Entropy and calculating entropy change

In a nutshell

Entropy measures how disordered a system is. Entropy increases when there's a favourable change in physical state or increased number of moles. Entropy change is calculated by finding the difference in entropy between the reactants and products. The total entropy changes is the total entropy change from the surroundings and system.

Equations

Word equation	symbol equation
$entropy\>change\>of\>a\>system=entropy\>of\>products\>-\>entropy\>of\>reactants$	$\triangle S_{system}=S_{products}-S_{reactants}$
$total\>entropy\>change=entropy\>change\>of\>a\>system\>+\>entropy\>change\>of\>surroundings$	$\triangle S_{total}=\triangle S_{system}+\triangle S_{surroundings}$
$entropy\>change\>of\>surroundings=-\dfrac{enthalpy\>change}{temperature}$	$\triangle S_{surroundings}=-\dfrac{\triangle H}{T}$

Variable units

quantity name	symbol	unit name	unit
$entropy\>change\>of\>a\>system$	$\triangle S_{system}$	$joules \>per\> kelvin \>per \>mole$	$J\>K^{-1}\>mol^{-1}$
$entropy\>of\>products$	$S_{products}$	$joules \>per\> kelvin \>per \>mole$	$J\>K^{-1}\>mol^{-1}$
$entropy\>of\>reactants$	$S_{reactants}$	$joules \>per\> kelvin \>per \>mole$	$J\>K^{-1}\>mol^{-1}$
$total\>entropy\>change$	$\triangle S_{total}$	$joules \>per\> kelvin \>per \>mole$	$J\>K^{-1}\>mol^{-1}$
$entropy\>change\>of\>surroundings$	$\triangle S_{surroundings}$	$joules \>per\> kelvin \>per \>mole$	$J\>K^{-1}\>mol^{-1}$
$enthalpy\>change$	$\triangle H$	$joules\>per\>mole$	$J\>mol^{-1}$
$temperature$	$T$	$kelvin$	$K$

Entropy

Entropy, $S$ , is the measure of how disordered a system is. The higher the entropy value, the more disorder there is in the system. This means there will be a lot of different arrangements for the position and energy shared between the particles.

The physical state of species gives an idea of the entropy. Solids have particles that are fixed in position. This means they are very ordered, with a low entropy. Liquids have particles with more freedom than solids. They have a higher entropy value than solids due to the increased disorder. Gases are very disordered with particles that are free to move around. Their entropy value will be very high.

Chemistry; Energetics II; KS5 Year 12; Entropy and calculating entropy change — 1. Solid, 2. Liquid, 3. Gas

When a reaction involves a change in physical state, this changes the entropy value.

Example

Sodium is oxidised to form sodium oxide. Will the entropy value increase or decrease?

$2Na(s)+\dfrac{1}{2}O_2(g)→Na_2O(s)$

The entropy value is lowered due to the gas species converting to a solid species.

Entropy will also increase if the number of moles in a reaction increases. Entropy decreases if the number of moles in a reaction decreases.

Example

The chemical equation for the dissociation of $PI_5$ is shown below. Does the entropy increase or decrease?

$PI_5(g) →PI_3 (g)+ I_2 (g)$

The entropy value increases. The gaseous moles increased from one reactant mole to two product moles.

Some reactions don't require additional energy to react if the entropy is very high. This is because particles will move to increase the disorder in the system to become stable.

Entropy change

The entropy change of a system is calculated by finding the difference in entropy between the products and reactants. A positive entropy change leaves a reaction likely to be feasible. A negative entropy change means the reaction may not be feasible. However, you can't rely on the entropy change alone to determine if a reaction will happen or not. Other factors such as kinetics, temperature and enthalpy affect whether a reaction is feasible too.

$entropy\>change\>of\>a\>system=entropy\>of\>products\>-\>entropy\>of\>reactants \newline \triangle S_{system}=S_{products}-S_{reactants}$

Procedure

1.	Calculate the total entropy of products.
2.	Calculate the total entropy of reactants.
3.	Insert the values into the equation below to find the entropy change of a system: $\triangle S_{system}=S_{products}-S_{reactants}$

Example

Calculate the entropy change for the production of carbon dioxide from carbon and oxygen. The entropy of carbon dioxide is $214\>J\>K^{-1}\>mol^{-1}$ , carbon is $158\>J\>K^{-1}\>mol^{-1}$ and oxygen is $205\>J\>K^{-1}\>mol^{-1}$ .

