Question

Identify from following the correct set of thermodynamic conditions for a reaction to be nonspontaneous at all temperatures.

• A. $\Delta H<0$ and $\Delta S<0$
• B. $\Delta H>0$ and $\Delta S>0$
• C. $\Delta H<0$ and $\Delta S>0$
• D. $\Delta H>0$ and $\Delta S<0$

Hint:

Nature always favors lower energy ($\Delta H < 0$) and higher chaos ($\Delta S > 0$). If a reaction tries to do the exact opposite of both ($\Delta H > 0$ AND $\Delta S < 0$), nature will absolutely refuse to let it happen!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We must determine the signs of enthalpy change ($\Delta H$) and entropy change ($\Delta S$) that universally guarantee a chemical reaction will never happen spontaneously, regardless of the temperature.

Step 2: Detailed Explanation:
The spontaneity of a reaction at constant temperature and pressure is dictated completely by the Gibbs Free Energy equation:
$\Delta G = \Delta H - T\Delta S$
For a reaction to be nonspontaneous, $\Delta G$ must be strictly positive ($\Delta G > 0$).
Let's evaluate the four possible thermodynamic scenarios:
Option (a) $\Delta H < 0, \Delta S < 0$: Spontaneous only at low temperatures. At high T, the positive $T\Delta S$ term takes over, making it nonspontaneous.
Option (b) $\Delta H > 0, \Delta S > 0$: Spontaneous only at high temperatures (where the favorable negative $-T\Delta S$ term outgrows the unfavorable $\Delta H$).
Option (c) $\Delta H < 0, \Delta S > 0$: Highly favorable. $\Delta G$ will always be negative. This reaction is spontaneous at all temperatures.
Option (d) $\Delta H > 0, \Delta S < 0$: Highly unfavorable. The reaction requires heat input ($\Delta H$ is positive), and results in a more ordered, less random state ($\Delta S$ is negative).
When substituting these into the equation:
$\Delta G = (+ \text{value}) - T(- \text{value})$
Because absolute temperature ($T$ in Kelvin) is always positive, the term $-T(- \text{value})$ becomes a positive quantity.
$\Delta G = (+ \text{value}) + (+ \text{value}) = \text{Always Positive}$.
Since $\Delta G$ can never be negative under these conditions, the reaction is strictly nonspontaneous at all temperatures.

Step 3: Final Answer:
The conditions are $\Delta H>0$ and $\Delta S<0$, matching option (d).