Chemical Kinetics: Reaction Rates, Collision Theory, and Mechanisms

Chemical Kinetics

Rate Constant and Temperature

Rate Law: The rate of a reaction is expressed as rate = k[A]^x[B]^y where:
- k is the rate constant.
- [A] and [B] are the concentrations of reactants A and B.
- x and y are the reaction orders with respect to A and B, respectively.
Dependence on Temperature:
- The rate of reaction generally increases with increasing temperature (T).
- The rate constant k does not depend on concentration.
- The rate constant k depends only on temperature (T).
- When T increases, k increases, and consequently, the rate of reaction increases.

Collision Theory of Chemical Reactions

Effective Collisions: Collisions between reactant molecules that result in a chemical reaction.
Factors Determining Reaction: For molecules to react upon collision, two conditions must be met:
1. Sufficient Energy: The collision must have enough energy to overcome the activation energy barrier (E_a).
2. Proper Orientation: The reacting molecules must collide in the proper geometric orientation for new bonds to form.
Example Reaction: Cl + NOCl \rightarrow Cl_2 + NO

Activated Complex (Transition State) and Activation Energy

Activated Complex: When molecules collide effectively, they form a short-lived, high-energy intermediate called an activated complex (or transition state).
Activation Energy (E_a): The minimum energy required for a reaction to occur (to form the activated complex).
- Represented in potential energy diagrams as the energy difference between reactants and the activated complex.
Relationship between E_a and \Delta H:
- For an endothermic reaction (\Delta H > 0):
  - E.g., if E_a (forward) = 45 \text{ kJ/mol} and \Delta H = +28.0 \text{ kJ/mol}.
  - E{a,rev} = Ea - \Delta H = 45 \text{ kJ/mol} - (+28.0 \text{ kJ/mol}) = 17 \text{ kJ/mol} (as demonstrated on page 8).
- For an exothermic reaction (\Delta H < 0):
  - E.g., if E_a (forward) = 45 \text{ kJ/mol} and \Delta H = -28.0 \text{ kJ/mol}.
  - E{a,rev} = Ea + |\Delta H| = 45 \text{ kJ/mol} + 28.0 \text{ kJ/mol} = 73 \text{ kJ/mol} (as demonstrated on page 6).

Temperature and Kinetic Energy Distribution

The rate of reaction increases with increasing temperature due to the increased kinetic energy of molecules.
At higher temperatures (T2 > T1), a larger fraction of colliding molecules possess kinetic energy equal to or greater than the activation energy (E_a), leading to more effective collisions and a faster reaction rate.

Arrhenius Equation: Dependence of Rate Constant on Temperature

The Arrhenius equation describes the quantitative relationship between the rate constant (k) and temperature (T): k = A e^{-E_a/RT} Where:
- k is the rate constant.
- A is the frequency factor (also known as the collision frequency), representing the number of times reactants approach the activation barrier per unit time. It also incorporates the orientation factor (fraction of collisions with proper orientation).
- E_a is the activation energy (in J/mol or kJ/mol).
- R is the ideal gas constant (8.314 \text{ J/(mol}\cdot\text{K)}).
- T is the absolute temperature (in Kelvin).
- e^{-E_a/RT} is the exponential factor, representing the fraction of successful collisions that surmount the activation barrier and form products.
Factors influencing the exponential factor:
- Increases with increasing temperature (T).
- Decreases with increasing activation energy (E_a).
Does Ea change with temperature? No, the activation energy (Ea) is independent of temperature. The rate constant increases with temperature because the fraction of molecules with sufficient energy to overcome E_a increases.

Orientation Factor

The orientation factor is the fraction of collisions that have an orientation suitable for a reaction to occur.
Smallest Orientation Factor: Reactions involving complex molecules or specific atomic alignments will have smaller orientation factors.
- Example from slide 12: O2(g) + O(g) \rightarrow O3(g) would likely have the smallest orientation factor among the given choices because of the need for a specific alignment to form ozone, while reactions like H(g) + I(g) \rightarrow HI(g) involve simpler atomic collisions with fewer orientation constraints.

Logarithmic Forms of the Arrhenius Equation

Linear Form: Taking the natural logarithm of the Arrhenius equation yields: \ln k = \ln A - \frac{Ea}{RT} This can be rearranged into a linear equation (y = mx + b): \ln k = -\frac{Ea}{R} \left(\frac{1}{T}\right) + \ln A
- Plotting \ln k (y-axis) versus \frac{1}{T} (x-axis) will yield a straight line.
- Slope (m): -E_a/R
- Y-intercept (b): \ln A
Two-Point Form: If rate constants (k1, k2) are known at two different temperatures (T1, T2), the activation energy can be calculated:
\ln \left(\frac{k2}{k1}\right) = -\frac{Ea}{R} \left(\frac{1}{T2} - \frac{1}{T1}\right) or \ln \left(\frac{k2}{k1}\right) = \frac{Ea}{R} \left(\frac{1}{T1} - \frac{1}{T2}\right)
This form allows calculation of Ea from two data points, or a rate constant at a new temperature if Ea and one (k, T) pair are known.

