91Ó°ÊÓ

The entropy change should be positive for the spontaneous process.

When $\mathbf{∆}\mathbf{S}\mathbf{>}\mathbf{0}$, the reaction is spontaneous.

When $\mathbf{∆}\mathbf{S}\mathbf{<}\mathbf{0}$, the reaction is non-spontaneous.

When $\mathbf{∆}\mathbf{S}\mathbf{=}\mathbf{0}$, the reaction is in equilibrium.

When $∆\mathrm{S}$is positive, the order of the system is decreased.

When $∆\mathrm{S}$is positive, the reaction is spontaneous at all temperatures.

When $∆\mathrm{H}$is negative, the reaction should be spontaneous.

Hence, the system will be giving off heat.

The reaction can be spontaneous at a lower temperature, where the value of $∆\mathrm{H}$is greater than $\mathrm{T}∆\mathrm{S}$, and at a high temperature, the value of $∆\mathrm{H}$is smaller than $\mathrm{T}∆\mathrm{S}$.

The relation between $∆\mathrm{G}$,$∆\mathrm{H}$, and $∆\mathrm{S}$ is:

$∆\mathrm{G}°=∆\mathrm{H}°-\mathrm{T}∆\mathrm{S}°$

Here,

$∆\mathrm{G}°$is Gibbs free energy (J or kJ)

$∆\mathrm{H}°$is enthalpy

T is temperature

$∆\mathrm{S}°$is entropy

$\mathbf{∆}\mathbf{G}$ is negative for spontaneous change

At constant pressure,

$∆\mathrm{G}=∆\mathrm{H}-\mathrm{T}∆\mathrm{S}$

finds the direction of a chemical reaction. If $∆\mathrm{G}<0$that is negative, the reaction is spontaneous. If $∆\mathrm{G}<0$ that is positive, the reaction is non-spontaneous.

If $∆\mathrm{G}=0$, the reaction is in equilibrium.