If gibbs is negative what is enthalpy
A negative Gibbs free energy change (\(\Delta G < 0\)) indicates thermodynamic favorability at the stated temperature and pressure, but it does not uniquely determine the enthalpy change (\(\Delta H\)). The enthalpy can be negative (exothermic) or positive (endothermic) because the entropy term contributes through \(T\Delta S\).
Core relationship between \(\Delta G\), \(\Delta H\), and \(\Delta S\)
\[ \Delta G = \Delta H - T\Delta S \]
When \(\Delta G < 0\) at temperature \(T\), \[ \Delta H - T\Delta S < 0 \quad \Longrightarrow \quad \Delta H < T\Delta S \]
The inequality \(\Delta H < T\Delta S\) is the only general statement forced by \(\Delta G < 0\). No single sign for \(\Delta H\) follows without additional information about \(\Delta S\) and \(T\).
Sign combinations and temperature dependence
| \(\Delta H\) | \(\Delta S\) | Expression \(\Delta G = \Delta H - T\Delta S\) | Consequence for \(\Delta G\) |
|---|---|---|---|
| negative | positive | negative \(-\) (positive) | \(\Delta G < 0\) at all \(T\) (both terms favor) |
| positive | negative | positive \(-\) (negative) \(=\) positive \(+\) | \(\Delta G > 0\) at all \(T\) (both terms oppose) |
| negative | negative | negative \(-\) (negative) \(=\) negative \(+\) | \(\Delta G\) becomes negative at sufficiently low \(T\) |
| positive | positive | positive \(-\) (positive) | \(\Delta G\) becomes negative at sufficiently high \(T\) |
Equivalent enthalpy statement when \(\Delta G\) and \(\Delta S\) are known
Rearrangement gives an explicit form for enthalpy:
\[ \Delta H = \Delta G + T\Delta S \]
A negative \(\Delta G\) can coexist with a positive \(\Delta H\) when \(T\Delta S\) is sufficiently positive, and it can coexist with a negative \(\Delta H\) even when \(\Delta S\) is negative, provided the magnitude of \(\Delta H\) dominates.
Illustrative numerical consistency
Endothermic yet \(\Delta G < 0\)
\[ \Delta G = -5\ \text{kJ·mol}^{-1},\quad T\Delta S = +30\ \text{kJ·mol}^{-1} \]
\[ \Delta H = \Delta G + T\Delta S = (-5) + 30 = +25\ \text{kJ·mol}^{-1} \]
Here \(\Delta H\) is positive, while the entropy term drives \(\Delta G\) negative.
Exothermic with \(\Delta S < 0\) and \(\Delta G < 0\)
\[ \Delta H = -40\ \text{kJ·mol}^{-1},\quad T\Delta S = -15\ \text{kJ·mol}^{-1} \]
\[ \Delta G = \Delta H - T\Delta S = (-40) - (-15) = -25\ \text{kJ·mol}^{-1} \]
Here the enthalpy term outweighs the unfavorable entropy contribution.
Geometric view of \(\Delta G < 0\): a boundary in the \((T\Delta S,\Delta H)\) plane
Common confusions
“Negative Gibbs” and “exothermic” are different statements. Exothermic behavior corresponds to \(\Delta H < 0\), while spontaneity at a given \(T\) corresponds to \(\Delta G < 0\). The entropy contribution \(T\Delta S\) is the link that allows \(\Delta H\) to be either sign when \(\Delta G\) is negative.