$C(s)+O_2(g)→CO_2(g)$

Write the equation to find the entropy change of a system:

$\triangle S_{system}=S_{products}-S_{reactants}$

Find the total entropy of products (carbon dioxide):

$S_{products}=S_{(CO_2)}=214\>J\>K^{-1}\>mol^{-1}$

Calculate the total entropy of reactants (carbon and oxygen):

$S_{reactants}=S_{(C)}+S_{(O_2)}=158+205=363\>J\>K^{-1}\>mol^{-1}$

Calculate the difference by inserting known values into the entropy change of system equation:

$\triangle S_{system}=(214)-(364)=-150\>J\>K^{-1}\>mol^{-1}$

The entropy change of the system is $\underline{-150\>J\>K^{-1}\>mol^{-1}}$ .

Total entropy change

The total entropy change is the total entropy changes from the surroundings and system.

$total\>entropy\>change=entropy\>change\>of\>a\>system\>+\>entropy\>change\>of\>surroundings \newline \triangle S_{total}=\triangle S_{system}+\triangle S_{surroundings}$

The equation to calculate the entropy change of surroundings is shown below.

$entropy\>change\>of\>surroundings=-\dfrac{enthalpy\>change}{temperature} \\ \ \\ \triangle S_{surroundings}=-\dfrac{\triangle H}{T}$

The procedure to calculate the total entropy change is given below.

Procedure

1.	Calculate the entropy change of a system using the equation: $\triangle S_{system}=S_{products}-S_{reactants}$
2.	Calculate the entropy change of surroundings using the equation: $\triangle S_{surroundings}=-\dfrac{\triangle H}{T}$
3.	Insert values into total entropy change equation: $\triangle S_{total}=\triangle S_{system}+\triangle S_{surroundings}$

Example

Calculate the total entropy change for the reaction forming ammonium bromide under standard conditions ( $293\>K$ ). The entropy of ammonium bromide is $113\>J\>K^{-1}\>mol^{-1}$ , ammonia is $193\>J\>K^{-1}\>mol^{-1}$ and hydrogen bromide is $92\>J\>K^{-1}\>mol^{-1}$ . The enthalpy change for the reaction is $-188\>kJ\>mol^{-1}$ .

$NH_3(g)+HBr(g)→NH_4Br(s)$

Write the equation for the entropy change of a system:

$\triangle S_{system}=S_{products}-S_{reactants}$ 

Find the total entropy change for products (ammonium bromide):

$S_{products}=S_{(NH_4Br)}=113\>J\>K^{-1}\>mol^{-1}$ 

Calculate the total entropy change for reactants (ammonia and hydrogen bromide):

$S_{reactants}=S_{(NH_3)}+S_{(HBr)}=193+92=285\>J\>K^{-1}\>mol^{-1}$

Calculate the entropy change of a system:

 $\triangle S_{system}=S_{products}-S_{reactants}=113-285=-172\>J\>K^{-1}\>mol^{-1}$

Write the equation for the entropy change of surroundings:

$\triangle S_{surroundings}=-\dfrac{\triangle H}{T}$

Insert known values to calculate entropy change of surroundings:

$\triangle S_{surroundings}=-\dfrac{(-188\times10^3\>J\>mol^{-1})}{293\>K}=641.6382\>J\>K^{-1}\>mol^{-1}$

Note: Make sure the enthalpy change units are converted into $J\>mol^{-1}$ and the temperature units are converted into $K$ .

Write the total entropy change equation:

$\triangle S_{total}=\triangle S_{system}+\triangle S_{surroundings}$

Insert calculated values to calculate the total entropy change equation:

$\triangle S_{total}=(-172\>J\>K^{-1}\>mol^{-1})+(-\dfrac{(-188\times10^3\>J\>mol^{-1})}{293\>K}J\>K^{-1}\>mol^{-1}) \newline \triangle S_{total}=470 \>J\>K^{-1}\>mol^{-1}(3\> s.f)$

The total entropy change is $\underline{470 \>J\>K^{-1}\>mol^{-1}}$ .