In the case of an irreversible process, the change in entropy between two equilibrium states is calculated by finding a reversible path between the two states and calculating the entropy change for that path. When the products of a reaction are less complex and more disordered than the reactants, the reaction is said to proceed with a gain in entropy.

In all natural processes, the entropy of the universe increases. Living organisms consist of collections of molecules much more highly organized than the surrounding materials from which they are constructed, and organisms maintain and produce order, seemingly oblivious to the second law of thermodynamics.

Change in

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Solved problems based on entropy are given below. ### Solved Examples

**Question 1: **Calculate $\Delta$S for the synthesis of ammonia at 25^{o}C.

N_{2} + 3H_{2} $\rightarrow$ 2NH_{3} $\Delta$H = -92.6kJmol^{-1}

** Solution: **

Here we use the formula

$\Delta$S = 2(NH_{3}) - [S(N_{2}) + 3S(H_{2})]

$\Delta$S = (2)(192.5 JK^{-1}mol^{-1}) - [191.6JK^{-1}mol^{-1} + (3)(130.6 JK^{-1}mol^{-1})]

$\Delta$S = -198.4 JK^{-1}mol^{-1}

**Question 2: **Calculate the entropy change per mole when ice melts at 0^{o}C. Remember that the enthalpy change equals the heat added to the system at constant pressure.

** Solution: **

The equation is

H_{2}O(s) $\rightarrow$ H_{2}O(l)

$\Delta$H (fusion) = +6.01KJmol^{-1}

So we use the equation

$\Delta$S = $\frac{6010Jmol-1}{273K}$

= +22.0JK^{-1}mol^{-1}

N

Here we use the formula

$\Delta$S = 2(NH

$\Delta$S = (2)(192.5 JK

$\Delta$S = -198.4 JK

The equation is

H

$\Delta$H (fusion) = +6.01KJmol

So we use the equation

$\Delta$S = $\frac{6010Jmol-1}{273K}$

= +22.0JK