Chemical Kinetics Notes

PZF

  • k is based on PZF.

  • P is the steric factor.

Steric Factor (P)

  • Only certain orientations of colliding molecules lead to a reaction.

  • The steric factor represents the fraction of collisions with the correct orientation:

    • Less than 1 (1 would be 100%).

    • Example: Iodide and hydrogen from hydrogen iodide.

      • Correct orientation: collision to break old bonds and form new bonds.

      • Incorrect orientations: collisions that glance or don't have enough energy.

Collision Frequency (Z)

  • How often molecular collisions overcome the energy barrier.

  • Number of collisions per unit time for a given reactant concentration.

  • Derived from the kinetic molecular theory.

  • Kinetic energy of particles is a function of temperature: higher temperature means higher movement, and lower temperature means lower movement.

  • Higher concentration leads to higher collision frequency because there are more particles.

  • Only a small fraction of collisions have enough energy to produce a product.

  • Collisions with the wrong orientation cannot produce a product.

  • This is especially important with complex reactions such as those involving biochemical molecules, proteins, and enzymes.

Factor F

*Statistical factor.

  • Relates to particles with the right orientation and energy for successful collisions.

  • f = e^{-\frac{E_a}{RT}}

Variables in Factor F

  • E_a = Energy of activation.

  • R = Gas constant.

    • R = 8.314 \frac{J}{mol \cdot K}, (joules per kelvin mole). Used for energy calculations.

    • Other R value for osmosis etc.

  • T = Temperature.

Implications of the Equation

  • If E_a (energy of activation) increases, k (rate constant) decreases because a larger energy barrier results in a smaller fraction of successful collisions.

  • A larger Ea will make e^{-\frac{Ea}{RT}} a very small number.

Simplified Model

  • Not all reactions will be successful.

  • Successful reactions require the right orientation and energy to break old bonds and form new bonds.

  • Example: chlorine radical collision with a complex molecule.

    • Collision with oxygen: bounces off because of the wrong orientation.

    • Collision with another chlorine: forms a transitory state.

Activated Complex and Transition State

  • Activated Complex: temporary state when overcoming the energy hilltop.

  • Transition State: the position at the top of the energy hilltop where the activated complex is formed.

Activation Energy

  • Minimum kinetic energy colliding particles must have for successful collisions.

  • Activation energy changes to potential energy as bonds rearrange.

  • Activation energies can be large.

  • Higher temperatures increase the average kinetic energy of reacting partners.

  • Example: Diamonds

    • Stable configuration of carbon is graphite (2D sheets attached by Van der Waals forces).

    • Diamond is a metastable state.

    • The activation energy to convert graphite to diamond is very high, so diamonds last a long time.

    • With enough energy, diamond can be turned back into graphite.

Temperature and Kinetic Energy

  • Higher temperatures mean higher kinetic energies.

  • Solid State: Molecules in fixed positions in a crystal lattice with oscillatory vibrational modes of energy.

  • Liquid State: Molecules can slide around, still vibrating with more energy, but no fixed positions.

  • Gaseous State: Molecules move freely with high energy.

  • Lowering temperature: gas condenses to liquid and then to solid.

Fraction of Collisions

  • The red line represents the fraction of collisions with enough kinetic energy to overcome the energy barrier.

  • Temperature is a function of kinetic energy.

  • Microwaves and Water: Microwaves resonate with water molecules, causing them to vibrate faster and heat up the food.

Temperature and Kinetic Energy Distribution

  • The higher the temperature, the more molecules have enough kinetic energy to overcome the energy barrier.

  • The shaded area under the curves of kinetic energy distribution represents the fraction of collisions with the minimum activation energy.

Transition State Theory

  • Describes what happens when reactant particles come together.

  • Potential energy diagrams visualize the relationship between activation energy and potential energy.

Model

  • Reactants with correct alignment and energy surmount the energy barrier.

  • The activated complex forms at the transition state as old bonds break and new bonds form.

  • At the transition state, reactive bonds are partially broken, and product bonds are partially formed.

  • The transition state has maximum potential energy.

  • The unstable chemical species at this state is called the activated complex.

Simplifying the Rate Constant (k)

  • The steric factor (p) and collision frequency (z) are combined into a frequency factor (A) or pre-exponential factor.

  • k = Ae^{-\frac{E_a}{RT}}

Arrhenius Equation

  • Natural log of the equation:

    • ln(k) = ln(A) - \frac{E_a}{R} \frac{1}{T}

Implications

  • Equation of a straight line: y = mx + b

    • ln(k) is the y factor.

    • ln(A) is the y-intercept.

    • -\frac{E_a}{R} is the slope.

Plotting

  • Plot ln(k) vs. \frac{1}{T} to get a straight line with a negative slope.

  • Activation energy can be related to the rate constant at two temperatures.

Clausius Clapeyron Equation

  • Analogous equation for vapor pressure of liquids involving partial pressures and enthalpy of the reaction.

  • ln(\frac{k2}{k1}) = -\frac{Ea}{R}(\frac{1}{T2} - \frac{1}{T_1})

  • The slope of the line is -\frac{E_a}{R}.

Calculating Activation Energy

*By multiplying R, the slope, you get the activation energy.
*Take the exponential of the y-intercept to get the frequency factor (A).

Arrhenius

  • Theoretical physicist who contributed to chemistry.

  • The Arrhenius equation is his work.

Importance:

  • Kinetics is experimentally determined.

  • The Arrhenius equation allows determination of reaction rates at different temperatures with minimal experimentation.

