Chapter 17

 Define key terms and concepts  Explain the three laws of thermodynamics.  Calculate the entropy for a system.  Determine if a reaction is spontaneous or nonspontaneous.  Calculate the Gibbs Free Energy for a reaction.

 Energy cannot be created or destroyed, only converted from one form to another. ΔH = ΔU + PΔV Δ H° rxn = ΣΔH° f(products) - ΣΔH° f(reactants)

 Spontaneous processes/reactions are favored over non- spontaneous processes.  Entropy  How dispersed the energy of a system is  The more dispersed, the greater the entropy

 Solids have less entropy than liquids  Liquids have less entropy than gases  As a solute dissolves in solution, entropy increases.  Where k = 1.38x J/K  W = the number of microstates S = k ln WΔS = S f - S i ΔS = k ln W f - k ln W i

 Melting aluminum  Iodine crystals subliming  A car accident  Water freezing  Dissolving sodium chloride in water

 The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.  For a spontaneous process, ΔS univ > 0  For an equilibrium process, ΔS univ = 0 ΔS univ = ΔS sys + ΔS surr

For the general reaction: aA + bB  cC + dD ΔS° rxn =[cS°(C) + dS°(D)] – [aS°(A) + bS°(B)]  If the entropy for a system increases, ΔS° is positive.  If the entropy for a system decreases, ΔS° is negative. ΔS° rxn = ΣnS°(products) - ΣnS°(reactants)

 If a reaction produces more gas molecules than it consumes, ∆S° is positive.  If the total number of gas molecules diminishes, ∆S° is negative.  If there is no net change in the total number of gas molecules, then ∆S° may be negative or positive, but the numerical value will be small.

C 6 H 12 O 6(s)  C 2 H 5 OH (l) + CO 2(g) NH 3(g) + CO 2(g)  NH 2 CONH 2(aq) + H 2 O (l) CO (g) + H 2 O (g)  CO 2(g) + H 2(g)

Calculate the standard entropy of reaction for the following reaction at 25°C. C 6 H 12 O 6(s)  C 2 H 5 OH (l) + CO 2(g)

Calculate the standard entropy of reaction for the following reaction at 25°C. NH 3(g) + CO 2(g)  NH 2 CONH 2(aq) + H 2 O (l)

Calculate the standard entropy of reaction for the following reaction at 25°C. CO (g) + H 2 O (g)  CO 2(g) + H 2(g)

Surroundings System 1System 2 HeatHeat HeatHeat Entropy

 If the temperature of the surroundings is low, the energy released by a system will increase the entropy.  The Third Law of Thermodynamics  The Entropy of a perfect crystalline substance is zero at absolute zero temperature. ∆S surr = -∆H sys T

The heat of vaporization, ∆H vap, of carbon tetrachloride at 25°C is 43.0 kJ/mole. If 1 mole of liquid carbon tetrachloride at 25°C has an entropy of 214 J/K, what is the entropy of 1 mole of the vapor in equilibrium with the liquid at this temperature?

Liquid ethanol at 25°C has an entropy of 161 J/moleK. If the heat of vaporization, ∆H vap, at 25°C is 42.3kJ.mole, what is the entropy of the vapor in equilibrium with the liquid at 25°C?

 The energy available to do work.  If ∆G<0, the reaction is spontaneous in the forward direction.  If ∆G >0, the reaction is nonspontaneous in the forward direction.  If ∆G=0, the system is at equilibrium and no net change in energy will be observed. ∆G= ∆H-T∆S

Using ∆H° and ∆S° values from Appendix 3, calculate the ∆G° for the following reaction at 25°C. N 2(g) + H 2(g)  NH 3(g)

Using ∆H° and ∆S° values from Appendix 3, calculate the ∆G° for the following reaction at 25°C. CaCO 3(s) ↔ CaO (s) + CO 2(g)

Using ∆H° and ∆S° values from Appendix 3, calculate the ∆G° for the following reaction at 25°C. KClO 3(s) ↔ KCl (aq) + O 2(g)

∆H∆S∆G ++ Reaction is spontaneous in the forward direction at high temperatures, but is spontaneous in the reverse direction at low temperatures. +-∆G is always positive. Non-spontaneous reaction. -+∆G is always negative. Spontaneous reaction. -- Reaction is spontaneous in the forward direction at low temperatures, but is spontaneous in the reverse direction at high temperatures.

 Standard Free Energy of a Reaction (∆G° rxn )  The change in free energy when 1 mole of a compound is synthesized from its elements in their standard state.  For the general reaction: aA + bB  cC + dD ΔS° rxn =[cS°(C) + dS°(D)] – [aS°(A) + bS°(B)] ΔG° rxn = ΣnG° f (products) - ΣnG° f (reactants)

Determine if the following reaction is spontaneous using Gibbs Free Energy if the reaction is conducted at 25°C. CaCO 3(s)  CaO (s) + CO 2(g)

Determine if the following reaction is spontaneous using Gibbs Free Energy if the reaction is conducted at 25°C. KClO 3(s)  KCl (aq) + O 2(g)

Determine if the following reaction is spontaneous using Gibbs Free Energy if the reaction is conducted at 25°C. NH 3(g) + CO 2(g)  NH 2 CONH 2(aq) + H 2 O (l)

 For calculating reactions where not all reactants and products are in their standard states.  Where R= J/Kmole  T is temperature in Kelvin  K is the equilibrium constant. ΔG° = -RTlnK

KInKΔG°Affect at Equilibrium > 1+-Products are favored = 100Neither side of the reaction is favored < 1-+Reactants are favored

What is the value of the equilibrium constant K at 25°C for the following reaction? Predict whether the products or reactants are favored at equilibrium. CaCO 3(s) ↔ CaO (s) + CO 2(g)

What is the value of the equilibrium constant K at 25°C for the following reaction? Predict whether the products or reactants are favored at equilibrium. KClO 3(s) ↔ KCl (aq) + O 2(g)

What is the value of ΔG° at 25°C for the following reaction? The K sp of magnesium hydroxide is 1.8x Predict whether the products or reactants are favored at equilibrium. Mg(OH) 2(s) ↔ Mg 2+ (aq) + OH - (aq)

What is the value of the equilibrium constant K at 25°C for the following reaction? The K sp of lead (II) iodide is 6.5x Predict whether the products or reactants are favored at equilibrium. PbI 2(s) ↔ Pb 2+ (aq) + I 2 - (aq)