# Entropy

## Key Questions

• Entropy increases when a system increases its disorder. Basically, a solid is pretty ordered, especially if it is crystalline. Melt it, you get more disorder because molecules can now slide past one another. Dissolve it, and you get another entropy increase, because the solute molecules are now dispersed among solvent.

Order of entropy from least to greatest:
Solid--> Liquid --> Gas

Example below explaining entropy and enthalpy

#### Explanation:

In a 4 stroke diesel engine, when piston takes the air into cylinder (sucks the air into cylinder) i.e. performing the first or inlet stroke while moving from top to bottom, it fully fills the cylinder with air then comes the turn of compressing this air and this compression is too much (about the ratio of 1:20) as compared with a 4 stroke petrol engine; in this process Entropy is decreased i.e. size of air is decreased without removing heat from the air (while Enthalpy remains same), this decrease in Entropy without decreasing Enthalpy i.e. without dissipating heat somewhere in the surrounding or nearby matter or environment causes a useful rise in the temperature of air about 600 degree centigrade which is more than enough to burn the diesel sprayed at that very time.

• Here are some formulas.

ΔS_"total" = ΔS_"univ" = ΔS_"sys" + ΔS_"surr" = q_"sys"/T_"sys" + q_"surr"/T_"surr"

where $q$ is the heat and $T$ is the Kelvin temperature.

Entropy Change for the System

ΔS_"sys" = ΔS_"rxn" = Sigma(n_pS_"products"^"o") – Sigma(n_rS_"reactants"^"o")

where ${n}_{p}$ and ${n}_{r}$ represent moles of products and reactions.

Entropy Change for the Surroundings

ΔS_"surr" = q_"surr"/T_"surr"

${q}_{\text{surr" = -q_"sys}}$

ΔS_"surr" = q_"surr"/T_"surr" = -q_"sys"/T_"surr"

Example 1:

What is ΔS_"surr" at 300 K for the reaction

reactants → products; ΔH = 75 kJ

Solution:

ΔS_"sys" = q/T = (75 000" J")/(300" K") = 250 J/K

ΔS_"surr" = q_"surr"/T_"surr" = -(ΔH_"sys")/T_"surr" = -(75 000" J")/(300" K") = -250 J/K

Example 2:

What is ΔS_"rxn"^"o" for the following reaction?

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

The ${S}^{\text{o}}$ values are NH₃ = 193 J·K⁻¹mol⁻¹; O₂ = 205 J·K⁻¹mol⁻¹; NO = 211 J·K⁻¹mol⁻¹;
H₂O = 189 J·K⁻¹mol⁻¹

ΔS_"sys" = Sigma(n_pS_"products"^"o") – Sigma(n_rS_"reactants"^"o")

ΔS_"rxn"^"o" = 4S_"NO"^"o" + 6S_"H₂O"^"o" – 4S_"NH₃" - 5S_"O₂"^"o"

ΔS_"rxn"^"o" = (4×211 + 6 × 189 - 4 × 193 + 5 × 205 ) J/K⁻¹ = 181 J•K⁻¹