Question

Which of the following is true for a reaction that is spontaneous only at high temperature?

• A. \( \Delta_r H^\circ<0, \Delta_r S^\circ>0, \Delta_r G^\circ<0 \)
• B. \( \Delta_r H^\circ>0, \Delta_r S^\circ>0, \Delta_r G^\circ>0 \)
• C. \( \Delta_r H^\circ>0, \Delta_r S^\circ>0, \Delta_r G^\circ<0 \)
• D. \( \Delta_r H^\circ>0, \Delta_r S^\circ<0, \Delta_r G^\circ<0 \)
• E. \( \Delta_r H^\circ<0, \Delta_r S^\circ<0, \Delta_r G^\circ<0 \)

Hint:

Remember, for a reaction to become spontaneous at high temperatures due to entropy considerations, both \(\Delta_r H^\circ\) and \(\Delta_r S^\circ\) need to be positive. This ensures that the thermal energy provided at high temperatures can overcome the enthalpic requirements of the reaction while benefiting from the increase in entropy.

Answer 1

The Correct Option is C

For a reaction to be spontaneous only at high temperatures, the enthalpy change (\(\Delta_r H^\circ\)) must be positive, indicating that the reaction absorbs heat. Additionally, the entropy change (\(\Delta_r S^\circ\)) must also be positive, meaning the disorder of the system increases. 
These factors influence the Gibbs free energy equation: \[ \Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ \] As the temperature \(T\) increases, the term \(T\Delta_r S^\circ\), which is subtracted from \(\Delta_r H^\circ\), becomes significant enough to make \(\Delta_r G^\circ\) negative, hence driving the reaction to spontaneity at higher temperatures.