Step 1: Understanding the Question:
A reaction is spontaneous if \( \Delta G<0 \). The Gibbs free energy is defined as \( \Delta G = \Delta H - T\Delta S \).
Step 2: Key Formula or Approach:
At equilibrium, \( \Delta G = 0 \), so \( T_{eq} = \frac{\Delta H}{\Delta S} \).
For a reaction to be spontaneous below this temperature (\( T<T_{eq} \)), we need to examine the signs of \( \Delta H \) and \( \Delta S \).
Step 3: Detailed Explanation:
- If \( \Delta H<0 \) (exothermic) and \( \Delta S<0 \) (decrease in entropy):
\[ \Delta G = (-\text{value}) - T(-\text{value}) = -|\Delta H| + T|\Delta S| \]
At low temperatures (below \( T_{eq} \)), the magnitude of the negative term (\( |\Delta H| \)) is greater than the positive term (\( T|\Delta S| \)). So \( \Delta G<0 \).
Thus, the reaction is spontaneous only at lower temperatures.
- If \( \Delta H>0 \) and \( \Delta S>0 \), the reaction becomes spontaneous above the equilibrium temperature.
- If \( \Delta H<0 \) and \( \Delta S>0 \), spontaneous at all temperatures.
- If \( \Delta H>0 \) and \( \Delta S<0 \), non-spontaneous at all temperatures.
Step 4: Final Answer:
The correct conditions are \( \Delta H<0 \) and \( \Delta S<0 \).