1. Principles of Transition State Theory
Transition State Theory (TST), developed by Henry Eyring, Michael Polanyi, and Eugene Wigner in 1935, describes chemical kinetics using classical and statistical thermodynamics. It assumes that as reactant molecules approach one another, their potential energy increases to a maximum. The configuration of atoms at this potential energy maximum is the activated complex or transition state.
The theory is built on three core postulates:
1. The activated complex is in a state of quasi-equilibrium with the reactants.
2. The rate of crossing the potential energy barrier is determined by the vibrational motion of the activated complex along the reaction coordinate (the breaking bond).
3. This vibrational motion can be treated as a loose translation, with a frequency given by \(\nu = \frac{k_B T}{h}\).
2. Derivation of the Eyring Equation
Consider the bimolecular elementary reaction: \[A + B \rightleftharpoons [X^\ddagger] \rightarrow \text{Products}\] The rate of the reaction is proportional to the concentration of the activated complex at the barrier peak: \[\text{Rate} = \nu [X^\ddagger]\] where \(\nu\) is the frequency at which the complex breaks apart to form products. Along the reaction coordinate, the vibrational partition function can be factored out. This yields: \[\nu = \frac{k_B T}{h}\] Using the equilibrium constant in terms of concentration (\(K^\ddagger\)) for the quasi-equilibrium: \[K^\ddagger = \frac{[X^\ddagger]}{[A][B]} \implies [X^\ddagger] = K^\ddagger [A][B]\] Substituting this into the rate equation gives: \[\text{Rate} = \frac{k_B T}{h} K^\ddagger [A][B]\] Since the rate law is also defined as \(\text{Rate} = k [A][B]\), we equate the expressions to obtain the Eyring Equation: \[k = \frac{k_B T}{h} K^\ddagger\] From thermodynamics, we know that standard Gibbs free energy of activation \(\Delta G^\ddagger\) is related to \(K^\ddagger\) by: \[\Delta G^\ddagger = -RT \ln K^\ddagger \implies K^\ddagger = e^{-\Delta G^\ddagger / RT}\] Since \(\Delta G^\ddagger = \Delta H^\ddagger - T\Delta S^\ddagger\), we can write: \[k = \frac{k_B T}{h} e^{\Delta S^\ddagger / R} e^{-\Delta H^\ddagger / RT}\]
3. Comparing Eyring and Arrhenius Parameters
The empirical Arrhenius equation is written as: \[k = A e^{-E_a / RT}\] Taking the derivative of \(\ln k\) with respect to temperature \(T\) for both equations allows us to establish relationships between experimental parameters (\(E_a\), \(A\)) and thermodynamic parameters (\(\Delta H^\ddagger\), \(\Delta S^\ddagger\)): \[\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}\] Applying this derivative to the Eyring equation: \[\ln k = \ln \left(\frac{k_B}{h}\right) + \ln T + \frac{\Delta S^\ddagger}{R} - \frac{\Delta H^\ddagger}{RT}\] \[\frac{d(\ln k)}{dT} = \frac{1}{T} + \frac{\Delta H^\ddagger}{RT^2} = \frac{\Delta H^\ddagger + RT}{RT^2}\] Equating the two derivatives: \[E_a = \Delta H^\ddagger + RT \implies \Delta H^\ddagger = E_a - RT \quad \text{(For unimolecular / solution reactions)}\] For a gas-phase bimolecular reaction, because there is a loss of one mole of gas molecules upon forming the transition state (decreasing volume), the relation is: \[\Delta H^\ddagger = E_a - 2RT\] Substituting these relations back into the Eyring equation allows us to express the Arrhenius pre-exponential frequency factor \(A\) in terms of activation entropy: \[A = e \left(\frac{k_B T}{h}\right) e^{\Delta S^\ddagger / R} \quad \text{(For unimolecular / solution reactions)}\] \[A = e^2 \left(\frac{k_B T}{h}\right) e^{\Delta S^\ddagger / R} \quad \text{(For gas-phase bimolecular reactions)}\]
4. Solved CSIR-NET & GATE Questions
Question 1 (CSIR-NET Chemical Sciences)
For a gas-phase bimolecular reaction, the Arrhenius activation energy (\(E_a\)) is measured to be \(58.2\text{ kJ/mol}\) at \(300\text{ K}\). Calculate the enthalpy of activation (\(\Delta H^\ddagger\)) in \(\text{kJ/mol}\). (\(R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}\)).
Detailed Solution:
- Identify the correct relationship for a gas-phase bimolecular reaction: \[\Delta H^\ddagger = E_a - 2RT\]
- Calculate the correction term \(2RT\): \[2RT = 2 \times (8.314\text{ J K}^{-1}\text{ mol}^{-1}) \times 300\text{ K} = 4988.4\text{ J/mol} = 4.988\text{ kJ/mol}\]
- Subtract this term from the activation energy: \[\Delta H^\ddagger = 58.2\text{ kJ/mol} - 4.988\text{ kJ/mol} = 53.21\text{ kJ/mol}\]
Answer: Enthalpy of activation \(\Delta H^\ddagger = 53.21\text{ kJ/mol}\).
Question 2 (GATE Chemistry)
A solution-phase reaction at \(27^\circ\text{C}\) (\(300\text{ K}\)) has an Arrhenius pre-exponential factor \(A = 2.718 \times 10^{13}\text{ s}^{-1}\). Determine the entropy of activation (\(\Delta S^\ddagger\)) in \(\text{J K}^{-1}\text{ mol}^{-1}\). (Given: \(k_B = 1.38 \times 10^{-23}\text{ J K}^{-1}\), \(h = 6.626 \times 10^{-34}\text{ J s}\)).
Detailed Solution:
- For solution-phase, the relationship between frequency factor \(A\) and activation entropy \(\Delta S^\ddagger\) is: \[A = e \left(\frac{k_B T}{h}\right) e^{\Delta S^\ddagger / R}\]
- Calculate the term \(e \left(\frac{k_B T}{h}\right)\) at \(300\text{ K}\): \[e \frac{k_B T}{h} = 2.718 \times \frac{(1.38 \times 10^{-23}) \times 300}{6.626 \times 10^{-34}}\] \[e \frac{k_B T}{h} = 2.718 \times (6.248 \times 10^{12}) = 1.698 \times 10^{13}\text{ s}^{-1}\]
- Set up the equation to solve for \(e^{\Delta S^\ddagger / R}\): \[2.718 \times 10^{13} = 1.698 \times 10^{13} \times e^{\Delta S^\ddagger / R}\] \[e^{\Delta S^\ddagger / R} = \frac{2.718 \times 10^{13}}{1.698 \times 10^{13}} \approx 1.60\]
- Take the natural logarithm of both sides: \[\frac{\Delta S^\ddagger}{R} = \ln(1.60) \approx 0.47\] \[\Delta S^\ddagger = 0.47 \times 8.314\text{ J K}^{-1}\text{ mol}^{-1} \approx +3.91\text{ J K}^{-1}\text{ mol}^{-1}\]
Answer: Entropy of activation \(\Delta S^\ddagger = +3.91\text{ J K}^{-1}\text{ mol}^{-1}\).
Muhammad Ijas M
Co-founder & Academic Administrator | IIT Madras Alumni | JRF AIR-9 | GATE AIR-69Muhammad Ijas M is an IIT Madras graduate who qualified CSIR NET JRF with AIR 9 and secured GATE AIR 69. He manages the academic curriculum and physical chemistry modules at Benzil Academy.