Why does entropy depend on temperature

entropy

In connection with reactions, we only know the term heat of reaction and we know whether energy is released or whether it has to be used. In the field of chemical reactions there is another important variable: entropy.

Entropy is a measure of the disorder that a system exhibits. The higher the disorder, the higher the entropy. If the disorder is increased by a chemical reaction, it is a positive entropy change ∆S; if more order is generated, there is a negative change in entropy ∆S.

How is the entropy increased?

  • If there are more particles on the product side than on the educts, the entropy increases.
  • If the reaction produces liquids or gases from a solid or gases from liquids, the entropy increases.

Direction of spontaneous processes

But what is entropy good for now? If we take the entropy change ∆S and the reaction energy ∆E (if ∆E < 0="" handelt="" es="" sich="" um="" eine="" exotherme="" reaktion="" und="" es="" wird="" energie="" frei;="" wenn="" ∆e=""> 0 it is an endothermic reaction and energy has to be expended), then we can use the Gibbs energy ∆G (also free energy) to determine whether a reaction takes place voluntarily or not. We can get Gibbs' energy with the equation

Calculate, with ∆E as reaction energy and T as temperature in Kelvin.

If we insert the values ​​for the heat of reaction, temperature and change in entropy into the equation, we get positive or negative values ​​for ∆G:
∆G < 0:="" exergonische="" reaktion="" )="" die="" reaktion="" läuft="" freiwillig="" ab="" ∆g=""> 0: endergonic reaction) the reaction does not take place voluntarily

Altogether, with the heat of reaction ∆E and the change in entropy ∆S, four different cases can be distinguished for the Gibbs energy:

  1. ∆E < 0="" und="" ∆s=""> 0:
    In this case we subtract a positive number from a negative one, which means that the Gibbs energy ∆G is always negative regardless of the temperature and such reactions always take place voluntarily. E.g.:

    \ begin {align *}
    \ Delta E & = - 10 \ {kJ}, T = 200 \ {K}, \ Delta S = 0 {,} 5 \ {kJ} / {K} \
    \ Delta G & = - 10 \ {kJ} - 200 \ {K} \ cdot 0 {,} 5 \ {kJ} / {K} = -110 \ {kJ} \ Rightarrow \ text {exergonic}
    \ end {align *}

  2. ∆E> 0 and ∆S < 0:="" hier="" ziehen="" wir="" eine="" negative="" zahl="" von="" einer="" positiven="" ab.="" durch="" die="" differenz="" einer="" negativen="" zahl="" ergibt="" sich="" insgesamt="" eine="" addition.="" daher="" ist="" die="" gibbs’sche="" energie="" ∆g="" unabhängig="" von="" der="" temperatur="" immer="" positiv="" und="" solche="" reaktionen="" laufen="" nie="" freiwillig="" ab.="" bsp.:="" \begin{align*}="" \delta="" e="" &="56" \="" {kj},="" t="200" \="" {k},="" \delta="" s="-0{,}2" \="" {kj}/{k}="" \\="" \delta="" g="" &="56" \="" {kj}="" -="" 200="" \="" {k}="" \cdot="" (-0{,}2="" \="" {kj}/{k})="96" \="" {kj}="" \rightarrow="" \text{endergonisch}="" \end{align*}="">
  3. Here it depends on the temperature whether the reaction takes place voluntarily or not. Reactions of this kind can only take place at low temperatures. E.g.:
    \ begin {align *}
    1. \ quad \ Delta E & = -90 \ {kJ}, T = 100 \ {K}, \ Delta S = -0 {,} 4 \ {kJ} / {K} \
    \ Delta G & = -90 \ {kJ} - 100 \ {K} \ cdot (-0 {,} 4 \ {kJ} / {K}) = -50 \ {kJ} < 0="" \rightarrow="" \text{exergonisch}="" \\="" 2.="" \quad="" \delta="" e="" &="-90" \="" {kj},="" t="300" \="" {k},="" \delta="" s="-0{,}4" \="" {kj}/{k}="" \\="" \delta="" g="" &="-90" \="" {kj}="" -="" 300="" \="" {k}="" \cdot="" (-0{,}4="" \="" {kj}/{k})="30" \="" {kj}="">< 0="" \rightarrow="" \text{endergonisch}="" \end{align*}="">
  4. ∆E> 0 and ∆S> 0:
    In this case, too, it depends on the temperature whether the reaction takes place voluntarily or not. Such reactions only take place voluntarily when the temperature is very high. E.g.:

    \ begin {align *}
    1. \ quad \ Delta E & = 55 \ {kJ}, T = 100 \ {K}, \ Delta S = 0 {,} 3 \ {kJ} / {K} \
    \ Delta G & = 55 \ {kJ} - 100 \ {K} \ cdot 0 {,} 3 \ {kJ} / {K} = 25 \ {kJ} < 0="" \rightarrow="" \text{endergonisch}="" \\="" 2.="" \quad="" \delta="" e="" &="55" \="" {kj},="" t="500" \="" {k},="" \delta="" s="0{,}3" \="" {kj}/{k}="" \\="" \delta="" g="" &="55" \="" {kj}="" -="" 500="" \="" {k}="" \cdot="" 0{,}3="" \="" {kj}/{k}="-95" \="" {kj}="">< 0="" \rightarrow="" \text{exergonisch}="" \end{align*}="">