Question

What is the use of standard molar entropy?

Answer 1

Hint: Standard molar entropy is basically the entropy that 1 mole of any substance will gain when its conditions change from 0K to the standard room temperature.

Complete step by step solution:
-First of all, let us see what standard molar entropy is.
Entropy is basically the measure of randomness or disorders. The values of entropy are given as the standard value of entropy. The standard molar entropy is the total (minimal) amount of entropy gained by 1 mole of a substance when its temperature is increased from 0K to the standard conditions. For any element in its standard state, the value of entropy has a definite, non-zero value.
-We will now discuss the uses or application of standard molar entropy:
1) The standard molar entropies of reactants and products of a reaction can be used to calculate $\Delta {S^ \circ }$ for a reaction. This is done by subtracting the standard molar entropies of all the reactants from that of the products. This is given as:
$\Delta {S^ \circ } = \sum {m{S^ \circ }(products)} - \sum {n{S^ \circ }(reactants} )$
Where, m and n are the coefficients of products and reactants in the balanced reaction.
Let us take a general example: $aA + bB \to cC + dD$; the value of $\Delta {S^ \circ }$ for this reaction will be:$\Delta {S^ \circ } = \left( {cS_C^ \circ + dS_D^ \circ } \right) - \left( {aS_A^ \circ + bS_B^ \circ } \right)$
2) The entropies of different substances or compounds can be under the same pressure and temperature using the standard molar entropy (${S^ \circ }$).
3) With the increasing complexity of the molecules, the value of standard molar entropy (${S^ \circ }$) also increases.
For example: Acetic acid ($C{H_3}COOH$) is more complex than methanol ($C{H_3}OH$) and so the standard molar entropy (${S^ \circ }$) of $C{H_3}COOH$ is more than that of $C{H_3}OH$ as given below:
${S^ \circ }$ for $C{H_3}COOH$= $160$ $J.{K^{ - 1}}.mo{l^{ - 1}}$
${S^ \circ }$ for $C{H_3}OH$= $127$ $J.{K^{ - 1}}.mo{l^{ - 1}}$
4) Heavier substances or compounds have higher value of standard molar entropy (${S^ \circ }$) as compared to lighter substances.
For example: let us take iodine and fluorine.
For ${I_{(g)}}$ the value of ${S^ \circ }$ is 180.7 $J.{K^{ - 1}}.mo{l^{ - 1}}$ because it is heavier while for ${F_{(g)}}$ it is 158.6 $J.{K^{ - 1}}.mo{l^{ - 1}}$.

Note: The changes in entropy occur due to phase transitions and chemical reactions. The standard entropy of a reaction determines the spontaneity of the reaction in the following manner.
If $\Delta {S_{total}}$ = positive (+ve); the process is spontaneous and a spontaneous process is thermodynamically irreversible.
If $\Delta {S_{total}}$ = negative (-ve); the process is nonspontaneous.
If $\Delta {S_{total}}$ = 0; the process is at equilibrium.