# Entropy

### Written by tutor Uneeb Q.

Like enthalpy, entropy is a thermodynamic term that is also a state function. More broadly we define a spontaneous process as one that occurs in the direction a chemical reaction is written (from left to right, from reactants to products). Mathematically the thermodynamic term used to describe entropy is S, as seen in formulas like Gibb’s Free energy where G = H – TS. It is a measure of the disorder of a system (though this definition does not fully describe what is happening it is a good approximation as an introductory concept). Recall from physics the term Δ as in ΔS which is the difference between the initial and final states (ΔS = Sf – Si).

In terms of reactions, if ΔS is positive, the process results in more disorder (more entropy). If ΔS is negative, the process results in more order (less entropy).

We use entropy as part of the second law of thermodynamics which states that the total entropy of a system and its surroundings always increases for a spontaneous process (in general the universe is always increasing its total entropy). So this law relates spontaneity to entropy. And therefore the effect of entropy on the surroundings must be taken into account.

For a spontaneous process, ΔS must be > q/T, and at equilibrium, ΔSsurroundings = -q/T = -ΔH/T. These mathematical definitions are needed to understand any concept that involves entropy. Keep in mind that although most thermodynamic terms are written in kJ and use the temperature system of degrees Celsius (C), the units of entropy are typically Joules per Kelvin (J/K, or J/K*mol) and thus requires conversion.

Question: What is the entropy of a chemical system which has lost 100 kJ as it changed over 1000 C? Is this increasing entropy or decreasing entropy?

Answer: To solve for ΔS use ΔSsurroundings = -q/T = -ΔH/T.

In this case we were given enthalpy of a chemical system’s reaction (-100 kJ/mol), and it is negative because it lost heat to the environment (exothermic). Convert kJ to Joules; 100 kJ * 1000 J/kJ = 100000 J

Also keep in mind that entropy requires the units of Kelvin not Celsius, convert this number also; K = 1000 C + 273 = 1273

Finally plug into the equation to solve;

ΔSsurroundings = -ΔH/T = -(-100000 J/mol) / 1273 K = 7.85 J/K*mol

Because this number is positive, entropy is increasing.

Keep this example in mind, because typically exothermic reactions contribute to the entropy of the universe (order from a chemical system is lost to its surroundings; -H values lead to +S values). However these values also influence Gibb’s free energy to determine if a reaction is spontaneous or not.

The third law of thermodynamics states that a substance that is perfectly crystalline at 0 K has an entropy value of zero. The change in entropy can be negative or positive, however. It is because of this third law that we can determine absolute zero; plotting the values of entropy over temperature will reveal a state in which there is no entropy at -273 Celsius, otherwise known as zero Kelvin (0 K). Experimentally (in the real world) it is impossible to reach absolute zero because it would require reducing entropy to zero.

Despite this we can use entropy just fine at any other temperature. In chemical reactions we define S° as the standard entropy at 25°C and 1 atm. The ΔS° for a reaction is the difference between the products’ entropies and the reactants’ entropies. Similar to how enthalpy is calculated.

ΔSrxn = ΣΔSproducts – ΣΔ Sreactants

Recall that sigma (Σ) is used for the mathematical operation of sums, and that products are found in the right side of the chemical equation (or the head of the arrow) and reactants are found at the left (or the tail of the arrow). For example

Reactants –> Products

And just like enthalpy there are values available from textbooks (constants) at various temperatures which can be applied to solving problems. Below are data values for diatomic gases.

Entropy (S) of certain diatomic gases [units = J/ K * mol]

 Compound Entropy H2 130.7 HCl 186.9 Cl2 223.1

A hypothetical reaction utilizing the above information can be written as follows:

H2 + Cl2 –> 2 HCl
Using the state function property of entropy
ΔSrxn= ΣΔSproducts – ΣΔSreactants
ΔSrxn= Σ(2 HCl) – Σ(H2 + Cl2)

ΔSrxn= (2*186.9) – (130.7 + 223.1)

ΔSrxn= 373.8 – 353.8
ΔSrxn= 20J/K

Notice that we did not need to write mol at the bottom, we took this unit into account when we balanced the equation and properly balanced it in the sum. Notice that there is a 2 in the balanced equation in front of HCl, this must be accounted for (e.g. 2 * HCl). The entropy of this reaction is increasing because S is positive.

Another way to think about this is if disorder is increasing, ΔSrxn is positive, then you may also be creating more particles. Some examples of processes with increasing disorder include more products produced than used up (see example below) A solid sublimes as a gas. And a solid dissolved in a solvent. If order is increasing, ΔSrxn is negative. Some examples of processes that have increasing order include less gas produced than used up; a liquid converted to a solid; and a solid precipitated from a solution.

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