# Calculating Entropy Changes

Entropy (S) of the universe is always increasing, and is denoted by S.

### Total Entropy Change

A system and surrounding are both components of the universe (total), which can be seen in this equation:

$ΔS_{total}=ΔS_{univ}= ΔS_{sys} + ΔS_{surr}>0 \tag{1}$

the system and surroundings collective represent the change in entropy. The entropy change of the surroundings is driven by heat flow and the heat flow determines the sign of ΔSsurr. Therefore if we had an exothermic reactions with a constant temperature the system would cause heat to flow into the surroundings, which causes ΔSsurr to become positive. The opposite can be see in an endothermic reaction.

### Entropy change for the System

We begin by first calculating the difference between the standard entropy values of the products and standard entropy values of the reactants:

$ΔS_{reaction}=Δn_pS_{products} - Δn_pS_{reactants} \tag{2}$

where S represents entropy, and np and nr represent moles of products and reactions. Once the you have calculated the entropy of the system, you can calculate the entropy change for the surroundings.

### Entropy change for the Surroundings

And from the Second Law of Thermodynamics, we know that the sign of ΔStotal determines whether the reaction under investigation happens spontaneously. For calculating the entropy for the surrounding, we can use the general definition of S:

$\Delta S_{surroundings} = \int dS_{surroundings} = \dfrac{q_{surroundings}}{T_{surroundings}} \tag{3}$

In which $$d_q$$ is the element of heat and $$T$$ is the absolute temperature (in K). From the conservation of energy, q surroundings and q system are related:

$q_{surroundings} = -q_{ system} \tag{4}$

In general, to do the integration, one needs to know $$T$$ along the path. As a common specific case, if the surrounding is big enough (e.g. the lab or the universe, with respect to the beaker in which the experiment is being done is too big), one can assume that temperature does not change during the course of the reaction, in which case the integral is simplified:

$\Delta S_{surroundings} = \dfrac{q_{surroundings}}{T_{surroundings}} = - \dfrac{q_{system }}{T_{surroundings}} \tag{5}$

in which the value for $$q_{system}$$ is the same as enthalpy change and internal energy (internal energy change) of the system at constant pressure and constant volume, respectively.

Example 1

For example, if one has a generic reaction

$\text{Reactants} \rightarrow \text{Products}$

with $$q_{rxn} = 75 \;kJ$$ at 300K

Then, for the surrounding, one has,

$ΔS_{surroundings} = - \dfrac{75 \times 10^3}{300} = 250\; J/K$

One should appreciate the fact that the aim of defining new state functions like Gibbs free energy and Helmholtz free energy is partly to bypass the calculation of the change in the entropy of the surrounding. Calculating the entropy change for the system in such thermodynamic potentials can substitute the calculation of the total entropy in determining the spontaneity of the reaction (under particular conditions (e.g., G, Gibbs free energy is useful at constant pressure (isobar) and constant temperature (isotherm) conditions).

### References

1. Zumdahl, Steven S. Chemistry. New York: Houghton Mifflin, 1997.
2. Kotz, John C., Paul M. Treichel and Grabriela C. Weaver. Chemistry and Chemical Reactivity. Belmont: Brooks Cole, 2005.
3. Greiner, Neise, Stocker - Thermodynamics and Statistical Mechanics (Springer, 1997)
4. Petrucci, Harwood, Herring, Madura. General Chemistry: Principles & Modern Applications, Ninth Ed. Upper Saddle River, NJ: Pearson Education, Inc., 2007.
5. McMurray, Fay. Chemistry, Third Ed. Upper Sadle River, NJ: Prentice-Hall, Inc., 2001.

### Contributors

• Cindy Ramirez (UCD), Zafir Javeed (UCD)