    • Only need data from two different temperatures to estimate the reaction rate at a third.

    • Plot the results on a graph through those two points.

Manipulating the Equation

*Multiplying across to switch the terms:
* ln(\frac{k2}{k1}) = \frac{Ea}{R}(\frac{1}{T1} - \frac{1}{T_2})

Solving for k_2

  • Take the exponential of both sides:

    • (\frac{k2}{k1}) = e^{\frac{Ea}{R}(\frac{1}{T1} - \frac{1}{T2})} *Isolate and solve for k2: k2 = k1 * e^{\frac{Ea}{R}(\frac{1}{T1} - \frac{1}{T2})} *Convert Ea to joules (if given in kilojoules) to match units, and convert it to kilojoules for your final answer.

Solving for E_a

Move the r terms to the other side and isolate and solve for Ea: Ea= R \frac{ln(\frac{k2}{k1})}{(\frac{1}{T1} - \frac{1}{T2}) }

Reaction Mechanisms

  • Most chemical reactions occur by a series of elementary steps.
    *Elementary steps: nitrogen dioxide becomes nitrogen trioxide gas plus nitrogen monoxide.

  • An intermediate forms in one step and gets used up in the subsequent step (not in the overall reaction).

  • Only bimolecular collisions are possible (collisions between two particles/species).

    • Three-body problem in physics: it's very unlikely for three particles to collide with the correct orientation and energy.

Rate Determining Step

  • A reaction is only as fast as the slowest step.

Commute analogy

*George Washington Bridge and traffic determining the rate of getting home.

  • The sum of the elementary steps gives the overall balanced equation.

  • Mechanisms must agree with the experimentally determined rate law.

  • Multiple possible mechanisms may exist.

Example

*Translating the math problem:
* Two nitrogen dioxides form nitrogen trioxide and nitrogen monoxide (balanced).
* Nitrogen trioxide reacts with carbon monoxide to give nitrogen dioxide and carbon dioxide (balanced).
*Following list guidelines:
* Most chemical reactions occur by a series of elementary steps.
* The intermediate forms are in one step, then used in the next.
* Only bimolecular collisions.

Overall Balanced Reaction

*Elementary steps result in: Nitrogen dioxide gas plus carbon monoxide gas results in nitrogen monoxide gas plus carbon dioxide gas.
*Considers similaries to Hess' law since you would have to manipulate equations around, adding them together and you add the heat of the reaction.

Additional points to note

*Reactions that start and are later consumed in the reaction are intermediates

Considerations for Proving Expression

  • Consider Reaction: Nitrogen monoxide plus hydrogen yields nitrogen plus water.

    • However, the rate law expression does not match the balanced equation above, (based on the initial rates method).
      *The rate of slowest step is the rate determining step, which can be the first step, so it is equal to the rate expression.

Elementary Steps

*There are three steps, with the second being the slowest reaction.
* Reaction one: 2NO double arrow N2 O2 (fast).
* Reaction two: N2 O2 + H2 yields N2 O + H2 O (slow step). * Reaction three: N2 O + H2 yields N2 + H_2 O
*Elementary steps only allows by molecular collisions (2 of each). However, the reactions below do not show that.

Rules for Reaction Step

*Follow reactions and apply a rate constant (k). This does not match because in this equation the hydrogen does not appear in the overall reaction.

Additional Rules

  • If there it doesn't fit use a bigger hammer, which equates to solving an alternate problem that's very common. *Apply all the following:

    • Use the first step and equation to solve for the value.
      *Write the expression of what is going with forward reaction and backward
      *Apply forward reactants and equals the reverse back.
      *Take the forward factor and do the expression over with the constant
      *Substitute those value in, and solve the equation, which is equal to initial experimental value.
      *As long as the constant is the same, the process holds, meaning it will work for experimental data set.

Homogeneous vs. Heterogeneous Reactions

  • Homogeneous: reactants and products are in the same phase (gas, aqueous solution, vapor, liquid).

  • Heterogeneous: reactants and products are in different phases (solid with liquid, liquid with gas, etc.).

Factors Affecting Reaction Rates

  • Nature of the reactants.

    • More bonds to break: more complex and slower.

    • Fewer bonds to break: less complex and faster.
      *Consider reaction: hydrogen lodide creates a fraction that reacts so it must be a fraction of a second.

  • Concentration.

    • Higher concentration means more particles to collide, and higher reaction rate.
      *Temperature (K).

    • Higher kinetic energy leads to a faster rate of rxn
      *The rule of thumb- for every 10 degree increase it doubles. Therefore temperature is a function
      *Apply these considerations when you place food in the fridge.
      *Surface Area.
      *More surface, then higher rate of chemical reaction.
      *This means there will be more action, in cases for pharmacies.
      *Catalyst. A catalyst can lower energy to activate a reaction.
      *Lower activation then catalytic reactions must be faster when lowering activation.

    • Lowers the energy of activation.

    • Catalytic reactions are faster.
      *Pressure- Effect of pressure impact is only on the gases - this pressure does not impact solids or liquids.
      *Consider pressure will impact with relation to liquid gas impact for conceptual points.

Tea Example with Sugar

  • Sugar cube is roughly 1 cm square:

    • Surface area: 6 cm^2

    • Water molecules surround the sugar cube and dissolve the surface, affecting the inner molecules later.
      *If you consider smaller sugar cubes (1mm = .1cm):

    • Then the sugar and cube would = .2 but multiplied by the surface area. Boom!! The number goes up to 600 from little cubes multiplied.
      *This smaller is able to dissolve faster due to the fact that the surface is larger - more area