Question

The enthalpy and entropy change for a chemical reaction are ${\text{ - 2}}{\text{.5 X 1}}{{\text{0}}^{\text{3}}}{\text{cal and 7}}{\text{.4cal}}{{\text{K}}^{{\text{ - 1}}}}$ respectively. The reaction at 298 K is:
A.Spontaneous
B.Reversible
C.Irreversible
D.Non-spontaneous

Answer 1

According to thermodynamics, the spontaneous and non-spontaneous nature of the reaction can be found with the help of Gibbs free energy or free energy.

Complete step by step answer:
Let's discuss the definition of Gibbs free energy, it is defined as the decrease in energy which is equal to the maximum useful work obtained from the system
If $\Delta G > 0$, i.e. the value of Gibbs free energy is positive, then the process will be non-spontaneous.
If $\Delta G < 0$, i.e. value of Gibbs free energy is negative then the process will be spontaneous.
The process will be in equilibrium, if the value of $\Delta G = 0$.
As per the above question, Enthalpy change is given as $(\Delta H) = - 2.5 \times {10^{ - 3}}cal$ and entropy change is given as. The temperature T=298K. So, for Gibbs free energy ($\Delta G$), we know $\Delta G = \Delta H - T\Delta S$
Here, $\Delta G$= Gibbs free energy
$\Delta H$= change in enthalpy
T= temperature
$\Delta S$= change in entropy
First we change temperature (T) from K to ℃
${X^o}C + 273 = K \\\ \Rightarrow {X^o}C = 298 - 273 \\\ \Rightarrow {25^o}C \\\$

On putting the value in the above equation we get:
$\Delta G = - 2.5 \times {10^{ - 3}} - (25 \times 7.4)$
$\therefore \Delta G = - 185Cal$
Hence the value $\Delta G$ is found to be negative, so we can say reaction will be spontaneous.
Hence the correct answer is option A.

Note:
Alternatively, for the spontaneous nature of the reaction there are two responsible factors:
A.Enthalpy ($\Delta H$)
B.Entropy ($\Delta S$)
In the above question, enthalpy given is negative hence the first factor favors the reaction and change in entropy is positive it also favors the reaction. So we can say the process is highly spontaneous and $\Delta G$ will be highly negative.