Foundations of Chemical Kinetics Lecture 16: Unimolecular reactions - PowerPoint PPT Presentation

Foundations of Chemical Kinetics Lecture 16: Unimolecular reactions in the gas phase: The Lindemann mechanism Marc R. Roussel Department of Chemistry and Biochemistry

Examples of unimolecular reactions in the gas phase • C N C H 3 • N C CH 3 Isomerizations: • • Decompositions: CH 2 O CH 3 CH 2 CH 2 + CH 2 CH CH 2 C C CH 3 O C H 3 CH 3 CH 3 CH 2 What makes these reactions happen?

General behavior of unimolecular reactions ◮ These reactions are generally carried out with a “bath gas” (M) for which [M] ≫ [A] ([A] = reactant). ◮ At low pressures, the rate law has partial orders with respect to A and M of 1 each, so an overall order of 2. ◮ At high pressures, first-order kinetics in [A] is observed. These observations are both 1. a clue as to the mechanism, and 2. a puzzle to be solved.

The Lindemann mechanism k 1 A ∗ + M ⇀ A + M − ↽ − − − k − 1 A ∗ k 2 → products − Implied potential energy profile: A * E A P x

The Lindemann mechanism Rate equations k 1 A ∗ + M ⇀ A + M − ↽ − − − k − 1 A ∗ k 2 → products − Rate equations: d [A] = − k 1 [A][M] + k − 1 [A ∗ ][M] dt d [A ∗ ] = k 1 [A][M] − k − 1 [A ∗ ][M] − k 2 [A ∗ ] dt ◮ There is no way to solve these equations exactly.

The Lindemann mechanism Steady-state approximation ◮ There is a low barrier for reaction of A ∗ , so we expect k − 1 ≫ k 1 and k 2 ≫ k 1 [M]. ◮ Expected time course for [A ∗ ]: 1 0.9 0.8 [A*] (arbitrary units) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 100 200 300 400 500 t (arbitrary units)

The Lindemann mechanism Steady-state approximation ◮ During the long, slow decline of [A ∗ ], d [A ∗ ] ≈ 0 dt ◮ This is called the steady-state approximation. d [A ∗ ] = k 1 [A][M] − k − 1 [A ∗ ][M] − k 2 [A ∗ ] ≈ 0 dt k 1 [A][M] ∴ [A ∗ ] ≈ k − 1 [M] + k 2 ∴ v = k 2 [A ∗ ] ≈ k 1 k 2 [A][M] k − 1 [M] + k 2

The Lindemann mechanism v = k 1 k 2 [A][M] k − 1 [M] + k 2 ◮ Low-pressure limit: ◮ High-pressure limit:

The Lindemann mechanism Interpretation of the high-pressure limit k 1 A ∗ + M was the only ◮ Suppose that the reaction A + M ⇀ − ↽ − − − k − 1 one occurring. ◮ When this reaction reached equilibrium, we would have da dt = da ∗ dt = 0 ∴ k 1 [A][M] = k − 1 [A ∗ ][M] ∴ [A ∗ ] [A] = k 1 = K 1 k − 1 where K 1 is the equilibrium constant for this reaction. ◮ The high-pressure limit of the Lindemann rate equation is therefore v ≈ K 1 k 2 [A]

The Lindemann mechanism A slight rewrite v = k 1 k 2 [A][M] = k L [A] k − 1 [M] + k 2 k 1 k 2 [M] with k L = . k − 1 [M] + k 2 Define k ∞ = k 1 k 2 / k − 1 (high-pressure limit of k L ) k ∞ [M] Then, k L = . [M] + k ∞ / k 1

Experimental determination of the Lindemann parameters k ∞ [M] k L = . [M] + k ∞ / k 1 ∴ 1 = 1 + 1 1 k L k ∞ k 1 [M] vs [M] − 1 therefore allows us to recover k ∞ and k 1 . A plot of k − 1 L

Example: cis-trans isomerization of 2-butene at 740 K CH 3 H CH 3 C H 3 C C C C H H CH 3 H [2-butene] / 10 − 5 mol L − 1 0.25 0.3 0.6 1.2 5.9 k / 10 − 5 s − 1 1.05 1.14 1.43 1.65 1.82 Note: The theory is the same even if the “bath gas” is the reactant itself.

Example: cis-trans isomerization of 2-butene at 740 K (continued) 1 0.9 0.8 k -1 /10 5 s 0.7 0.6 0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 [2-butene] -1 /10 5 L mol -1

Example: cis-trans isomerization of 2-butene at 740 K (continued) slope = 0 . 106 mol s L − 1 1 slope = 9 . 4 L mol − 1 s − 1 ∴ k 1 = intercept = 0 . 53 × 10 5 s ∴ k ∞ = 1 . 9 × 10 − 5 s − 1

The value of k 1 ◮ If we vary the temperature, we can get the preexponential factor and activation energy corresponding to k 1 . ◮ We might guess that k 1 (rate constant for A + M → A ∗ + M) is collision limited. We should therefore be able to predict the pre-exponential factor from collision theory.

Collision-theory preexponential factor for the collisional activation of cyclopropane at 760 K ◮ The hard-sphere radius of cyclopropane is 2.2 ˚ A. ◮ σ = π (2 r ) 2 = 6 . 1 × 10 − 19 m 2 ◮ The mean relative speed of cyclopropane molecules is � 8 RT v r = ¯ πµ m � 8(8 . 314 472 J K − 1 mol − 1 )(760 K) = π (70 . 134 × 10 − 3 kg mol − 1 ) = 479 m s − 1 . v r L = 1 . 75 × 10 8 m 3 mol − 1 s − 1 ≡ 1 . 75 × 10 11 L mol − 1 s − 1 ◮ A ct = σ ¯ ◮ Experimental value: 9 × 10 18 L mol − 1 s − 1

Curved Lindemann plots 0.4 0.35 0.3 0.25 k -1 /10 5 s 0.2 0.15 0.1 0.05 0 0 50 100 150 200 250 300 350 400 p -1 /torr -1 Lindemann plot for the decomposition of cyclobutane to ethene at 449 ◦ C Source: Butler and Ogawa, JACS 85 , 3346 (1963).

Summary: the two problems with Lindemann theory 1. The rate constant k 1 exceeds the collision theory value, which should be an upper limit according to the theory studied so far. 2. Plots of k − 1 vs p − 1 , which should be straight, deviate from linearity at low pressures.