Question

What does it mean when a reaction is spontaneous?

Answer 1

Hint: As we know that a spontaneous reaction is a reaction that favors the formation of products at the conditions under which the reaction occurs. Spontaneity of a reaction is determined by the change in Gibbs Free Energy that is the maximum amount of non pressure or volume work which can be done by the system.

Complete answer:
Let us first understand about spontaneity of a reaction as follows:-
-Spontaneity: It is a measure of how feasible the reaction is in terms of energy (assuming that it would occur quickly if it can occur). This means that if the kinetics is fast enough for us to observe and the reaction is spontaneous as well, then we can see it happen.
-Spontaneity is determined by the change in the Gibbs' free energy (denoted by$\Delta G$) which is the maximum amount of non pressure or volume work that can be done by the system.
-The general equation which is true at constant temperature is: $\Delta G=\Delta H-T\Delta S$ where,
$\Delta H$= change in enthalpy (heat flow at constant pressure)
$\Delta S$= change in entropy (tendency of energy dispersal) at a certain temperature T in Kelvin.
-Change in enthalpy ($\Delta H$) and change in entropy ($\Delta S$) commonly tabulated at T=298.15 K and 1 atm (standard temperature and pressure conditions) as standard enthalpies of formation $\Delta H_{f}^{o}$ and standard molar entropies $\Delta S^{o}$, and thus, the spontaneity of known reactions can be predicted at 298.15 K and 1 atm.
-The following are the conditions for various reactions:-
If$\Delta {{G}^{o}}<0$, the reaction is spontaneous at 298.15 K and 1 atm
If$\Delta {{G}^{o}}>0$, the reaction is non spontaneous at 298.15 K and 1 atm
If$\Delta {{G}^{o}}=0$, the reaction is neither of them and is at dynamic equilibrium at 298.15 K and 1 atm
-Hence the reaction is said to be spontaneous when $\Delta {{G}^{o}}<0$at 298.15 K and 1 atm.

Note:
-If the reaction is at equilibrium then in general $\Delta G\ne \Delta {{G}^{o}}$ and$\Delta {G}=0$. Therefore we can write the following equation:-
$\begin{align}
  & \Delta G=\Delta {{G}^{o}}+RT\ln Q \\
 & \Rightarrow \Delta {{G}^{o}}=-RT\ln K \\
\end{align}$
where K is the dissociation constant.