13.5-13.6: Potential Energy Diagrams-Arrhenius Equation

Potential Energy Diagrams:

Example:

*Endothermic Reaction

E_a = Activation Energy, the minimum energy of collision required for two molecules to react.
Activated Complex = ǂ
Reaction Pathway = progress of reaction
ΔH = Heat of Reaction, the difference between the energy of the products and the energy of reactants. In this example, the energy of the products is higher than the energy of the reactants, which means energy was put in, and the reaction is endothermic.

Simply: The potential energy of the reactants increases as the reaction progresses until it increases to the value for the activated complex. The activated complex is unstable and breaks down into products, the products go to lower potential energy.

*Exothermic Reaction


In this example, the energy of the reactants is higher than that of the products, so heat energy is released when the reaction progresses, and the reaction is exothermic.

Arrhenius Equation:

epic mustache man - Svante August Arrhenius

Arrhenius Equation allows us to find values for k at different temperatures (because k is temperature dependent).

ln (k₂ / k₁) = (E_a / R) [(1 / T₁) – (1 / T₂)]

Here, brackets don't relate to any concentration, they just denote a second pair of parentheses for algebraic purposes.
ln = natural log
k₁ = rate constant at T₁
k₂ = rate constant at T₂
(units for k defined by reaction order, should cancel each other in Arrhenius Equation, should be given)
E_a = Activation Energy in J / mol
R = The Ideal Gas Constant converted to different units of measure. Equal to 8.31 J / (K . mol)
T₁ = Temperature 1 in degrees K (temp. associated with k₁)
T₂ = Temperature 2 in degrees K (temp. associated with k₂)
Degrees K = Degrees Celsius + 273.15

Example:

*The rate constant for the formation of hydrogen iodide from the elements Hydrogen and Iodine is 2.7x10^-4 L / (mol . s) at 600 K and 3.5x10^-3 L / (mol . s) at 650 K. Find the activation energy.

Establish which values are given and which value is unknown:
(It doesn't matter which k becomes k₁ or k₂, as long as k₁ matches T₁ and k₂ matches T₂. This is also the part where anything not already in the proper units should be converted to the units of the Arrhenius Equation. When everything is in the appropriate units, the units don't have to be carried through the calculations, units will cancel to what needs to be left over for the unknown)
k₂ = 3.5x10^-3 L / (mol . s)
k₁ = 2.7x10^-4 L / (mol . s)
E_a = unknown
R = 8.31 J / (K . mol)
T₁ = 600 K
T₂ = 650 K

ln ( 3.5x10^-3 / 2.7x10^-4) = (E_a / 8.31) [(1 / 600) – (1 / 650)]

Perform operations, E_a = 1.66x10⁵ J / mol

*Find the rate constant at 700 K.

To establish the values for the new equation, choose either previous value for k to be the new k₁, remember its corresponding temperature is the new T₁.
k₂ = unknown
k₁ = 2.7x10^-4 L / (mol . s)
E_a = 1.66x10⁵ J / mol
R = 8.31 J / (K . mol)
T₁ = 600 K
T₂ = 700 K

ln (k₂ / 2.7x10^-4) = (1.66x10⁵ / 8.31) [(1 / 600) – (1 / 700)]

Perform operations, k₂ = 3.2x10^-2 L / (mol . s)

Note: to “undo” the ln on the left hand side, apply the inverse. The inverse of the natural log is e. Take e to the power of both sides. e^ln (k₂ / 2.7x10^-4) = k₂ / 2.7x10^-4