Chemistry > Thermodynamics and Thermochemistry > 19.0 Enthalpy, Entropy, Free Energy Changes and The Nature of Process

  Thermodynamics and Thermochemistry
    1.0 Introduction to Thermodynamics
    2.0 Thermodynamic Processes
    3.0 Heat $(Q)$
    4.0 Work $(W)$
    5.0 Internal Energy
    6.0 Zeroth Law of Thermodynamic
    7.0 First Law of Thermodynamics
    8.0 Heat Capacity
    9.0 Enthalpy $(H)$ or Heat Content
    10.0 Significance of $?H$ and $?E$
    11.0 Relationship Between $?H$ and $?E$
    12.0 Standard Enthalpy of Reaction
    13.0 Law of Thermodynamics
    14.0 Bond Energy or Bond Enthalpies
    15.0 Second Law of Thermodynamic
    16.0 Entropy (S)
    17.0 Spontaneous Processes
    18.0 Free Energy (G)
    19.0 Enthalpy, Entropy, Free Energy Changes and The Nature of Process
    20.0 Standard Free Energy Change
    21.0 Coupled Reactions
    22.0 Third Law of Thermodynamic
    23.0 Important Facts

19.1 Characteristics of Free Energy
• It is a thermodynamic state function

• It is represented by $‘G’.$
$$G=H-TS$$

Gibbs free energy change at constant temperature
$$\Delta G = \Delta H - T\Delta S$$

• It is an extensive property since its value depends upon the quantity of substance.

• Criteria of spontaneity based on constant $T$ and $P.$

(a) If $\Delta G < 0,$ the process is spontaneous

(b) If $\Delta G > 0,$ the process is non-spontaneous

(c) If $\Delta G = 0$ the system is at equilibrium

• $\Delta G$ is equal to maximum possible useful work other than the $P-V$ work that is obtainable from the system.

• $\Delta G = - NF\;{E_{cell}}$

$\Delta G^\circ = - nF\;E{^\circ _{cell}}$ where $E_{cell}^\circ $ $=$ standard cell potential.

• The relation between Gibb’s free energy change and equilibrium constant.
$$\Delta G = \Delta {G^\circ } + \frac{{RT}}{{\ln Q}}$$

At equilibrium $?G = 0$

$$\therefore \quad \Delta {G^\circ } = - RT\;\ln \;K = - 2.303\;RT\;\log \;K\quad ....(1)$$

Again
$$\therefore \quad \Delta {G^\circ } = \Delta {H^\circ } - T\Delta {S^\circ }$$

• At equilibrium $?G = 0$

So $$\Delta H - T\Delta S = 0\quad or\quad T\Delta S = \Delta H$$
$$T = \frac{{\Delta H}}{{\Delta S}}$$
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