At a Basic Level
Entropy, whilst a mathematical concept can be described as a measure of disorder. The more disordered, then the greater the entropy. Correspondingly, a perfect crystal at 0K (absolute zero) has an entropy of 0, liquids will have considerably more and gases even more still. As an example, water in the solid state (ice) has a standard entropy of approximately 48 JK-1mol-1, water (liquid) approximately 70 JK-1mol-1 and water as steam, approximately 189 JK-1mol-1.
In Chemistry, Entropy is given the symbol S and the change in entropy ΔS; the units being JK-1mol-1. To calculate the entropy change for a reaction, one simply needs to utilise the following equation.
ΔS system = ∑S products - ∑S reactants (note the ∑ sign means the sum of)
In the calculation above, the entropy values are usually given as standard molar entropies (ΔS), that is the entropy at 298K (°C + 273 to convert to Kelvin).
Entropy can also be separated into 2 terms, the system (i.e. the reaction) and everything else or the universe.
ΔS total (or universe) = ΔS surroundings + ΔS system
Furthermore, the entropy change of the surroundings can also be calculated using:
ΔS surroundings = -ΔH reaction or system/T
For example, in an exothermic reaction (Enthalpy is negative). Energy, in the form of heat (kinetic energy) is given out to the surroundings. The result of this is to increase the disorder of the surroundings and thus the entropy of the surroundings increases. Even if the entropy of the system decreases, if the entropy of the surroundings increases to a greater extent, then the reaction will be feasible. This is a slightly long winded statement of the second law of thermodynamics
“for a reaction to occur then “ΔS total (or universe) > 0”
Entropy is at play in all chemical and physical processes: A gas has higher entropy than a solid, when a solid lattice dissolves entropy increases and when 1 mole of reagent makes 2 moles of products; entropy again increases.
Going back to the beginning with our definition of a perfect crystal having an entropic value of 0, we can now maybe think about this in a more detailed way (note this is advanced and for interest only). For a perfect crystal with no thermal energy (no movement), the atoms in the lattice are only arranged in one way or state. By using a famous equation (related to the Third Law), we can see how this arises. As W represents ways of arranging the particles (in this case W=1) and that ln(1)=0, we can see that S must therefore equal zero.
S = k*lnW