1. Kinetics Pt. 2 Temperature Dependence, Reaction Mechanisms, and Reaction Coordinates

2. Temperature and Rate
The rates of most chemical reactions increase with temperature.
How is this temperature dependence reflected in the rate expression?
Rates increase with temperature because rate constants increase with temperature. An example is the 2nd order reaction:
O3+ NO  O2 + NO2
𝐫𝐚𝐭𝐞=𝐤[ 𝐎 𝟑 ] 𝟏 [𝐍𝐎] 𝟏
Variation in k with temperature

3. Collision Model
The collision model makes sense of this. This model is based on kinetic theory
We’ve seen that the thermal energy of a molecule is converted to kinetic energy in order to facilitate motion, and we’ve seen that the velocity of a molecule increases with T.
The central idea of the collision model is that molecules must collide to react. The more collisions per second, the faster the reaction goes.
This model also rationalizes the concentration dependence. The more molecules present, the more collisions you have. 𝑉 𝑉

4. Activation Energy
Of course, there is more to a chemical reaction than just collisions of molecules.
Molecules must have some minimum kinetic energy in order to overcome the energy barrier of reaction. The kinetic energy of molecules is used to stretch, bend, and break bonds in order to cause a reaction.
If the molecules don’t have enough kinetic energy, they simply bounce off one another and nothing happens.
This minimum energy that molecules must have is called the activation energy,(Ea)

5. Ex. O3 + NO  NO2 + O2
1. Effective collision, activation energy absorbed, high energy intermediate forms (slow)
2. Intermediate decomposes and products form (fast)

6. Reaction Coordinate
To the right is a reaction coordinate plot of the reaction. The y-axis represents enthalpy. The x-axis represents the progress of reaction from reactants to products.
The activation energy is represented by the upward bump in the curve. The intermediate is the highest energy state of a reaction. Once the intermediate forms, the reaction is rapid and is energetically “downhill”
This reaction is exothermic because the products have lower enthalpy than the reactants.

7. Arrhenius Equation
Arrhenius noted that reaction-rate data depended on three aspects
(1) the fraction of molecules possessing an energy Ea or greater
(2) the number of collisions per second
(3) the fraction of molecules oriented in the right way for a reaction to occur
These factors are incorporated into the Arrhenius Equation
where k is the rate constant, R is the gas constant, T is temperature, Ea is the activation energy, and A is a value called the frequency factor, which is related to (2) and (3) above.
Higher Ea values result in much lower values of k, which directly alters reaction rate.
𝐥𝐧 𝒌=− 𝑬 𝒂 𝑹𝑻 + 𝐥𝐧 𝑨

8. Arrhenius Equation
The energy of molecules in a system is distributed. Not every molecule possesses the same energy at a given temperate and time. The fraction of molecules in a system with enough energy to react (E > Ea) is given by:
A plot of f vs Ek for reactant molecules is shown to the right. At higher values of T, a greater fraction of the curve is beyond Ea.
f = e − E a RT

9. The Arrhenius Equation is a Linear Equation
y=ln k
𝐥𝐧 𝒌=− 𝑬 𝒂 𝑹𝑻 + 𝐥𝐧 𝑨
ln k vs 1/T yields a linear plot with slope –Ea/R and a y-intercept of ln A
𝑚 𝑠𝑙𝑜𝑝𝑒 =− 𝐸 𝑎 𝑅
𝑥= 1 𝑇
𝑠𝑙𝑜𝑝𝑒=− 𝐸 𝑎 𝐽 𝑚𝑜𝑙 𝐾

10. Calculating Changes in k Mathematically
The Arrhenius equation can be rearranged once again to compare k values at different temperatures
Remember:
ln 𝑘 2 𝑘 1 = 𝐸 𝑎 𝑅 𝑇 2 − 𝑇 1 𝑇 1 𝑇 2
𝑒 ln⁡(𝑥) =𝑥

11. Example
A reaction has an activation energy of 43.5 kJ/mol. The reaction occurs at 298 K with a rate constant of 110 s-1. What will the rate constant be if we increase the temperature to 308 K?
ln 𝑘 2 𝑘 1 = 𝐸 𝑎 𝑅 𝑇 2 − 𝑇 1 𝑇 1 𝑇 2
ln 𝑘 = 𝐽 𝑚𝑜 𝑙 − 𝐽 𝑚𝑜 𝑙 −1 𝐾 − −
ln 𝑘 =.57
𝑒 𝑙𝑛 𝑘 = 𝑒 .57
𝑘 =1.77
𝑘 2 = 𝑠 −1

