Questions 36-38 refer to the following information. [VISUAL] Two molecules of the amino acid glycine — Thermodynamics Chemistry Question

STEPS:

1. Understand the thermodynamic criterion for favorability:
A chemical reaction is thermodynamically favored (spontaneous) under standard conditions when its standard Gibbs free energy change is negative ($\Delta G^\circ < 0$).
2. Recall the Gibbs^-Helmholtz equation:
The relationship between Gibbs free energy ($\Delta G^\circ$), enthalpy ($\Delta H^\circ$), entropy ($\Delta S^\circ$), and absolute temperature ($T$, in Kelvin) is given by:
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$
3. Analyze the signs of the thermodynamic parameters from the table:
Looking at the data provided on page 25 of the exam booklet:
* Enthalpy change ($\Delta H^\circ_{298}$): $+12\text{ kJ/mol}_{\text{rxn}}$ (positive, meaning the reaction is endothermic).
* Entropy change ($\Delta S^\circ_{298}$): $-10\text{ J/(K mol}_{\text{rxn}})$ (negative, meaning the system becomes more ordered).
4. Evaluate the sign of $\Delta G^\circ$ at any temperature ($T > 0\text{ K}$):
Substitute the signs of $\Delta H^\circ$ ($+$) and $\Delta S^\circ$ ($-$) into the Gibbs^-Helmholtz equation:
$$\Delta G^\circ = (+) - T(-)$$
$$\Delta G^\circ = (+) + T(+)$$
Since absolute temperature $T$ in Kelvin must always be positive, the term $-T\Delta S^\circ$ is always positive. Adding two positive values ($\Delta H^\circ$ and $-T\Delta S^\circ$) will always yield a positive standard Gibbs free energy change ($\Delta G^\circ > 0$).
5. Conclude:
Because $\Delta G^\circ$ is positive at all temperatures, the reaction is not thermodynamically favored at any temperature, confirming Option D as the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: For a reaction to be favored *only* at high temperatures, both $\Delta H^\circ$ and $\Delta S^\circ$ must be positive. In that scenario, a high temperature makes the $-T\Delta S^\circ$ term large and negative enough to overcome the positive $\Delta H^\circ$ term. Here, since $\Delta S^\circ$ is negative, increasing the temperature only makes $\Delta G^\circ$ more positive and even less favored.
• Option B is incorrect: For a reaction to be favored *only* at low temperatures, both $\Delta H^\circ$ and $\Delta S^\circ$ must be negative. In that case, a low temperature keeps the unfavorable $+T\Delta S^\circ$ term small, allowing the favorable negative $\Delta H^\circ$ term to make $\Delta G^\circ$ negative. Here, the positive $\Delta H^\circ$ prevents the reaction from being favored at low temperatures.
• Option C is incorrect: For a reaction to be favored at *all* temperatures, $\Delta H^\circ$ must be negative (exothermic) and $\Delta S^\circ$ must be positive (entropy-favored). This reaction has the exact opposite set of parameters, making it favored at no temperatures.