Applications of Arrhenius Equation (Examples)

Determining E_a from Plot:
- Given data for \ln k and \frac{1}{T} for the reaction CO(g) + NO2(g) \rightarrow CO2(g) + NO(g).
- A graph of \ln k vs. \frac{1}{T} yields a slope of -5.6 \times 10^3 \text{ K}.
- Since slope = -E_a/R
- E_a = -(slope) \times R = -(-5.6 \times 10^3 \text{ K}) \times (8.314 \text{ J/(mol}\cdot\text{K)}) = 46.56 \text{ kJ/mol} (rounded to 46 \text{ kJ/mol} in slide 20).
Determining E_a from Linear Regression:
- Given a trendline equation: y = -3.2549x + 14.427 for \ln k vs. 1/T \times 1000
- The slope is -3.2549 \times 10^3 \text{ K} (since x-axis is 1/T \times 1000).
- E_a = - (slope) \times R = - (-3.2549 \times 10^3 \text{ K}) \times (8.314 \text{ J/(mol}\cdot\text{K)}) \approx 27060 \text{ J/mol} = 27 \text{ kJ/mol}.
Refrigerator Principle: Food is kept in a refrigerator because lowering the temperature significantly decreases reaction rates (e.g., spoilage reactions).
Doubling Rate for 10^{\circ}C Increase: If the rate of reaction doubles with a 10^{\circ}C increase around room temperature (T1 = 25^{\circ}C = 298 \text{ K}, T2 = 35^{\circ}C = 308 \text{ K} and k2/k1 = 2).
- \ln (2) = \frac{E_a}{8.314 \text{ J/(mol}\cdot\text{K)}} \left(\frac{1}{298 \text{ K}} - \frac{1}{308 \text{ K}}\right)
- 0.693 = \frac{Ea}{8.314} (0.003355 - 0.003247) = \frac{Ea}{8.314} (0.000108)
- E_a = \frac{0.693 \times 8.314}{0.000108} \approx 53360 \text{ J/mol} = 53.4 \text{ kJ/mol}.
Calculating T2: Given Ea = 55 \text{ kJ/mol}, T1 = 20^{\circ}C = 293 \text{ K}, and rate2 = 1.7 \times rate1 (so k2/k_1 = 1.7).
- \ln (1.7) = \frac{55000 \text{ J/mol}}{8.314 \text{ J/(mol}\cdot\text{K)}} \left(\frac{1}{293 \text{ K}} - \frac{1}{T_2}\right)
- 0.5306 = 6615.35 \left(\frac{1}{293} - \frac{1}{T_2}\right)
- 8.021 \times 10^{-5} = 0.003413 - \frac{1}{T_2}
- \frac{1}{T_2} = 0.003413 - 8.021 \times 10^{-5} = 0.003333
- T_2 = 1/0.003333 \approx 300 \text{ K} = 27^{\circ}C.

Reaction Mechanisms

Definition: The sequence of elementary steps that sum to give the overall balanced chemical reaction.
- A balanced equation (e.g., 2NO + O2 \rightarrow 2NO2) does not show how the reaction occurs at a molecular level.
Intermediates: Chemical species that are produced in one step of a reaction mechanism and consumed in a subsequent step. They do not appear in the overall balanced chemical equation.
- Example: For 2NO + O2 \rightarrow 2NO2
  - Step 1: NO + NO \rightarrow N2O2
  - Step 2: N2O2 + O2 \rightarrow 2NO2
  - Overall: 2NO + O2 \rightarrow 2NO2
  - N2O2 is an intermediate.

Elementary Reactions and Molecularity

Elementary Reaction: A reaction that occurs in a single collision of the reactant molecules.
Molecularity: The number of reactant molecules involved in an elementary reaction/collision.
- Unimolecular: One reactant molecule (e.g., A \rightarrow products). Rate law: rate = k[A] (first order).
- Bimolecular: Two reactant molecules (e.g., A + B \rightarrow products or A + A \rightarrow products). Rate laws: rate = k[A][B] or rate = k[A]^2 (both second order).
- Termolecular: Three reactant molecules. These are rare due to the low probability of three molecules colliding simultaneously in the correct orientation and with sufficient energy.
Rate Laws for Elementary Reactions: For elementary reactions only, the exponents in the rate law are equal to the stoichiometric coefficients of the reactants.
- Disclaimer: This is valid only for elementary reactions. The rate law of a multi-step reaction can only be determined experimentally (e.g., by the method of initial rates).
- Example: For an elementary step A + 2B \rightarrow products, its rate law is rate = k[A][B]^2.

