Question

The enthalpy and entropy change for a chemical reaction $- 2.5 \times {10^{ - 3}}{\text{cal}}$ and $7.4{\text{cal}}{\deg ^{ - 1}}$respectively. Predict what will be the nature of the reaction at $298K$
A.Spontaneous
B.Reversible
C.Irreversible
D.Non-spontaneous

Answer 1

To answer this question, you should recall the concept of free energy. Also, you should know that a reaction is spontaneous if the energy of a reaction is negative. Substitute the appropriate values in the formula given below.

Formula used: $\;\Delta G = \Delta H - T\Delta S$ where $\Delta G$ is free energy, $\Delta H$ is enthalpy, $\Delta S$ is entropy and $T$ is temperature.

Complete step by step solution:
Gibbs energy, or free enthalpy, is a quantity that is used to measure the maximum amount of work done in a thermodynamic system when the temperature and pressure are kept constant. Gibbs free energy is denoted by the symbol ‘G’. Gibbs free energy can be defined as the maximum amount of work that can be extracted from a closed system.
The relation can be written as $\Delta G = \Delta H - T\Delta S\;$.
Substituting these values in the equation
$\Rightarrow \Delta G = - 2.5 \times {10^3} - 298 \times 7.4$.
After solving:
$\Rightarrow \Delta G = - 4705.2\;{\text{cal}}$.
As the value of $\Delta G$ is $- 4705.2\;{\text{cal}}$, which is a negative value, so this reaction will be spontaneous.

Hence, the correct answer for this question is option A.

Note: You should know about the Properties of Gibbs free energy:
1.It is a thermodynamic state function which means that it depends on the state of the system and not the path that is followed.
2.It is represented by the symbol $G$ but in the standard state, it is represented by $G^\circ$.
3.It’s SI unit is ${\text{Joules}}$
4.It’s CGS unit is ${\text{calories}}$.
5.It is related to the reduction potential of a reaction by the formula: $\Delta {{\text{G}}^{\text{o}}} = {\text{ }} - {\text{nFE}}_{{\text{cell}}}^{\text{o}}$ where n is the electrons transferred, F is Faraday’s constant.