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chapter_2014 - CHAPTER 14 CHEMICAL KINETICS 14.5 In general...

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CHAPTER 14 CHEMICAL KINETICS 14.5 In general for a reaction a A + b B c C + d D 1[ A ] B ] 1[ C D ] rate ΔΔ Δ Δ =− = = Δ Δ tt t t ab c d (a) 22 [H ] [I ] H I ] rate 2 Δ = ΔΔΔ ttt (b) 22 2 [H ] [O ] [H O] 11 rate = Δ t (c) 3 2 [BrO ] [Br ] B r] H] 1 rate 56 3 −+ Δ Δ = Δ Δ t t Note that because the reaction is carried out in the aqueous phase, we do not monitor the concentration of water. 14.6 Strategy: The rate is defined as the change in concentration of a reactant or product with time. Each “change in concentration” term is divided by the corresponding stoichiometric coefficient. Terms involving reactants are preceded by a minus sign. 3 [NH ] [N ] [H ] rate = = 32 Δ Δ t Solution: (a) If hydrogen is reacting at the rate of 0.074 M /s, the rate at which ammonia is being formed is 3 2 [H ] = 23 Δ Δ or 3 2 [H ] 2 = 3 Δ Δ 3 2 ( 0.074 /s) 3 Δ =− − = Δ 0.049 /s M t M (b) The rate at which nitrogen is reacting must be: == ( 0 . 0 7 4 / s ) 33 −= 0.025 /s M M Will the rate at which ammonia forms always be twice the rate of reaction of nitrogen, or is this true only at the instant described in this problem? 14.15 rate = k [NH 4 + ][NO 2 ] = (3.0 × 10 4 / M s)(0.26 M )(0.080 M ) = 6.2 × 10 6 M /s
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CHAPTER 14: CHEMICAL KINETICS 288 14.16 (a) Assume the rate law has the form: rate = k [F 2 ] x [ClO 2 ] y To determine the order of the reaction with respect to F 2 , find two experiments in which the [ClO 2 ] is held constant. Compare the data from experiments 1 and 3. When the concentration of F 2 is doubled, the reaction rate doubles. Thus, the reaction is first-order in F 2 . To determine the order with respect to ClO 2 , compare experiments 1 and 2. When the ClO 2 concentration is quadrupled, the reaction rate quadruples. Thus, the reaction is first-order in ClO 2 . The rate law is: rate = k [F 2 ][ClO 2 ] (b) The value of k can be found using the data from any of the experiments. If we take the numbers from the second experiment we have: 3 11 22 rate 4.8 10 /s =1 . 2 s [F ][ClO ] (0.10 )(0.040 ) × == M kM MM Verify that the same value of k can be obtained from the other sets of data. (c) Since we now know the rate law and the value of the rate constant, we can calculate the rate at any concentration of reactants. rate = k [F 2 ][ClO 2 ] = (1.2 M 1 s 1 )(0.010 M )(0.020 M ) = 2.4 × 10 4 M /s 14.17 By comparing the first and second sets of data, we see that changing [B] does not affect the rate of the reaction. Therefore, the reaction is zero-order in B. By comparing the first and third sets of data, we see that doubling [A] doubles the rate of the reaction. This shows that the reaction is first-order in A. rate = k [A] From the first set of data: 3.20 × 10 1 M /s = k (1.50 M ) k = 0.213 s 1 What would be the value of k if you had used the second or third set of data? Should k be constant? 14.18 Strategy: We are given a set of concentrations and rate data and asked to determine the order of the reaction and the initial rate for specific concentrations of X and Y. To determine the order of the reaction, we need to find the rate law for the reaction. We assume that the rate law takes the form rate = k [X] x [Y] y How do we use the data to determine x and y ?
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This note was uploaded on 05/22/2008 for the course CHEM 122 taught by Professor Bellew during the Spring '07 term at New Mexico.

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chapter_2014 - CHAPTER 14 CHEMICAL KINETICS 14.5 In general...

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