Rate-Determining Step (RDS)

Definition: In a multi-step mechanism, the rate-determining step is the slowest elementary step in the sequence. It limits the overall rate of the reaction.
Significance:
- The overall rate of reaction cannot be faster than the rate of the slowest step.
- The rate law of the rate-determining step determines the rate law of the overall reaction.
- The slowest step usually has the largest activation energy (E_a).
Potential Energy Diagram: A diagram showing an intermediate formed, and the slow step having a higher E_a than the fast step.

Validating a Reaction Mechanism

To propose and validate a mechanism, two criteria must be met:

Sum to Overall Reaction: The elementary steps of the mechanism must add up to the overall balanced chemical reaction.
Consistent Rate Law: The rate law predicted by the rate-determining step of the mechanism must be consistent with the experimentally observed rate law.

Examples of Mechanism Validation and Rate Law Derivation

Mechanism with a Slow First Step:
- Mechanism: NO2(g) + Cl2(g) \rightarrow ClNO2(g) + Cl(g) (Slow) NO2(g) + Cl(g) \rightarrow ClNO_2(g) (Fast)
- Overall reaction: 2NO2(g) + Cl2(g) \rightarrow 2ClNO_2(g)
- Intermediates: Cl(g)
- Predicted rate law: Since the first step is slow, its rate law determines the overall rate law. rate = k[NO2][Cl2].
Mechanism with a Fast Equilibrium First Step (e.g., 2NO(g) + H2(g) \rightarrow 2H2O(g) + N_2(g)):
1. 2NO(g) \rightleftharpoons N2O2(g) (Fast equilibrium, forward rate k1[NO]^2, reverse rate k{-1}[N2O2])
2. H2(g) + N2O2(g) \rightarrow H2O(g) + N_2O(g) (Slow)
3. H2(g) + N2O(g) \rightarrow H2O(g) + N2(g) (Fast)
- Overall reaction: 2H2(g) + 2NO(g) \rightarrow 2H2O(g) + N_2(g)
- Intermediates: N2O2(g) and N_2O(g).
- 1. Write the rate law of the slowest step: rate = k2[H2][N2O2].
- 2. Express the concentration of the intermediate (N2O2) from the fast equilibrium step. Since it's an equilibrium, the forward and reverse rates are equal:
  k1[NO]^2 = k{-1}[N2O2]
  [N2O2] = \frac{k1}{k{-1}} [NO]^2 = K_{eq}[NO]^2
- 3. Substitute the intermediate's concentration into the slow step's rate law:
  rate = k2[H2] \left( \frac{k1}{k{-1}} [NO]^2 \right) = \left( \frac{k1 k2}{k{-1}} \right) [H2][NO]^2
  rate{obs} = k{total}[H2][NO]^2 (where k{total} = k1 k2 / k_{-1}).
Detailed Example from Slide 53-54:
1. Cl2(g) \rightleftharpoons 2Cl(g) (Fast equilibrium, constants k1, k_{-1})
2. Cl(g) + CHCl3(g) \rightarrow HCl(g) + CCl3(g) (Slow, constant k_2)
3. Cl(g) + CCl3(g) \rightarrow CCl4(g) (Fast, constant k_3)
- Overall reaction: Cl2(g) + CHCl3(g) \rightarrow HCl(g) + CCl_4(g)
- Intermediates: Cl(g) and CCl_3(g).
- Predicted rate law:
  1. Slow step rate law: rate = k2[Cl][CHCl3]
  2. Express intermediate [Cl] from rapid equilibrium (Step 1):
    k1[Cl2] = k{-1}[Cl]^2 \Rightarrow [Cl]^2 = \frac{k1}{k{-1}}[Cl2] \Rightarrow [Cl] = \sqrt{\frac{k1}{k{-1}}}[Cl_2]^{0.5}
  3. Substitute into slow step rate law:
    rate = k2 \left( \sqrt{\frac{k1}{k{-1}}}[Cl2]^{0.5} \right) [CHCl3] = k^*[Cl2]^{0.5}[CHCl_3]

