# Measurement of Entropy Change

For a reversible change at constant temperature the change in entropy is equal to heat absorbed or evolved divided by the constant temperature in Kelvin.

Thus,

$\Delta S = \dfrac{q_{rev.}}{T}$

The unit of entropy in $J K mol^{-1}$. The value of AS is positive if heat is absorbed and negative if heat is evolved. The entropy change in melting a solid can be calculated if enthalpy of fusion $(\Delta H_{fusion})$ is known.

$\Delta S_{melt} = \dfrac{\Delta H_{fusion}}{T_{melt}}$

Similarly, the entropy change for vaporisation of a liquid into its vapour at its boiling point can be calculated if enthalpy of vapourisation $(\Delta H_{Vap.})$ is known.

$\Delta S_{vap.} = \dfrac{\Delta H_{vap.}}{T_{boiling}}$

Similarly the entropy of sublimation is given as

$\Delta S_{sub} = S_{Vapour}- S_{solid} = \dfrac{\Delta H_{Sub}}{T}$

It is important to note that, the entropy change for a reaction carried out reversibly is different from a reaction carried out irreversibly. In other words, for an irreversible spontaneous change,

$\Delta S > \dfrac{q}{T} \\[3mm] \Delta S > \dfrac{\Delta E + P \Delta V}{t} \\[3mm] \Delta S > 0$

For a reversible change,

$\Delta S = \dfrac{q}{T} = \dfrac{\Delta E + P \Delta }{T} \\[3mm] or \Delta S = 0$

Entropy changes for Ideal gases:

(A)   For change of state: (Initial to final)

$\Delta S = 2.303 n C_v \log_{10}[ \dfrac{T_2}{T_1} ] + 2.303 n R \log_{10} [ \dfrac{V_2}{V_1} ]$

$\Delta S = 2.303 n C_p \log_{10} [ \dfrac{T_2}{T_1} ] + 2.303 n R \log_{10}[ \dfrac{P_1}{P_2}]$

(B)   For isothermal process:

$\Delta S = 2.303 n R \log_{10} [\dfrac{V_2}{V_1}] or [\dfrac{P_1}{P_2}]$

(C)   For Isochoric process:

$\Delta S = 2.303 n C_v \log_{10}[\dfrac{T_2}{T_1}] or [\dfrac{P_1}{P_2}]$

(D)   For Isobaric process:

$\Delta S = 2.303 n C_p \log_{10} [\dfrac{T_2}{T_1} or [\dfrac{V_2}{V_1}]$

Some examples of increase or decrease of entropy:

On stretching a rubber band the entropy decreases as the coiled macro molecule become un-coiled i.e. randomness in structure decreases.

Increase of entropy losses:

On boiling an egg entropy increases as due to denaturation the helical structure of protein become more complicated and random coiled structure.

In a solution of $H_2O + C_2H_5OH; \Delta S_{vap}$ is very high due to extensive H-bonding.

Trouton’s rule: According to .it $\Delta S_{Vap}$ of most of the liquids is $88 \pm 5J$ / mole K at normal B.P.

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