Question

At zero absolute temperature, \(\Delta G^\circ =\)

• A. \(\Delta S^\circ\)
• B. \(T \cdot \Delta S^\circ\)
• C. \(\Delta H^\circ\)
• D. RT

Hint:

The Gibbs Free Energy equation, \(\Delta G = \Delta H - T \Delta S\), is one of the most important in thermodynamics. Remember that the \(T \Delta S\) term represents the entropic contribution to free energy, which diminishes to zero as the temperature approaches absolute zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The question asks for the value of the standard Gibbs Free Energy change (\(\Delta G^\circ\)) at absolute zero temperature (\(T = 0\) K). This requires using the fundamental equation that relates Gibbs Free Energy, enthalpy, and entropy.

Step 2: Key Formula or Approach:
The definition of Gibbs Free Energy change is: \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \] where \(\Delta G^\circ\) is the standard Gibbs Free Energy change, \(\Delta H^\circ\) is the standard enthalpy change, \(T\) is the absolute temperature in Kelvin, and \(\Delta S^\circ\) is the standard entropy change.

Step 3: Detailed Explanation:
We need to evaluate this equation at absolute zero temperature, which is \(T = 0\) K.
Substituting \(T = 0\) into the Gibbs Free Energy equation: \[ \Delta G^\circ = \Delta H^\circ - (0) \cdot \Delta S^\circ \] \[ \Delta G^\circ = \Delta H^\circ - 0 \] \[ \Delta G^\circ = \Delta H^\circ \] This result is also consistent with the Third Law of Thermodynamics, which states that the entropy of a perfect crystal at absolute zero is zero. For a reaction, this implies that \(\Delta S^\circ\) approaches zero as \(T\) approaches 0 K for reactions involving perfect crystalline solids. Regardless, the \(T \Delta S^\circ\) term definitively becomes zero because \(T=0\).
Therefore, at absolute zero, the change in Gibbs Free Energy is equal to the change in enthalpy.

Step 4: Final Answer:
By substituting T=0 into the Gibbs-Helmholtz equation (\(\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ\)), we find that \(\Delta G^\circ = \Delta H^\circ\).