ENTROPY CHANGES IN CHEMICAL REACTIONS CHAPTER 19 SECTION 4 & 5 ENTROPY CHANGES IN CHEMICAL REACTIONS AND GIBBS FREE ENERGY

19.4 ENTROPY CHANGES IN CHEMICAL REACTIONS Absolute entropies of substances are based on the reference point of zero entropy for perfect crystalline solids @ 0 K. Entropies are tabulated as standard molar entropies (So) for substances in their standard state and have units of J/mol·K The standard state for any substance is defined as the pure substance at 1 atm & 25oC

Unlike enthalpies of formation (ΔHf) for elements which are always zero, the standard molar entropies of elements at 25oC are not zero. Standard molar entropies of gases are greater than those of liquids and solids. Standard molar entropies increase with increasing molar mass or with an increase in the number of atoms in the chemical formula of a substance.

*The entropy change in a reaction equals the sum of the entropies of the products less the sum of the entropies of the reactants: ΔSo = ∑ nSo (products) - ∑ mSo (reactants) (n and m are the coefficients in the balanced chemical equation) * Study Sample Exercise 19.5 for the application of the equation above (simple math using values from Appendix C).

ENTROPY CHANGES IN THE SURROUNDINGS “Surroundings” are essentially a large constant temperature heat source (or heat “sink” if the heat flows from the system to the surroundings). The ΔSsurr will depend on how much heat is absorbed or released by the system: ΔSsurr = - qsys / T = - ΔHsys / T

For the reaction in Sample Exercise 19 For the reaction in Sample Exercise 19.5, the formation of NH3(g) from H2(g) and N2(g) , the qsys is the enthalpy change for the reaction under standard conditions (ΔHo): ΔHorxn = 2ΔHof [NH3(g)] - 3ΔHof [H2(g)] - ΔHof [N2(g)] = 2( - 46.19 kJ) - 3(0 kJ) - (0 kJ) = - 92.38 kJ The formation of ammonia from hydrogen and nitrogen is exothermic.

Absorption of the heat given off by the system results in an increase in the entropy of the surroundings: ΔSosurr = 92.38 kJ/298 K = 0.0310 kJ/K = 310 J/K The magnitude of the entropy gained by the surroundings is greater than that lost by the system ( - 198.3 J/K): ΔSouniv = ΔSosys + ΔSosurr = - 198.3 J/K + 310 J/K = 112 J/K

Because ΔSouniv is positive for any spontaneous reaction, the above calculation indicates that when ammonia, hydrogen, and nitrogen are together at 298 K (room temp.) in their standard states (all gases at 25oC), the reaction system will move spontaneously toward the formation of NH3(g). At some higher temperature the reverse reaction would be favored. This is why heat is removed when manufacturing ammonia. *If a process is exothermic, the entropy of the surroundings always increases (unless the system is thermally isolated from the surroundings).

SECTION 19.5 GIBBS FREE ENERGY Spontaneous processes can be either exothermic or endothermic. A spontaneous process that is endothermic (+ ΔH) must be accompanied by an increase in the entropy (+ΔS) of the sytem: example: NH4NO3(aq)  NH4+ + NO3- A spontaneous process that results in a decrease in entropy ( - ΔS) must be exothermic ( - ΔH): example: Na(s) + Cl2(g)  NaCl(s)

Thus, the spontaneity of a reaction involves two thermodynamic concepts, enthalpy and entropy (ΔH and ΔS). The Gibbs free energy (G) of a state is the potential energy of a system that can become available to do work…(Are you “free” to help me do some work?). The change in free energy of the system (ΔG) is given by the expression: ΔG = ΔH - TΔS

Under standard conditions, the equation becomes: ΔGo = ΔHo - TΔSo The Gibbs free energy change (ΔG) in a system is equal to - TΔSuniv For spontaneous processes, ΔSuniv is always positive and therefore - TΔSuniv will always be negative. Thus the sign of ΔG can determine the spontaneity of a process: 1. If ΔG is negative, the reaction is spontaneous in the forward direction (ΔSuniv is positive).

2. If ΔG is zero, the reaction is at equilibrium. 3 2. If ΔG is zero, the reaction is at equilibrium. 3. If ΔG is positive, the reaction in the forward direction is non-spontaneous (work must be supplied to the system to make it occur) and ΔSuniv is negative. However, the reverse reaction will be spontaneous. * Study Figure 19.18

Sample Exercise 19.6 Calculating Free-Energy Change from ΔHo, T, and ΔSo Calculate the standard free energy change (ΔGo) for the formation of NO(g) from N2(g) and O2(g) at 298 K: N2(g) + O2(g)  2NO(g) given that ΔHo = 180.7 kJ and ΔSo = 24.7 J/K is the reaction spontaneous under these circumstances? Key Equation: ΔGo = ΔHo - TΔSo

ΔGo = ΔHo - TΔSo = 180. 7 kJ - (298 K)(0. 0247 kJ/K) (. note: 24 ΔGo = ΔHo - TΔSo = 180.7 kJ - (298 K)(0.0247 kJ/K) (*note: 24.7 J/K was converted to kJ/K) = 180.7 kJ - 7.4 kJ = 173.3 kJ *Because ΔGo is positive, the reaction is not spontaneous (work must be done to the system for the reaction to proceed).

Standard Free Energy of Formation The standard free energies of formation (Appendix C) are useful in calculating ΔGo for chemical processes: ΔGo = ∑nΔGof (products) - ∑mΔGo (reactants) * (n and m are the coefficients in the balanced equation) The sign of ΔGo can predict the spontaneity of the reaction.

Sample Exercise 19.8 Estimating ΔGo Looking at the following reaction, the combustion of propane gas at 298 K: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) ΔHo = - 2220 kJ …can you predict whether ΔGo for this reaction is more negative or less negative than ΔHo ? Recall: ΔGo = ΔHo - TΔSo Solve: What would the value of ΔSo be and how will it affect the value of ΔGo ?

The sign of ΔSo can be determined by looking at the reaction The sign of ΔSo can be determined by looking at the reaction. There is a decrease in entropy in the reaction: 6 moles of gas on the left side of the reaction with 3 moles of gas and 4 moles of liquid on the right side indicating and increase in order! So, ΔSo must be negative making - TΔSo positive and making ΔGo less negative (more positive) than ΔHo. The phases of the reactants and products are important in a reaction. If H2O had been in its gaseous (vapor) state in the reaction, then entropy (ΔSo) would be positive and ΔGo would be more negative than ΔHo…

…this would violate the fact that a spontaneous exothermic reaction must be accompanied by an increase in the entropy of the Universe. The key to understanding this is that the reaction takes place at standard conditions (298 K) and water is liquid at this temperature! This is why standard conditions are specified for the calculations that give standard thermodynamic values.