Thermodynamics 1 st law (Cons of Energy) Deals with changes in energy Energy in chemical systems Total energy of an isolated system is constant Total energy = Potential energy + kinetic energy E p mgh E k = ½ mv 2 Energy is transferred between a system and its surroundings by heat and/or work Energy is defined as the capacity to do work Work = opposing force X distance moved Measured in Joules 1 J = 1kg x m 2 /s Thermodynamics of Chemical Reactions Heat of Reaction The quantity of energy released or absorbed as heat during a reaction Every reaction is accompanied by transfer of energy as heat Thermochemical Equation Equation that includes the heat of reaction ex: CH 4 (g) + 2O 2 (g) CO 2( g) + 2H 2 O (l) H = -890 kj Enthalpy - H The quantity of energy released or absorbed as heat during a reaction at a constant pressure H Change in enthalpy Enthalpy change H = (sum of the H f reactants) (sum of the H f products) When H is >0 the reaction is endothermic When H is <0 the reaction is exothermic Endothermic equations show heat as a reactant Exothermic equations show heat as a product 1
Identifying reactions 2H 2 O (g) 2H 2(g) +O 2(g) H = +483.6 kj/mol 2H 2 O (g) + 483.6 kj/mol 2H 2(g) +O 2(g) 2H 2(g) +O 2(g) 2H 2 O (g) H = -483.6 kj/mol 2H 2(g) +O 2(g) 2H 2 O (g) + 483.6 kj/mol How reactions proceed This shows how energy changes as a reaction proceeds from reactants to products The activated complex is formed when the reactant particles have collided. This is when the bonds between the reactant atoms are broken and the bonds between the product particles are formed. The activation energy is the energy the reactants need to form the activated complex How do the energy of the reactants and products compare? Exothermic reactions How does the energy of the reactants compare with the energy of the products? Is energy released or absorbed? 2
Writing Balanced Thermochemical Equations The coefficients represent the numbers of moles of reactants and products, never the numbers of molecules. This allows us to write these at fractions rather than whole numbers when necessary This reaction releases 241.8 kj of heat when one mole of water vapor is produced. The heat of formation of water vapor is -241.8 kj. (Why is this negative?) Since heat is released, where should it be placed in the equation? H 2 (g) + 1/2 O 2 (g) H 2 O (g) + 241.8 kj Standard Enthalpy Values H o is measured under standard conditions P = 1 atmosphere T = usually 25 o C with all species in standard states Standard state of an element is zero under standard conditions (e.g., C = graphite and O 2 = gas) Hess s Law The overall reaction enthalpy is the sum of the reaction enthalpies of the steps into which the reaction can be divided Enthalpy is dependant on state Enthalpy is dependant on state Therefore the value of H is independent of the path between initial and final states The reaction enthalpy thus, can be calculated from any sequence of reactions that can be balanced to add up to the reaction of interest 3
Consider the oxidation of solid carbon graphite to carbon dioxide C(gr) + O 2 (g) CO 2 (g) This reaction can be thought of as the outcome of 2 steps first the oxidation of carbon to carbon monoxide C(gr) + ½ O 2 (g) CO(g) H = -110.5 kj Then the oxidation of carbon monoxide to crbon dioxide CO(g) + ½ O 2 (g) CO 2 (g) H = -283.0 kj This 2 step process is an example of a reaction sequence The net outcome of the sequence is the sum of the steps C(gr) + ½ O 2 (g) CO(g) H = -110.5 kj CO(g) + ½ O 2 (g) CO 2 (g) H = -283.0 kj C(gr) + O 2 (g) CO 2 (g) H = -393.5 kj Step by Step Select one of the reactants in the overall reaction and choose a chemical reaction in which it also is a reactant Select one of the products in the overall reaction and choose a chemical reaction in which it also is a product Add these reactions and cancel species that appear on both sides of the equation Cancel unwanted species in the sum obtained by adding an equation that has the same substance(s) on the opposite side of the arrow Once the sequence is complete, combine the standard reaction enthalpies In each step we may need to reverse the equation or multiply it by a factor we must do the same to the reaction enthalpy 4
Consider the synthesis of propane (C 3 H 8 ) Use these three equations: (a) C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) H = -2220 kj (b) C(gr) + O 2 (g) CO 2 (g) H = -394 kj (c) H 2 (g) + ½O 2 (g) H 2 O(l) H =-286 kj To treat carbon as a reactant there must be 3 carbons 3C(gr) + 3O 2 (g) 3CO 2 (g) H =(3) -394 kj = -1182 kj Reverse (a) 3CO 2 (g) + 4H 2 O(l) C 3 H 8 (g) + 5O 2 (g) H = +2220 kj Add 3CO 2 (g) + 4H 2 O(l) + 3C(gr) + 3O 2 (g) C 3 H 8 (g) + 5O 2 (g) + 3CO 2 (g) H = (-1182 kj) + 2200 kj = 1038 kj Simplify by cancelling those on both sides 4H 2 O(l) + 3C(gr) C 3 H 8 (g) + 2O 2 (g) H = 1038 kj To cancel unwanted water and oxygen, multiply (c) by 4 4H 2 (g) + 2O 2 (g) 4H 2 O (l) H = 4 x (-286 kj)= -1144kJ Add 4H 2 O(l) + 3C(gr) + 4H 2 (g) + 2O 2 (g) C 3 H 8 (g) + 2O 2 (g) 4H 2 O(g) H = 1038 kj + (-1144kJ) = -106 kj Simplify 4H 2 (g) + 2O 2 (g) C 3 H 8 (g) H = -106 kj H f heat of formation The energy released or absorbed as heat when one mole of a compound is formed by the combination of its elements o\in the standard d state H f Change in heat of formation H reaction = H f products - H f reactants Natural trend is toward decreased enthalpy (negative H) 5
