Foundations of Chemical Kinetics Lecture 12: Transition-state - PowerPoint PPT Presentation

Foundations of Chemical Kinetics Lecture 12: Transition-state theory: Examples Marc R. Roussel Department of Chemistry and Biochemistry

Eyring plot example: Internal rotation of N,N-dimethylnicotinamide O C H 3 N CH 3 N This elementary process can be studied by NMR. The following data were obtained: T / ◦ C 10.0 15.7 21.5 27.5 33.2 38.5 45.7 k / s − 1 2.08 4.57 8.24 15.8 28.4 46.1 93.5 We want to calculate the enthalpy, entropy, and free energy of activation.

Eyring plot example: Internal rotation of N,N-dimethylnicotinamide (continued) For each point, we calculate ln( kh / k B T ) and plot this vs T − 1 : -24.5 -25 -25.5 -26 ln( kh / k B T ) -26.5 -27 -27.5 -28 -28.5 -29 3.1 3.15 3.2 3.25 3.3 3.35 3.4 3.45 3.5 3.55 T -1 /10 -3 K -1 Slope = − 9185 K Intercept = 3 . 829

Example: Eyring plot (continued) ∆ ‡ H ◦ m = − R (slope) = − (8 . 314 472 J K − 1 mol − 1 )( − 9185 K) = 76 kJ mol − 1 ∆ ‡ S ◦ m = R (intercept) = (8 . 314 472 J K − 1 mol − 1 )(3 . 829) = 32 J K − 1 mol − 1 ∆ ‡ G ◦ m = ∆ ‡ H ◦ m − T ◦ ∆ ‡ S ◦ m = 76 kJ mol − 1 − (298 . 15 K)(32 × 10 − 3 kJ K − 1 mol − 1 ) = 67 kJ mol − 1 Interpret ∆ ‡ S ◦ m . Do the sign and size of this quantity make sense? What does the reaction profile look like?

Statistical factors ◮ The reaction AB + C → A + BC should, all other things being equal, be slower than A 2 + C → A + AC because in the latter case, C can attack either end of the molecule and a reaction can still occur. ◮ We deal with this in the statistical approach to transition state theory by adding a statistical factor ( L ‡ ) to the rate constant: Q ‡ � − ∆ E 0 � k = L ‡ k B T N exp c ◦ h Q X Q Y RT (second-order reaction) ◮ The statistical factor is the number of distinct, equivalent transition states that can be formed if we label equivalent atoms in the reactants.

Statistical factors: examples 1. H + D 2 → HD + D 2. CH 3 + C 2 H 6 → CH 4 + C 2 H 5

Calculation of the rate constant for the reaction D + H 2 → HD + H Data: H 2 : ◮ R = 0 . 741 ˚ A ◮ ˜ ν = 4400 cm − 1 Transition state: ◮ Linear with R DH = R HH = 0 . 930 ˚ A ◮ Classical barrier height: 40 . 2 kJ mol − 1 ◮ Vibrational frequencies: ˜ ν sym stretch = 1764, ν bend = 870 cm − 1 with g bend = 2 ˜ To do: Calculate the rate constant at 450 K.

D + H 2 → HD + H (continued) ◮ L ‡ = 2 ◮ The factor involving the partition functions can be factored into terms for each type of motion: � Q ‡ � Q ‡ Q ‡ Q ‡ Q ‡ � � � � � � = Q D Q H 2 Q D Q H 2 Q D Q H 2 Q H 2 Q H 2 elec tr rot vib ◮ The first excited states of both D and H 2 are very high in energy, so their partition functions at moderate temperatures are just their respective ground-state degeneracies. The same turns out to be true for the transition state. g elec , H 2 = 1, g elec , D = 2, g elec , TS = 2 (one unpaired electron) Q ‡ � � = 1 ∴ Q D Q H 2 elec

D + H 2 → HD + H (continued) ◮ Recall Q tr = V h 3 (2 π mk B T ) 3 / 2 � 3 / 2 Q ‡ = h 3 � � � m TS ∴ 2 π k B Tm D m H 2 Q D Q H 2 V tr ◮ The masses are readily computed: m D = 3 . 344 × 10 − 27 kg, m H 2 = 3 . 347 × 10 − 27 kg, m TS = m D + m H 2 = 6 . 692 × 10 − 27 kg. ◮ Putting in all the constants, we get = 5 . 513 × 10 − 31 m 3 Q ‡ � � Q D Q H 2 V tr

D + H 2 → HD + H (continued) ◮ The vibrational partition function is ν/ k B T )] − 1 , so Q vib = [1 − exp( − hc ˜ � Q ‡ � 1 − exp( − hc ˜ ν H 2 / k B T ) = ν bend / k B T )] 2 . Q H 2 [1 − exp( − hc ˜ ν sym / k B T ] [1 − exp( − hc ˜ vib ◮ Putting in all the data, we get � Q ‡ / Q H 2 � vib = 1 . 140.

D + H 2 → HD + H (continued) ◮ For a linear triatomic molecule, R R 2 1 m m m 1 2 3 I = m 1 m 3 ( R 1 + R 2 ) 2 + m 2 m ( m 1 R 2 1 + m 3 R 2 2 ) m with m = m 1 + m 2 + m 3 . ◮ For the transition state of this reaction, m 1 = m D , m 2 = m 3 = m H and R 1 = R 2 = R ‡ : I TS = m H m ( R ‡ ) 2 (5 m D + m H ) = 3 . 979 × 10 − 47 kg m 2 ◮ For a diatomic molecule, I = µ R 2 , which gives I H 2 = 4 . 595 × 10 − 48 kg m 2 .

D + H 2 → HD + H (continued) ◮ The rotational partition function of a linear molecule is Q rot = 8 π 2 Ik B T h 2 ◮ The rotational term in the transition-state equation is therefore � Q ‡ � = I TS = 8 . 661 . Q H 2 I H 2 rot ◮ Putting it all together now, we have � 5 . 513 × 10 − 31 m 3 Q ‡ � = (1) (8 . 661)(1 . 192) Q D Q H 2 V = 5 . 693 × 10 − 30 m 3 V

D + H 2 → HD + H (continued) ◮ The zero-point energy of H 2 is ν H 2 = 4 . 370 × 10 − 20 J ≡ 26 . 32 kJ mol − 1 . 1 2 hc ˜ ◮ The zero-point energy of the transition state is the sum of the zero-point energies for each of its vibrational modes added to the height of the barrier: ZPE TS = L 2 hc (˜ ν sym stretch + 2˜ ν bend ) + E b = 61 . 16 kJ mol − 1 ◮ Therefore ∆ E 0 = 61 . 16 − 26 . 32 kJ mol − 1 = 34 . 84 kJ mol − 1

D + H 2 → HD + H (continued) ◮ If we want the rate constant in L mol − 1 s − 1 , pick V = 10 − 3 m 3 , N = 6 . 022 141 29 × 10 23 , and c ◦ = 1 mol L − 1 . ◮ Plugging everything in, we get k = 5 . 6 × 10 6 L mol − 1 s − 1 ◮ The experimental value is k = 9 × 10 6 L mol − 1 s − 1 ◮ Main source of error: tunneling, which is very significant for light particles