Although the rate constants k 1 and k 2 900c the rate

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The activation energy can be calculated using the Arrhenius equation. Although the rate constants, k 1 and k 2 , 90.0°C, the rate of reaction is 1 egg/4.8 min while at T 2 = 100.0°C, the rate is 1 egg/4.5 min. rate = 1 egg/4.8 min and rate = 1 egg/4.5 min are substituted for k 1 and k 2 , respectively. are not expressly stated, the relative times give an idea of the rate. The reaction rate is proportional to the rate constant. At T 1 = Therefore, 1 2 Solution: T 1 = 90.0°C + 273.2 = 363.2 K T 2 = 100.0°C + 273.2 = 373.2 K k 1 = 1 egg/4.8 min k 2 = 1 egg/4.5 min E a = ? 2 1 ln k k = a 1 E 2 1 1 T The number of eggs (1) is exact, and has no bearing on the significant figures. R T E a = 2 1 2 1 T T ln 1 1 k R k = 1 egg/4.5min J 8.314 ln mol•K 1 egg/4.8min 1 1 373 .2 K 363.2 K 3 J/mol 16.107 E = 7.2730x10 3 J/mol = 7.3x10 a Plan: The overall reaction can be obtained by adding the three steps together. The overall rate law for ed from the slowest step (the rate-determining step). An overall rate law can only include reactants and ; intermediates cannot be included in the rate law. Express [intermediate] in terms of [reactant]. Solution: the mechanism is determin products (1) 2H 2 SO 4 H 3 O + + HSO + SO 4 3 [fast] (2) SO + )SO [slow] + C 6 H 6 H(C H 3 6 5 3 (3) H(C 6 H 5 + + H C )SO SO H SO 3 4 6 5 3 + (4) C H 2 SO 4 [fast] 6 H 5 SO 3 + H O 3 + C H SO H + H O 6 5 3 2 [fast] a) Add the steps together and cancel: nt on k and [H SO ]: 16.108 C 6 H 6 + H 2 SO 4 C 6 H 5 SO 3 H + H 2 O b) Initially: Rate = k 2 [SO 3 ][C 6 H 6 ] (from the slow step) SO is an intermediate and cannot be included in the overall rate law. SO 3 is produced in step 1 and 3 its concentration is depende 1 2 4 [SO ] = k [H SO ] 3 1 2 4 Substituting for [SO 2 3 ] in the rate law from the slow step: Rate = k [ k [H SO ] 2 1 2 4 6 6 Rate = k [H 2 ][C H ] 2 SO 4 ] 2 [C 6 H 6 ] Plan: Starting with the fact that rate of formation of O (rate of step 1) equals the rate of consumption of O (rate of f k 1 , k 2 , [NO 2 ], and [O 2 ]. step 2), set up an equation to solve for [O] using the given values o Solution: R = rate a) Rate = k 1 ate 1 [NO 2 ] Rate 2 = k 2 [O][O 2 } 1 2 k [NO ] = k [O][O 2 ] 1 2 2 [O] = 1 2 2 2 NO k = 3 1 9 6.0x10 s 4.0x10 M O k 6 2 1.0x10 L/mol•s 1.0x10 = 2.4x10 M M ps is equal, either can be used to dete –15 rmine rate of formation of ozone. b) Since the rate of the two ste Rate 2 = k 2 [O][O 2 ] = (1.0x10 6 L/mol•s)(2.4x10 –15 M )(1.0x10 –2 M ) = 2.4x10 –11 mol/L · s 16-35