Standard Molar Enthalpy of Formation, ΔH o f Calculate the heat given off when one mole of B 5 H 9 reacts with excess oxygen according to the following reaction: Compound H f (kj/mol-k) B 5 H 9 (g) 73.2 2B 5 H 9 (g) + 12O 2 (g) 5B 2 O 3 (g) + 9H 2 O(g) Note: The heat of formation of any element under standard conditions is zero. B 2 O 3 (g) -1272.77 77 O 2 (g) 0 H 2 O(g) -241.82 Calculating Enthalpy Change Is this reaction endothermic or exothermic? Enthalpy Calculation #2 Problem: Given that the standard heat of formation, H f, of H 2 O and CO are - 242 kj/mol and - 111 kj/mol respectively calculate l enthalpy change for the reaction: H 2 O(g) + C(graphite) H 2 (g) + CO(g) Why aren t we given H f for C and H 2? 6
Entropy Entropy is a measure of the chaos or disorder of a system Entropy is represented by S The greater the entropy, the more disorder Nature trends toward greater disorder 2 nd law of thermodynamics S solid < S liquid < S gas Entropy Standard entropy is entropy of a system at 1.0 atm and 25 ºC S = heat transferred/absolute temperature at which transfer took place Thus unit for entropy is J/K S = S f S i Positive entropy indicates increasing disorder S > 0 Negative entropy indicates decreasing disorder S < 0 Entropy example This is an example of thermal disorder As the system is heated, the supply of energy increases the the thermal motion and thus the disorder 7
Entropy Example The decomposition of N 2 O 4 into NO 2 is also accompanied by an increase in randomness. (N 2 O 4 2NO 2 ) Whenever molecules break apart, randomness increases This is an example of an increase in positional disorder Increasing particle number Solution process When NaCl dissolves in water, the crystal breaks up, and the ions are surrounded by water molecules. 8
Salt dissolves in water to form Na + & Cl - Less entropy More entropy Is S positive or negative for the system below? Entropy of Chemical Reactions Standard entropy of reaction: S= ns (products) ns (reactants) means the sum of n = moles given by the coefficent 9
Calculating Entropy Predict the entropy change, and then calculate the standard entropy change for the following reactions at 25 C. 2 CO(g) + O 2 (g) 2 CO 2 (g) 3 O 2 (g) 2 O 3 (g) 2 NaHCO 3 (s) Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) Examples of standard molar entropies Reaction Spontaneity A reaction that does occur under specific conditions is called a spontaneous reaction. A reaction that does not occur under specific conditions is called a nonspontaneous reaction. If a reaction is spontaneous in one direction, it will be non-spontaneous in the opposite direction 10
Gibbs Free energy Gibbs free energy tells us whether a reaction is spontaneous or not spontaneous. It combines the concepts of entropy and enthalpy. All reactions have by nature enthalpy change and entropy change. All reactions occur at a certain temperature. A negative H would be a spontaneous reaction A positive S would be a spontaneous reaction G = H T S (temperature must be in Kelvin). If G < 0 The reaction is spontaneous. Any compound with a spontaneous formation reaction is thermodynamically stable If G > 0 The reaction is not spontaneous. Any compound with a non-spontaneous formation reaction is thermodynamically unstable Signs of G, S, and H Gibbs Free Energy Example Iron metal can be produced by reducing iron(iii) oxide with hydrogen: Fe 2 O 3 (s) + 3 H 2 (g) 2 Fe(s) + 3 H 2 O(g) H = +98.8 kj; S = +141.5 J/K Note: You have J and kj given above you must convert Is this reaction spontaneous at 25 C? 11
Example Problem Is the following reaction spontaneous under standard conditions? 4KClO 3 (s) 3KClO 4 (s) + KCl(s) H f (kj/mol) S (J/mol K) KClO 3-397.7 143.1 KClO 4-432.8 151.0 KCl -436.7 82.6 Solution: Calculating ΔH for the reaction Calculating H rxn 4KClO 3 (s) 3KClO 4 (s) + KCl(s) H rxn = [3 H f (KClO 4 ) + H f (KCl)] [4 H f (KClO 3 )] = [3( 432.8kJ) + ( 436.7kJ)] [4(397.7kJ)] = 144kJ Is this reaction endothermic or exothermic? Calculating ΔS for the reaction Calculating S rxn S rxn = [3S (KClO 4 )] + [S (KCl)] [4S (KClO 3 3)] = [3(151.0JK)+(82.6JK)] ( )] [4(143.1JK)] = 36.8JK Is entropy increasing or decreasing? 12
Calculating ΔG Calculating G rxn G rxn = H rxn -T S -144kJ [298K (-38.6 J/K)(1kJ/1000J)] = -133 kj G rxn < 0; therefore, reaction is spontaneous. A negative G rxn is an exergonic reaction. A positive G rxn is an endergonic reaction. 13