Question

A reaction is spontaneous at low temperature, but non-spontaneous at high temperature. Which of the following conditions is true for the chemical reaction?

• A. \(\Delta H < 0, \Delta S < 0\)
• B. \(\Delta H > 0, \Delta S = 0\)
• C. \(\Delta H < 0, \Delta S > 0\)
• D. \(\Delta H > 0, \Delta S > 0\)

Hint:

If a reaction is spontaneous only at low temperatures, it is exothermic (\(\Delta H<0\)) and accompanied by a decrease in entropy (\(\Delta S<0\)).

Answer 1

The Correct Option is A

To determine the spontaneity of a chemical reaction at different temperatures, we must consider the Gibbs free energy change (\(\Delta G\)). The equation for \(\Delta G\) is given by: 

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

Where:

• \(\Delta G\) = Gibbs free energy change
• \(\Delta H\) = Enthalpy change
• \(T\) = Temperature in Kelvin
• \(\Delta S\) = Entropy change

A reaction is spontaneous when \(\Delta G < 0\) and non-spontaneous when \(\Delta G > 0\).

For a reaction to be spontaneous at low temperature but non-spontaneous at high temperature, the signs of \(\Delta H\) and \(\Delta S\) must lead to the following behavior:

• At low temperatures, the reaction must be spontaneous, meaning \(\Delta G\) is negative.
• At high temperatures, the reaction becomes non-spontaneous, meaning \(\Delta G\) is positive.

Analyzing the options:

1. \(\Delta H < 0, \Delta S < 0\): Here, \(\Delta H\) is negative, contributing to a negative \(\Delta G\). At low temperatures, the \(- T\Delta S\) term is small (since \(\Delta S\) is negative), keeping the \(\Delta G\) negative. At high temperatures, the \(- T\Delta S\) term increases in magnitude (more negative), potentially overcoming the negative \(\Delta H\) and making \(\Delta G\) positive, thus non-spontaneous. This condition fits the question and is correct.
2. \(\Delta H > 0, \Delta S = 0\): This condition is independent of temperature changes because the entropy term is zero. If \(\Delta H > 0\), \(\Delta G\) is always positive, and the reaction is always non-spontaneous.
3. \(\Delta H < 0, \Delta S > 0\): Here, both terms contribute to a decrease in \(\Delta G\), making the reaction spontaneous at any temperature.
4. \(\Delta H > 0, \Delta S > 0\): For this, \(\Delta G\) only becomes negative at high temperatures, when \(T\Delta S\) surpasses \(\Delta H\), leading to incorrect analysis of the question conditions.

Thus, for the reaction to be spontaneous at low temperatures and non-spontaneous at high temperatures, the correct condition is \(\Delta H < 0\) and \(\Delta S < 0\).