Question

The correct thermodynamic conditions for the spontaneous reaction at all temperatures is

• 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

Answer 1

The Correct Option is B

 To determine the correct thermodynamic conditions for a spontaneous reaction at all temperatures, we must consider the Gibbs free energy change \( \Delta G \). The spontaneity criterion for a reaction is that the change in Gibbs free energy must be negative, i.e., \( \Delta G < 0 \). The relation between Gibbs free energy change, enthalpy change (\( \Delta H \)), and entropy change (\( \Delta S \)) is given by:

\[\Delta G = \Delta H - T\Delta S\]

1. If a reaction is to be spontaneous at all temperatures, \( \Delta G \) should be negative for all values of temperature \( T \).
2. For \( \Delta G \) to always be negative, irrespective of the temperature:
   • The term \( \Delta H \) must be negative (exothermic reaction), contributing a negative value.
   • The term \( T\Delta S \) must be positive, which requires \( \Delta S > 0 \) (increase in entropy).
3. Therefore, the condition \(\Delta H < 0\) and \(\Delta S > 0\) ensures that the reaction remains spontaneous at all temperatures.

Now, let's evaluate the provided options:

• \(\Delta H > 0\) and \(\Delta S < 0\): This condition cannot result in spontaneous reactions at all temperatures because both terms contribute to making \(\Delta G\) positive.
• \(\Delta H < 0\) and \(\Delta S > 0\): With this condition, both terms contribute to decreasing \(\Delta G\), making the reaction spontaneous at all temperatures.
• \(\Delta H < 0\) and \(\Delta S < 0\): Here, the spontaneity depends on \( T \); at very high temperatures, the negative entropy term might outweigh and make \(\Delta G > 0\).
• \(\Delta H < 0\) and \(\Delta S = 0\): This reduces to \(\Delta G = \Delta H\), which is negative, but there's no positive contribution from the entropy term at high temperatures.

Hence, the correct answer is: \(\Delta H < 0\) and \(\Delta S > 0\).