Catalysis

Catalyst Definition: A substance that increases the rate of a chemical reaction without being consumed in the reaction itself.
Mechanism of Action: A catalyst speeds up a reaction by providing an alternative reaction pathway (a new set of elementary steps) with a lower activation energy (E_a).
- It does not change the overall enthalpy (\Delta H) of the reaction.
- It increases the rate of both the forward and reverse reactions equally, thereby helping the system reach equilibrium faster but not changing the position of equilibrium (the equilibrium constant, K_{eq}).
- The catalyst typically participates in one step of the mechanism and is regenerated in a later step.
Effects of a Catalyst:
- Activation energy of forward reaction: Decreased
- Activation energy of reverse reaction: Decreased
- Rate of forward reaction: Increased
- Rate of reverse reaction: Increased
- \Delta H_{rxn}: Not affected
- The overall reaction: Not affected (reactants and products remain the same)
- The mechanism: Changed (catalyst provides an alternative pathway)

Quantifying Catalysis Effect

If you know the activation energies before (E{a1}) and after (E{a2}) catalysis, you can compare the rate constants (k1, k2) at a given temperature (T):
\ln \left(\frac{k2}{k1}\right) = \frac{1}{R T} (E{a1} - E{a2})
Where k1 is the rate constant before catalysis and k2 is the rate constant after catalysis.

Catalysis Examples

Calculating Uncatalyzed Rate Constant: Given catalyzed k2 = 0.20 \text{ 1/s}, E{a1} = 100 \text{ kJ/mol}, E_{a2} = 20 \text{ kJ/mol}, and T = 27^{\circ}C = 300 \text{ K}.
- \ln \left(\frac{0.20}{k_1}\right) = \frac{1}{8.314 \text{ J/(mol}\cdot\text{K)} \times 300 \text{ K}} (100000 \text{ J/mol} - 20000 \text{ J/mol})
- \ln \left(\frac{0.20}{k_1}\right) = \frac{1}{2494.2} (80000) = 32.074
- \frac{0.20}{k_1} = e^{32.074} \approx 9.49 \times 10^{13}
- k_1 = \frac{0.20}{9.49 \times 10^{13}} \approx 2.11 \times 10^{-15} \text{ 1/s}.
Calculating Catalyzed Activation Energy: Given E{a1} = 55 \text{ kJ/mol}, k2 = 10^8 \times k_1, and T = 298 \text{ K}.
- \ln (10^8) = \frac{1}{8.314 \text{ J/(mol}\cdot\text{K)} \times 298 \text{ K}} (55000 \text{ J/mol} - E_{a2})
- 18.42 = \frac{1}{2477.57} (55000 - E_{a2})
- 18.42 \times 2477.57 = 55000 - E_{a2}
- 45638.6 = 55000 - E_{a2}
- E_{a2} = 55000 - 45638.6 = 9361.4 \text{ J/mol} \approx 9.36 \text{ kJ/mol}.

Enzymatic Catalysis

Enzymes: Biological catalysts, typically proteins.
Efficiency: Enzymes can increase the rate of biochemical reactions by factors of 10^6 to 10^{18}.
Selectivity: Enzymes are highly selective; each enzyme typically acts only on a specific biochemical compound (substrate) or a specific type of reaction.
- Living cells contain thousands of different enzymes, each specialized for particular reactions.
Mechanism: Enzymes bind to reactants (substrates) at an active site, forming an enzyme-substrate (ES) complex. This complex then facilitates the reaction, lowering the activation energy for the transformation to products (P), which are then released from the enzyme (E).
- E + S \rightleftharpoons ES \rightarrow E + P
Potential Energy Diagram: A catalyzed reaction pathway exhibits a lower activation energy (E_a) compared to the uncatalyzed reaction, but the overall energy difference between reactants and products (\Delta H) remains the same. The catalyzed pathway might involve multiple, smaller activation energy barriers for the elementary steps catalyzed by the enzyme, as opposed to a single, much larger barrier for the uncatalyzed reaction.

Practice Problem: Overall Reaction and Rate Law

Consider the mechanism:
1. 2NO \rightleftharpoons N2O2 (Fast equilibrium)
2. N2O2 + O2 \rightarrow N2O_4 (Slow)
3. N2O4 \rightarrow 2NO_2 (Fast)
Overall Reaction: Summing the steps and canceling intermediates (N2O2, N2O4):
2NO + O2 \rightarrow 2NO2
Intermediates: N2O2 and N2O4
Rate Law:
1. Rate law for the slow step: rate = k2[N2O2][O2]
2. Express intermediate [N2O2] from fast equilibrium (Step 1):
  k1[NO]^2 = k{-1}[N2O2] \Rightarrow [N2O2] = \frac{k1}{k{-1}}[NO]^2
3. Substitute into the slow step's rate law:
  rate = k2 \left( \frac{k1}{k{-1}}[NO]^2 \right) [O2] = k{total}[NO]^2[O2]
  where k{total} = k1 k2 / k{-1}.