Kinetics and Thermodynamics: Puzzling questions

Two questions

Would a reaction occur if

(1) Ea is not met but dG is negative

(2) Ea is met but dG is positive

Case1

Can a reaction proceed if dG is negative but Ea is not met?

dG = Free energy change and Ea = activation energy

The answer is no.

Free energy change is a thermodynamic parameter that tells us if a reaction can occur spontaneously in the given conditions. When dG is negative, it tells that the reaction is thermodynamically favorable and can proceed spontaneously. However, the activation energy (Ea) is the minimum energy that a reaction must possess to convert the reactants into products.

If the activation energy is not met, the reaction will not proceed, even if dG is negative. This is because, despite the reaction being energetically favorable or thermodynamically meeting the conditions for a reaction to occur, the molecules must still acquire sufficient energy to overcome the activation barrier.

Therefore, the following are important points

Negative dG:

The reaction is thermodynamically favorable. Activation energy not met:

The reaction will not happen. Therefore, for a reaction to occur, both thermodynamic and kinetics conditions must be satisfied: a negative dG and sufficient energy to overcome the activation energy barrier.

Case 2

Can a reaction occur if Ea is met and dG is positive?

The answer is yes.

On the other hand, if Ea is met but the free energy change dG is not favorable like in the following equation dG is not negative, the reaction can still be made to occur by increasing the term TdS in the equation below to make dG negative. dG = dH - TdS.

We find such cases typically in endothermic reactions in which the enthalpy change dH is positive. Such reactions generally require high temperatures to increase the TdS term.

Example:

Splitting of water

The splitting of water is not a thermodynamically favorable reaction at standard conditions. It's an endothermic reaction.

The reaction for water splitting can be represented as:

2H2O(l) -> 2H2(g) + O2(g)

At standard conditions (STP: 25°C, 1 atm pressure), the standard enthalpy change (ΔH°) for this reaction is +483.6 kJ/mol, indicating that the reaction is endothermic.

The standard entropy change (ΔS°) for the reaction is +163.4 J/mol*K, meaning that the products have higher entropy than the reactants.

Using the equation:ΔG° = ΔH° - TΔS°

where T is the temperature in Kelvin, we can calculate the standard Gibbs free energy change (ΔG°) at standard conditions:ΔG° = +483.6 kJ/mol - (298K)*(+163.4 J/mol*K)ΔG° = +483.6 kJ/mol - 48.682 kJ/molΔG° = +434.9 kJ/mol

Since ΔG° is positive at standard conditions, the reaction for water splitting is not thermodynamically favorable under these conditions. It would require an input of energy to drive the reaction in the forward direction.

This added energy cost makes H2 gas-making by splitting water a cost-intensive process.