12. Reaction Mechanisms Consider the following reaction:
In looking at this equation, and applying your understanding of collision theory, you would assume that this reaction proceeds when
NO2 and CO collide
The activation energy is exceeded
The atoms rearrange to form NO and CO2
𝑵 𝑶 𝟐 𝒈 +𝑪𝑶 𝒈 →𝑵𝑶(𝒈)+𝑪 𝑶 𝟐 (𝒈)
Thus, we would expect the rate of the reaction to depend on both [NO2] and [CO]

13. Reaction Mechanisms
𝑵 𝑶 𝟐 𝒈 +𝑪𝑶 𝒈 →𝑵𝑶(𝒈)+𝑪 𝑶 𝟐 (𝒈)
However, according to experimental data, the reaction is 2nd order with respect to NO2 and zero order with respect to CO. The rate law is:
𝑟𝑎𝑡𝑒=𝑘[ 𝑁𝑂 2 ] 2 [𝐶𝑂 ] 0 =𝑘[ 𝑁𝑂 2 ] 2
This means that the rate of the reaction does not depend on [CO] at all. How can this be? What does this tell us about the reaction mechanism?
This particular reaction must proceed through multiple steps, and the rate-determining step must only depend on NO2

14. Reaction Mechanisms For this particular reaction, there are two steps:
𝑵 𝑶 𝟐 𝒈 +𝑪𝑶 𝒈 →𝑵𝑶(𝒈)+𝑪 𝑶 𝟐 (𝒈)
For this particular reaction, there are two steps:
1 𝑁 𝑂 2 𝑔 + 𝑁𝑂 2 𝑔 →𝑁 𝑂 3 𝑔 +𝑁𝑂 𝑔 (𝑠𝑙𝑜𝑤)
2 𝑁 𝑂 3 𝑔 +𝐶𝑂 𝑔 →𝑁 𝑂 2 𝑔 +𝐶 𝑂 2 𝑔 (𝑣𝑒𝑟𝑦 𝑓𝑎𝑠𝑡)
3 𝑁 𝑂 2 𝑔 +𝐶𝑂 𝑔 →𝑁𝑂 𝑔 +𝐶 𝑂 2 𝑔
This step-by-step description of the molecular pathway is the reaction mechanism. From reaction (1), we can physically understand why the reaction is 2nd order with respect to [NO2].
The 1st step is substantially slower than the 2nd. Therefore, step (1) acts as the “reaction bottleneck”, and the overall reaction speed can only be as fast as its slowest step.
Thus, step (1) is the rate determining step.

15. Each step in a reaction mechanism is called an elementary reaction.
An elementary reaction is any reaction that proceeds in a single step. Mechanisms consist of sums of elementary reactions.
Going back to rate laws, each elementary reaction must have a corresponding rate constant
*** For elementary reactions only, you can assume that the rate law depends directly on the number of species present. Thus, the rate law of an elementary reaction follows the stoichiometry.
intermediate k1
1 𝑁 𝑂 2 𝑔 + 𝑁𝑂 2 𝑔 𝑁 𝑂 3 𝑔 +𝑁𝑂 𝑔 (𝑠𝑙𝑜𝑤) k2
2 𝑁 𝑂 3 𝑔 +𝐶𝑂 𝑔 𝑁 𝑂 2 𝑔 +𝐶 𝑂 2 𝑔 (𝑣𝑒𝑟𝑦 𝑓𝑎𝑠𝑡)
𝑟𝑎𝑡 𝑒 1 = 𝑘 1 𝑁 𝑂 2 𝑁 𝑂 2 = 𝑘 1 [𝑁 𝑂 2 ] 2
rate of overall reaction

16. Catalysis The rates of reactions can be increased by using catalysts.
Catalysts work by providing a different reaction path (mechanism) between reactants and products with a lower activation energy
The catalyst itself is not consumed in the reaction.

17. Example: Oxidation of Sulfur Dioxide Pt-Catalyzed
uncatalyzed
2SO2(g) + O2(g)  2SO3(g)
(1) 2SO2(g) ---> 2S(s) + 2O2(g) ΔHo(1) = kJ/mol (slow step)
(2) 2S(s) + 3O2(g) ---> 2SO3(g) ΔHo (2) = kJ/mol
∆ 𝐻 𝑟𝑥𝑛 𝑜 =−197.8 𝑘𝐽 𝑚𝑜𝑙
catalyzed
2SO2(g) + O2(g)  2SO3(g) (1) O2(g) ---> 2O ΔHorxn (1) = kJ/mol (slow step) (2) SO2(s) + O(g) ---> SO3(g) ΔHorxn (2) = kJ/mol Pt
∆ 𝐻 𝑟𝑥𝑛 𝑜 =−197.8 𝑘𝐽 𝑚𝑜𝑙

18. Example: Oxidation of Sulfur Dioxide Pt-Catalyzed