Question

Gibbs-Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as $(\Delta G)_{ PT }=\Delta H-T \Delta S$ The magnitude of $\Delta H$ does not change much with the change in temperature but the entropy factor $T \Delta S$ changes appreciably. Thus, spontaneity of a process depends very much on temperature
When $CaCO _3$ is heated to a high temperature, it undergoes decomposition into $CaO$ and $CO _2$ whereas it is quite stable at room temperature. The most likely explanation of it is

• A. The enthalpy of reaction $(\Delta H)$ overweigths the term $T \Delta S$ at high temperature
• B. The term $T \Delta S$ overweigths the enthalpy of reaction at high temperature
• C. At high temperature, both enthalpy of reaction and entropy change become negative
• D. None of these

Answer 1

Answer: (B)

$\Delta G=\Delta H-T \Delta S$, at high temperature $T \Delta S$ factor dominates of $\Delta H$ and hence $\Delta G$ becomes negative and reaction occurs spontaneously