Chapter 16 Thermodynamics

Objectives 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.

The First Law of Thermodynamics 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)

Chemistry is Spontaneous 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

Chemistry is Spontaneous

Entropy Solids have less entropy than liquids Liquids have less entropy than gases As a solute dissolves in solution, entropy increases. Where k = 1.38x10-23 J/K W = the number of microstates ΔS = Sf - Si S = k ln W ΔS = k ln Wf - k ln Wi

This system has 16 microstates

Would entropy increase or decrease in each of the following system? Melting aluminum Iodine crystals subliming A car accident Water freezing Dissolving sodium chloride in water

The Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. For a spontaneous process, ΔSuniv > 0 For an equilibrium process, ΔSuniv = 0 ΔSuniv = ΔSsys + ΔSsurr

The Second Law of Thermodynamics 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)

The Second Law of Thermodynamics

Predicting the Sign of ∆S° 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.

Predict the sign of ∆S° in the following reactions. C6H12O6(s)  C2H5OH(l) + CO2(g) NH3(g) + CO2(g)  NH2CONH2(aq) + H2O(l) CO(g) + H2O(g)  CO2(g) + H2(g)

Calculate the standard entropy of reaction for the following reaction at 25°C. Standard molar enthalpy of glucose = 209.2 J/(mol*K) C6H12O6(s)  C2H5OH(l) + CO2(g)

Calculate the standard entropy of reaction for the following reaction at 25°C. Standard Entropy of Urea is 174 J/(mol*K) NH3(g) + CO2(g)  NH2CONH2(aq) + H2O(l)

Calculate the standard entropy of reaction for the following reaction at 25°C. CO(g) + H2O(g)  CO2(g) + H2(g)

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Entropy and Temperature Heat Heat Surroundings Entropy System 1 System 2 Entropy

Entropy and Temperature 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. ∆Hsys T ∆Ssurr =

The heat of vaporization, ∆Hvap, of carbon tetrachloride at 25°C is 43 The heat of vaporization, ∆Hvap, 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/mole•K Liquid ethanol at 25°C has an entropy of 161 J/mole•K. If the heat of vaporization, ∆Hvap, at 25°C is 42.3kJ.mole, what is the entropy of the vapor in equilibrium with the liquid at 25°C?

Gibbs Free Energy 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. N2(g) + H2(g)  NH3(g)

CaCO3(s) ↔ CaO(s) + CO2(g) Using ∆H° and ∆S° values from Appendix 3, calculate the ∆G° for the following reaction at 25°C. CaCO3(s) ↔ CaO(s) + CO2(g)

KClO3(s) ↔ KCl(aq) + O2(g) Using ∆H° and ∆S° values from Appendix 3, calculate the ∆G° for the following reaction at 25°C. KClO3(s) ↔ KCl(aq) + O2(g)

Factors Affecting the Sign of ∆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.

ΔG°rxn = ΣnG°f(products) - ΣnG°f(reactants) Gibbs Free Energy 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)

Gibbs Free Energy

Determine if the following reaction is spontaneous using Gibbs Free Energy if the reaction is conducted at 25°C. CaCO3(s)  CaO(s) + CO2(g)

Determine if the following reaction is spontaneous using Gibbs Free Energy if the reaction is conducted at 25°C. KClO3(s)  KCl(aq) + O2(g)

Determine if the following reaction is spontaneous using Gibbs Free Energy if the reaction is conducted at 25°C. NH3(g) + CO2(g)  NH2CONH2(aq) + H2O(l)

Free Energy and Chemical Equilibrium ΔG° = -RTlnK For calculating reactions where not all reactants and products are in their standard states. Where R= 8.314 J/K•mole T is temperature in Kelvin K is the equilibrium constant.

Free Energy and Chemical Equilibrium K InK ΔG° Affect at Equilibrium > 1 + - Products are favored = 1 Neither 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. CaCO3(s) ↔ CaO(s) + CO2(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. KClO3(s) ↔ KCl(aq) + O2(g)

What is the value of ΔG° at 25°C for the following reaction What is the value of ΔG° at 25°C for the following reaction? The Ksp of magnesium hydroxide is 1.8x10-11. Predict whether the products or reactants are favored at equilibrium. Mg(OH)2(s) ↔ Mg2+(aq) + OH-(aq)

What is the value of the equilibrium constant K at 25°C for the following reaction? The Ksp of lead (II) iodide is 6.5x10-9. Predict whether the products or reactants are favored at equilibrium. PbI2(s) ↔ Pb2+(aq) + I2-(aq)

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