Eyring Equation

Eyring Equation

Eyring Equation, developed by Henry Eyring in 1935, is based on transition state theory and is used to describe the relationship between reaction rate and temperature. It is similar to the Arrhenius Equation, which also describes the temperature dependence of reaction rates. However, whereas Arrhenius Equation can be applied only to gas-phase kinetics, the Eyring Equation is useful in the study of gas, condensed, and mixed-phase reactions that have no relevance to the collision model.
The equation of this theory of absolute reaction rate can be expressed in terms of thermodynamical functions instead of partition functions. Eyring and his coworkers derived the thermodynamical formulations of reaction reate.
Consider the following reaction-

where [AB]^* is the activated complex and K^* is the equilibrium constant between the reactants and activated complex.
From classical mechanics, the energy of vibration is given by-
RT/N_A (or k_BT where k_B is the Boltzmann constant) whereas from quantum mechanics, it is given by hν so that-
hν = RT/N_A
ν = RT/N_Ah.

The vibrational frequency ν is the rate at which the activated complex molecules move across the energy barrier. Thus, the rate constant k₂ can be identified with ν.
The reaction rate is given by-
Rate = -d[A]/dt = Kk₂[AB]^*
         = k(k_BT/h)[AB]^*          -----Equation-1
where the factor K, called the transmission coefficient, is a measure of the probability that a molecule, once it passes over the barrier, will keep on going ahead and not return. The value of K is taken to be unity; it is thus omitted from the rate expression. The concentration of the activated complex, [AB]^*, can be obtained by writing the equilibrium expression-
K^* = [AB]^*/[A][B]
whence, [AB]^* = K^*[A][B]          -----Equation-2

From equation-1 and equation-2, we have-
-d[A]/dt = (k_BT/h).K^*[A][B]
Thus the rate constant K₂ may be expressed as-
K₂ = (k_BT/h).K^*          -----Equation-3
The equilibrium constant K^* may be expressed in terms of (ΔG^o)^*, called standard Gibbs free energy of activation.
(ΔG^o)^* = − RTlnK^*          -----Equation-4
and (ΔG^o)^* = (ΔH^o)^* − T(ΔS^o)^*          -----Equation-5

From equation-3 and equation-4, we have-
K^* = e^{−(ΔGo)*/RT}          -----Equation-6
or, K^* = e^(ΔSo)*/R. e^{− (ΔHo)*/RT}          -----Equation-7
Now from equation-3 and equation-7, we have-

K₂ = (k_BT/h).e^(ΔSo)*/R. e^{− (ΔHo)*/RT}          -----Equation-8

The equation-8 is known as Eyring equation.

The Eyring equation describes the changes in the rate of a chemical reaction against temperature. The equation follows from the

• A. collision theory.
• B. transition state theory.
• C. molecularity of the reaction.
• D. kinetic theory of matter.

Correct Answer: B. Transition state theory

Explanation:

The Eyring equation is a cornerstone of Transition State Theory (TST). Unlike collision theory, which views molecules as hard spheres, TST provides a superior framework by assuming:

• • A chemical equilibrium exists between reactants and a high-energy transition state (activated complex).
• • The reaction rate is proportional to the concentration of this complex and the frequency of its passage over the energy barrier.

The equation connects the rate constant to thermodynamic quantities like enthalpy (Δ*H) and entropy (Δ*S), allowing chemists to understand the structural changes during a reaction.

In the linear form of the Eyring equation, a plot of ln(k/T) versus 1/T is used. What does the y-intercept of this plot represent?

• A. $-\Delta^\ddagger H / R$
• B. $\ln(k_B / h) + \Delta^\ddagger S / R$
• C. $\ln(A)$
• D. $-\Delta^\ddagger G / RT$

Correct Answer: B. $\ln(k_B / h) + \Delta^\ddagger S / R$

Explanation:

To find the enthalpy and entropy of activation, we rearrange the Eyring equation into a linear form:

$\ln\left(\frac{k}{T}\right) = -\frac{\Delta^\ddagger H}{R} \cdot \frac{1}{T} + \ln\left(\frac{k_B}{h}\right) + \frac{\Delta^\ddagger S}{R}$

When plotting $\ln(k/T)$ on the y-axis against $1/T$ on the x-axis:

• • Slope: corresponds to $-\Delta^\ddagger H / R$.
• • Y-intercept: corresponds to $\ln(k_B / h) + \Delta^\ddagger S / R$.

This is a powerful tool because a positive entropy of activation suggests the transition state is more disordered than the reactants, while a negative entropy suggests a more ordered, constrained transition state (common in bimolecular reactions).