🔬 Gibbs Free Energy Change, \(\Delta G^{\ominus}\): The Ultimate Decision Maker in Chemistry

Welcome to one of the most important concepts in Physical Chemistry! In your AS Level studies, you learned about Enthalpy change (\(\Delta H\))—whether a reaction gives out heat or takes it in. You then learned about Entropy change (\(\Delta S\))—how disordered the system becomes.

But here's the catch: a reaction can be exothermic (\(\Delta H\)) but still not happen, or it can be endothermic but happen spontaneously! Why?

We need one grand master quantity that combines both energy ($\Delta H$) and disorder ($\Delta S$). This is where the Gibbs Free Energy Change (\(\Delta G\)) comes in. It determines the true feasibility (or spontaneity) of a reaction.

Section 1: The Components of Feasibility (\(\Delta H\) and \(\Delta S\))

Before diving into \(\Delta G\), let's quickly recap the two factors that influence whether a reaction "wants" to happen.

Enthalpy Change (\(\Delta H\))

Enthalpy is all about energy transfer. Systems naturally prefer to lose energy (become more stable).

  • Exothermic Reaction: \(\Delta H\) is negative (-). This contributes to feasibility. (Good!)
  • Endothermic Reaction: \(\Delta H\) is positive (+). This works against feasibility. (Bad!)

Entropy Change (\(\Delta S\))

Entropy measures the dispersal of energy and matter (disorder). Systems naturally tend towards maximum disorder.
Analogy: Imagine your bedroom. It takes energy to keep it tidy (low entropy); it takes no effort for it to become messy (high entropy)!

  • Increase in disorder: \(\Delta S\) is positive (+). This contributes to feasibility. (Good!)
  • Decrease in disorder: \(\Delta S\) is negative (-). This works against feasibility. (Bad!)

We calculate \(\Delta S^{\ominus}\) for a reaction in the same way we calculate \(\Delta H^{\ominus}\):
\[\Delta S^{\ominus} = \Sigma S^{\ominus}\ (products) - \Sigma S^{\ominus}\ (reactants)\]

Quick Review: For a reaction to be truly 'easy' or 'spontaneous', we want \(\Delta H\) to be negative and \(\Delta S\) to be positive. But what if they conflict? That’s the job of \(\Delta G\).

Section 2: The Gibbs Equation: Combining Energy and Disorder

Definition and the Master Equation (LO 23.4.1)

The Gibbs Free Energy Change (\(\Delta G\)) is the maximum amount of non-expansion work extractable from a thermodynamically closed system. In simple terms, it tells us how much energy is free (available) to drive the reaction.

The key equation that links enthalpy and entropy is the Gibbs Equation:
\[\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\]

Where:

  • \(\Delta G^{\ominus}\) is the Standard Gibbs Free Energy Change ($J\,mol^{-1}$ or $kJ\,mol^{-1}$).
  • \(\Delta H^{\ominus}\) is the Standard Enthalpy Change ($J\,mol^{-1}$ or $kJ\,mol^{-1}$).
  • \(T\) is the absolute temperature ($K$, Kelvin).
  • \(\Delta S^{\ominus}\) is the Standard Entropy Change ($J\,K^{-1}\,mol^{-1}$).

⚠ Accessibility Alert: Units are CRUCIAL! ⚠

This is the number one mistake students make in \(\Delta G\) calculations.
\(\Delta H\) is almost always given in kilojoules ($kJ\,mol^{-1}$), but \(\Delta S\) is almost always given in joules per Kelvin ($J\,K^{-1}\,mol^{-1}$).
You must make them consistent!
Memory Aid: Convert \(\Delta S\) into $kJ\,K^{-1}\,mol^{-1}$ by dividing by 1000 OR convert \(\Delta H\) into $J\,mol^{-1}$ by multiplying by 1000. It is often safest to work entirely in Joules ($J$).

Don't worry if this seems tricky at first—just remember: Temperature (T) must be in Kelvin, and all energy terms (\(\Delta G\), \(\Delta H\), \(T\Delta S\)) must be in the same unit. (Standard conditions ($\ominus$) usually mean $T = 298\,K$.)

Section 3: Predicting Feasibility using \(\Delta G\) (LO 23.4.3)

The Feasibility Rule

The sign of \(\Delta G\) tells us whether a process is thermodynamically feasible (spontaneous) under the given conditions.

  • If \(\Delta G\) is NEGATIVE (\(\Delta G < 0\)): The process is Feasible (Spontaneous). It will proceed on its own (though it might still need activation energy).
  • If \(\Delta G\) is POSITIVE (\(\Delta G > 0\)): The process is Not Feasible (Non-spontaneous). It will only proceed if energy is continuously supplied.
  • If \(\Delta G\) is ZERO (\(\Delta G = 0\)): The system is at Equilibrium. The reaction is reversible, and the rates of the forward and reverse reactions are equal.

Memory Aid: "G" stands for "Go!" If \(\Delta G\) is negative, the reaction can "go".

Analogy: The Battle Between Order and Energy

Think of the Gibbs equation: \(\Delta G = \Delta H - T\Delta S\). It represents a tug-of-war between two main driving forces:
1. Enthalpy (\(\Delta H\)): The system wants to minimize its internal energy (prefers \(\Delta H < 0\)).
2. Entropy (\(T\Delta S\)): The system wants to maximize disorder (prefers \(\Delta S > 0\)).
The term \(T\Delta S\) is the "energy wasted on disorder" or the energy used to satisfy the universe's need for randomness.

