the atomic mass. Add it up if you have a molecule. (O=15.999, so O 2 =31.998)

Stoichiometry Notes. Equations to MEMORIZE 1 mol = (molar mass) g – Get it from the atomic mass. Add it up if you have a molecule. (O=15.999, so O 2 =31.998)

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