Calculation of Entropy Changes

We have seen that the energy given off (or absorbed) by a reaction, and monitored by noting the change in temperature of the surroundings, can be used to determine the enthalpy of a reaction (e.g. by using a calorimeter)

Tragically, there is no comparable easy way to measure the change in entropy for a reaction.

• However, one thing to think about is the following: suppose we know that energy is going into a system (or coming out of it), and yet we do not observe any change in temperature. What is going on in such a situation?
• Changes in internal energy, that are not accompanied by a temperature change, might reflect changes in the entropy of the system

For example, consider water at °0C at 1atm pressure

• This is the temperature and pressure condition where liquid and solid phases of water are in equilibrium (also known as the melting point of ice)

H₂O(s) ó H₂O(l)

• At such a temperature and pressure we have a situation (by definition) where we have some ice and some liquid water
• If a small amount of energy is input into the system the equilibrium will shift slightly to the right (i.e. in favor of the liquid state)
• Likewise if a small amount of energy is withdrawn from the system, the equilibrium will shift to the left (more ice)

• Since the quantitative term that relates the amount of heat energy input vs. the rise in temperature is the heat capacity, it would seem that in some way, information about the heat capacity (and how it changes with temperature) would allow us to determine the entropy change in a system
• In fact, values for the "standard molar entropy" of a substance have units of J/mol K. The same units for molar heat capacity.

Standard Molar Entropy, S⁰

The entropy of a substance has an absolute value of 0 entropy at 0K

• Standard molar entropies are listed for a reference temperature (like 298K) and 1 atm pressure (i.e. the entropy of a pure substance at 298K and 1 atm pressure). A table of standard molar entropies at 0K would be pretty useless because it would be 0 for every substance (duh!)
• When comparing standard molar entropies for a substance that is either a solid, liquid or gas at 298K and 1atm pressure, the gas will have more entropy than the liquid, and the liquid will have more entropy than the solid
• Unlike enthalpies of formation, standard molar entropies of elements are not 0.

The entropy change in a chemical reaction is given by the sum of the entropies of the products minus the sum of the entropies of the reactants

• As with other calculations related to balanced equations, the coefficients of each component must be taken into account in the entropy calculation (the n, and m, terms below are there to indicate that the coefficients must be accounted for)

Calculate the change in entropy associated with the Haber process for the production of ammonia from nitrogen and hydrogen gas.

At 298K as a standard temperature:

N₂(g) + 3H₂(g) ó 2NH₃(g)

From the balanced equation we can write the equation for D^S0 (the change in the standard molar entropy for the reaction):

DS⁰ = 2*S⁰(NH₃) - [S⁰(N₂) + (3*S⁰(H₂))]

DS⁰ = 2*192.5 - [191.5 + (3*130.6)]

DS⁰ = -198.3 J/mol K