A reaction is feasible if the energy released (\(\Delta H\)) or the positive disorder created (\(T\Delta S\)) outweighs the negative contribution from the other factor. When \(\Delta G\) is negative, it means the overall system (reaction + surroundings) has increased in stability/disorder.

Key Takeaway: \(\Delta G\) measures the balance between minimizing energy (\(\Delta H\)) and maximizing disorder (\(\Delta S\)). A negative value means the reaction is feasible.

Section 4: Calculations and Temperature Effects (LO 23.4.2 & 23.4.4)

Step-by-Step Calculation Guide (LO 23.4.2)

To calculate \(\Delta G^{\ominus}\) for a reaction, follow these steps:

  1. Find \(\Delta H^{\ominus}\): Look up or calculate the standard enthalpy change of the reaction (usually in $kJ\,mol^{-1}$).
  2. Find \(\Delta S^{\ominus}\): Calculate the standard entropy change of the reaction (usually in $J\,K^{-1}\,mol^{-1}$).
  3. Check Units: Convert \(\Delta S^{\ominus}\) to $kJ\,K^{-1}\,mol^{-1}$ (divide by 1000).
  4. Determine T: Use the temperature required (often $298\,K$ for standard conditions).
  5. Calculate the \(T\Delta S^{\ominus}\) term: Multiply $T$ by the converted \(\Delta S^{\ominus}\).
  6. Calculate \(\Delta G^{\ominus}\): Substitute values into the equation: \(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\).

Example Calculation Check:
If \(\Delta H^{\ominus} = -100\,kJ\,mol^{-1}\) and \(\Delta S^{\ominus} = +50\,J\,K^{-1}\,mol^{-1}\) at $T = 300\,K$.
1. Convert \(\Delta S^{\ominus}\): \(50\,J\,K^{-1}\,mol^{-1} = 0.050\,kJ\,K^{-1}\,mol^{-1}\)
2. Calculate \(T\Delta S^{\ominus}\): \(300\,K \times 0.050\,kJ\,K^{-1}\,mol^{-1} = 15\,kJ\,mol^{-1}\)
3. Calculate \(\Delta G^{\ominus}\): \(\Delta G^{\ominus} = (-100) - (15) = -115\,kJ\,mol^{-1}\)
Result: Since \(\Delta G^{\ominus}\) is negative, the reaction is feasible.

Predicting the Effect of Temperature (LO 23.4.4)

The feasibility of a reaction often depends heavily on the temperature $T$, because $T$ is the multiplier for the entropy term (\(T\Delta S\)).

The Four Combinations of \(\Delta H\) and \(\Delta S\)

| Case | \(\Delta H\) (Enthalpy) | \(\Delta S\) (Entropy) | Feasibility (\(\Delta G = \Delta H - T\Delta S\)) | Temperature Dependence | |---|---|---|---|---| | 1 | Negative (-) | Positive (+) | Always Negative (\(\Delta G < 0\)). Both terms drive the reaction forward. | Feasible at ALL temperatures. | | 2 | Positive (+) | Negative (-) | Always Positive (\(\Delta G > 0\)). Both terms drive the reaction backward. | Not Feasible at ANY temperature. (Requires continuous energy input.) | | 3 | Negative (-) | Negative (-) | Feasibility depends on magnitude. Favorable at low T. | Feasible only below a certain temperature T. | | 4 | Positive (+) | Positive (+) | Feasibility depends on magnitude. Favorable at high T. | Feasible only above a certain temperature T. |

Focusing on Temperature-Dependent Cases (3 and 4):
Feasibility changes when the reaction reaches equilibrium, where \(\Delta G = 0\). We can use this to find the critical temperature ($T_{crit}$) at which spontaneity switches.
If \(\Delta G = 0\): \[0 = \Delta H^{\ominus} - T_{crit}\Delta S^{\ominus}\]
This rearranges to: \[T_{crit} = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}\]

Note: Remember to use consistent units (e.g., both in $J\,mol^{-1}$) when calculating $T_{crit}$!

Case 3 (Negative $\Delta H$, Negative $\Delta S$): We want the exothermic term (\(\Delta H\)) to dominate the unfavorable entropy term (\(T\Delta S\)). This happens when $T$ is small. Feasible when $T < T_{crit}$.

Case 4 (Positive $\Delta H$, Positive $\Delta S$): We need the favorable entropy term (\(T\Delta S\)) to outweigh the unfavorable endothermic term (\(\Delta H\)). Since $T$ is the multiplier for \(\Delta S\), we need $T$ to be large. Feasible when $T > T_{crit}$.

Did You Know? The reason water boils at $100\,^{\circ}C$ (or $373\,K$) is because that is the critical temperature ($T_{crit}$) where the process of liquid \(\to\) gas becomes feasible. Above $373\,K$, \(\Delta G\) for boiling is negative.

KEY TAKEAWAY: Gibbs Free Energy is the central concept for predicting if a reaction will "go." If \(\Delta G\) is negative, it's feasible. Temperature is important because it amplifies the effect of entropy ($\Delta S$) on the overall free energy.