solut15 - CHAPTER FIFTEEN CHEMICAL KINETICS Reaction Rates...

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438 CHAPTER FIFTEEN CHEMICAL KINETICS Reaction Rates 10. The coefficients in the balanced reaction relate the rate of disappearance of reactants to the rate of production of products. From the balanced reaction, the rate of production of P 4 will be 1/4 the rate of disappearance of PH 3 and the rate of production of H 2 will be 6/4 the rate of disappearance of PH 3 . By convention, all rates are given as positive values. Rate = t ] PH [ - 3 = s L) 2.0 mol/ (-0.0048 - = 2.4 × 10 -3 mol L -1 s -1 t ] PH [ 4 1 - = t ] P [ 3 4 = 2.4 × 10 -3 /4 = 6.0 × 10 -4 mol L -1 s -1 t ] PH [ 4 6 - = t ] H [ 3 2 = 6(2.4 × 10 -3 )/4 = 3.6 × 10 -3 mol L -1 s -1 11. Using the coefficients in the balanced equation to relate the rates: dt ] N d[ 2 - = dt ] NH d[ and dt ] N d[ 3 = dt ] H d[ 2 3 2 2 ; So, dt ] NH d[ 2 1 - = dt ] H d[ 3 1 3 2 or dt ] H d[ 3 2 - = dt ] NH d[ 2 3 Ammonia is produced at a rate equal to 2/3 of the rate of consumption of hydrogen. 12. a. The units for rate are always mol L -1 s -1 . b. Rate = k; k has units of mol L -1 s -1 .
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439 c. Rate = k[A], ÷ ø ö ç è æ L mol k = s L mol d. Rate = k[A] 2 , ÷ ø ö ç è æ L mol k = s L mol 2 k must have units of s -1 . k must have units L mol -1 s -1 . e. L 2 mol -2 s -1 13. Rate = k[Cl] 1/2 [CHCl 3 ], ÷ ø ö ç è æ ÷ ø ö ç è æ L mol L mol k = s L mol 2 / 1 ; k must have units of L 1/2 mol -1/2 s -1 . 14. mol molecules 10 _ 6.022 _ cm 1000 L 1 _ s molecules cm 10 _ 1.24 23 3 3 -12 = 7.47 × 10 8 L mol -1 s -1 Rate Laws from Experimental Data: Initial Rates Method 15. a. In the first two experiments, [NO] is held constant and [Cl 2 ] is doubled. The rate also doubled. Thus, the reaction is first order with respect to Cl 2 . Or mathematically: Rate = k[NO] x [Cl 2 ] y ) (0.10 ) (0.20 = ) (0.10 ) k(0.10 ) (0.20 ) k(0.10 = 0.18 0.36 y y y x y x , 2.0 = 2.0 y , y = 1 We can get the dependence on NO from the second and third experiments. Here, as the NO concentration doubles (Cl 2 concentration is constant), the rate increases by a factor of four. Thus, the reaction is second order with respect to NO. Or mathematically: ) (0.10 ) (0.20 = (0.20) ) k(0.10 (0.20) ) k(0.20 = 0.36 1.45 x x x x , 4.0 = 2.0 x , x = 2; So, Rate = k[NO] 2 [Cl 2 ] Try to examine experiments where only one concentration changes at a time. The more variables that change, the harder it is to determine the orders. Also, these types of problems can usually be solved by inspection. In general, we will solve using a mathematical approach, but keep in mind, you probably can solve for the orders by simple inspection of the data. b. The rate constant k can be determined from the experiments. From experiment 1:
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CHAPTER 15 CHEMICAL KINETICS 440 ÷ ø ö ç è æ ÷ ø ö ç è æ L mol 0.10 L mol 0.10 k = min L mol 0.18 2 , k = 180 L 2 mol -2 min -1 From the other experiments: k = 180 L 2 mol -2 min -1 (2nd exp.); k = 180 L 2 mol -2 min -1 (3rd exp.) The average rate constant is k mean = 1.8 × 10 2 L 2 mol -2 min -1 16. a. Rate = k[I - ] x [S 2 O 8 2- ] y ; ) (0.040 ) k(0.040 ) (0.040 ) k(0.080 = 10 _ 6.25 10 _ 12.5 y x y x 6 - -6 , 2.00 = 2.0 x , x = 1 ) (0.020 ) k(0.080 ) (0.040 ) k(0.080 = 10 _ 6.25 10 _ 12.5 y x y x 6 - -6 , 2.00 = 2.0 y , y = 1; Rate = k[I - ][S 2 O 8 2- ] b. For the first experiment: ÷ ø ö ç è æ ÷ ø ö ç è æ L mol 0.040 L mol 0.080 k = s L mol 10 _ 12.5 -6 , k = 3.9 × 10 -3 L mol -1 s -1 The other values are: Initial Rate k (mol L -1 s -1 )( L m o l -1 s -1 ) 12.5 × 10 -6 3.9 × 10 -3 6.25 × 10 -6 3.9 × 10 -3 6.25 × 10 -6 3.9 × 10 -3 5.00 × 10 -6 3.9 × 10 -3 7.00 × 10 -6 3.9 × 10 -3 k mean = 3.9 × 10
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solut15 - CHAPTER FIFTEEN CHEMICAL KINETICS Reaction Rates...

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