Question

What can be said about an Endothermic reaction with a negative entropy change?

The reaction is
a. spontaneous at all temperatures.
b. spontaneous at high temperatures.
c. spontaneous at low temperatures.
d. spontaneous in the reverse direction at all temperatures.
e. nonspontaneous in either direction at all temperatures.



What can be said about an Exothermic reaction with a negative entropy change?

The reaction is
a. spontaneous at all temperatures.
b. spontaneous at high temperatures.
c. spontaneous at low temperatures.
d. spontaneous in the reverse direction at all temperatures.
e. nonspontaneous in either direction at all temperatures.

Answer 1

Concepts and reason

The concepts used to solve this question are the characteristics of endothermic and exothermic reactions, and the Gibbs free energy equation to determine a spontaneous reaction.

Fundamentals

The change in Gibb's free energy at constant temperature and pressure is given as;

${\rm{\Delta G}} = {\rm{\Delta H}} - {\rm{T\Delta S\;\;}}$

Where, ${\rm{\Delta G}}$is Gibb's free energy change; ${\rm{\Delta H}}$is change in enthalpy of the reaction; T is temperature of the reaction; ${\rm{\Delta S}}$is change in the entropy of the reaction.

Enthalpy (H) in thermodynamics is defined as the heat content of a reaction. Entropy (S) is termed as the quantification of disorder or randomness of a reaction. Gibb's free energy change helps to determine the direction of the reaction.

Part 1

The question states that it is an endothermic reaction with negative entropy. This means that the change in enthalpy will be positive and change in entropy will be negative. That is-

${\rm{\Delta G}} = {\rm{\Delta H}} - {\rm{T\Delta S\;\;}}$

The change in Gibb's free energy at constant temperature and pressure is given as-

${\rm{\Delta G}} = {\rm{\Delta H}} - {\rm{T\Delta S\;\;}}$

Now, the change in enthalpy and change in entropy can be written as follows,

$\begin{array}{l}\\{\rm{\Delta H > 0 }}\left( {{\rm{positive}}} \right)\\\\{\rm{\Delta S < 0\;}}\left( {{\rm{negative}}} \right)\\\end{array}$

Substitute these values in the above equation;

$\begin{array}{c}\\{\rm{\Delta G}} = {\rm{\Delta H}} - {\rm{T}} - ({\rm{\Delta S)}}\\\\{\rm{ = \;\Delta H}} + {\rm{T\;\Delta S}}\\\end{array}$

${\rm{\Delta G}} > {\rm{0}}$, thus, the reaction is spontaneous in the reverse direction at all temperatures.

Part 2

The question states that it is an exothermic reaction with negative entropy. This means that both the change in enthalpy and change in entropy will be negative. That is;

$\begin{array}{l}\\{\rm{\Delta H > 0 }}\left( {{\rm{positive}}} \right)\\\\{\rm{\Delta S < 0\;}}\left( {{\rm{negative}}} \right)\\\end{array}$

The change in Gibb's free energy at constant temperature and pressure is given as-

${\rm{\Delta G}} = {\rm{\Delta H}} - {\rm{T\Delta S\;\;}}$

Now, the change in enthalpy and change in entropy can be written as follows,

$\begin{array}{l}\\{\rm{\Delta H > 0 }}\left( {{\rm{positive}}} \right)\\\\{\rm{\Delta S < 0\;}}\left( {{\rm{negative}}} \right)\\\end{array}$

Substitute these values in the above equation as follows:

$\begin{array}{c}\\{\rm{\Delta G}} = {\rm{\Delta H}} - {\rm{T}} - ({\rm{\Delta S)}}\\\\{\rm{ = \;\Delta H}} + {\rm{T\;\Delta S}}\\\end{array}$

${\rm{\Delta G}}$will be negative, only when the conditions are such that the value of${\rm{\Delta H}}$is greater than${\rm{T\Delta S}}$.

Ans: Part 1

The reaction is spontaneous in the reverse direction at all temperatures.