Thermodynamics Review

Energy Ability to do work Units– Joules (J), we will use “kJ” Can be converted to different types Energy change results from forming and breaking chemical bonds in reactions

System vs. Surroundings

Heat (q) Energy transfer between a system and the surroundings due to a temperature change Transfer is instantaneous from high----low temperature until thermal equilibrium Temperature— Measure of heat, “hot/cold”

Heat (q) continued Kinetic theory of heat Increase in heat results in Heat increase resulting in temperature change causes an increase in the average motion of particles within the system. Increase in heat results in Energy transfer Increase in both potential and kinetic energies

1st Law of Thermodynamics (Conservation of Energy) Energy cannot be created or destroyed. With physical and chemical changes, energy can be transferred or converted. Total energy = Σenergy of its components ΔU = q + w , ΔEtotal = ΔEsys + ΔEsurr = 0

Enthalpy

Thermodynamics 101 First Law of Thermodynamics Energy is conserved in a reaction (it cannot be created or destroyed)---sound familiar??? Math representation: ΔEtotal = ΔEsys + ΔEsurr = 0 Δ= “change in” ΔΕ= positive (+), energy gained by system ΔΕ= negative (-), energy lost by system Total energy = sum of the energy of each part in a chemical reaction

Enthalpy (H) Measures 2 things in a chemical reaction: Energy change Amount of work done to or by chemical reaction 2 types of chemical reactions: Exothermic—heat released to the surroundings, getting rid of heat, -ΔΗ Endothermic—heat absorbed from surroundings, bringing heat in, +ΔΗ **Enthalpy of reaction—heat from a chemical reaction which is given off or absorbed, units = kJ/mol Enthalpy of reaction Heat from a chemical reaction which is given off or absorbed At constant pressure Units = kJ/mol

Exothermic Temperature increase (--isolated system) Heat is released to surroundings (--open/closed system) q = - value Chemical  Thermal Energy

Endothermic Temperature decrease (--isolated system) All energy going into reaction, not into surroundings Heat absorbed by system, surroundings have to put energy into reaction q = + value Thermal  Chemical Energy

Methods for Calculating Enthalpy-- Stoichiometric Calculations using Balanced Chemical Equation Calorimetry (lab based method) Hess’s Law Enthalpy of Formation Bond Enthalpies **Which method is the LEAST accurate?

Calorimetry How do we find the change in energy/heat transfer that occurs in chemical reactions???

Calorimetry Experimentally “measuring” heat transfer for a chemical reaction or chemical compound Calorimeter Instrument used to determine the heat transfer of a chemical reaction Determines how much energy is in food Observing temperature change within water around a reaction container ** assume a closed system, isolated container No matter, no heat/energy lost Constant volume

Specific Heat Capacity Amount of heat required to increase the temperature of 1g of a chemical substance by 1°C Units--- J/g°K Unique to each chemical substance Al(s) = 0.901J/g°K H2O(l) = 4.18 J/g°K

q = smΔT

“Coffee Cup” calorimeter (cont.) qchemical = -qwater

Δqrxn ΔHrxn Heat gained/lost in experiment with calorimeter Heat gained/lost in terms of the balanced chemical equation

Example 2: Using the following data, determine the metal’s specific heat. Metal mass = 25.0g Water mass = 20.0g Temperature of large water sample = 95°C Initial temperature in calorimeter = 24.5°C Final temperature in calorimeter = 47.2°C Specific heat of water = 1.00 cal/g°C OR 4.184 J/g°K (KNOW!!!!) 0.380 cal/g°C

Bond Energy Energy required to make/break a chemical bond Endothermic reactions Products have more energy than reactants More energy to BREAK bonds Exothermic reactions Reactants have more energy than products More energy to FORM bonds

Bond Enthalpy Focuses on the energy/heat between products and reactants as it relates to chemical bonding Amount of energy absorbed to break a chemical bond--- amount of energy released to form a bond. Multiple chemical bonds take more energy to break and release more energy at formation Amount of energy absorbed = amount of energy released to break chemical bond to form a chemical bond

Calculating ΔHrxn. by bond enthalpies (4th method) Least accurate method ΔH = ΣBE (bonds broken) - ΣBE (bonds formed)

Example 1: Using average bond enthalpy data, calculate ΔH for the following reaction. CH4 + 2O2  CO2 + 2H2O ΔH = ? Bond Average Bond Enthalpy C-H 413 kJ/mol O=O 495 kJ/mol C-O 358 kJ/mol C=O 799 kJ/mol O-H 467 kJ/mol

Hess’ Law Enthalpy change for a chemical reaction is the same whether it occurs in multiple steps or one step ΔHrxn = ΣΔHA+B+C (sum of ΔH for each step) Allows us to break a chemical reaction down into multiple steps to calculate ΔH Add the enthalpies of the steps for the enthalpy for the overall chemical reaction

Example 1: H2O(l)  H2O (g) ΔH° = ? Based on the following: H2 + ½ O2  H2O(l) ΔH° = -285.83 kJ/mol H2 + ½ O2  H2O(g) ΔH° = -241.82 kJ/mol

Enthalpy of Formation (ΔHf°) Enthalpy for the reaction forming 1 mole of a chemical compound from its elements in a thermodynamically stable state. Elements present in “most thermodynamically stable state” 25°C°, 1atm

Example 5 Isopropyl alcohol (rubbing alcohol) undergoes a combustion reaction 2(CH3)2CHOH + 9O2  6CO2 + 8H2O ΔH° = -4011 kJ/mol Calculate the standard enthalpy of formation for isopropyl alcohol.

Example 2: Calculate the ΔH for the following reaction when 12.8 grams of hydrogen gas combine with excess chlorine gas to produce hydrochloric acid. H2 + Cl2  2HCl ΔH = -184.6 kJ

Entropy

Spontaneous vs. Nonspontaneous Spontaneous Process Occurs WITHOUT help outside of the system, natural Many are exothermic—favors energy release to create an energy reduction after a chemical reaction Ex. Rusting iron with O2 and H2O, cold coffee in a mug Some are endothermic Ex. Evaporation of water/boiling, NaCl dissolving in water

Spontaneous vs. Nonspontaneous 2) Nonspontaneous Process REQUIRES help outside system to perform chemical reaction, gets aid from environment Ex. Water cannot freeze at standard conditions (25°C, 1atm), cannot boil at 25°C **Chemical processes that are spontaneous have a nonspontaneous process in reverse **

Entropy (S) Measure of a system’s disorder Disorder is more favorable than order ΔS = S(products) - S(reactants) ΔS is (+) with increased disorder State function Only dependent on initial and final states of a reaction Ex. Evaporation, dissolving, dirty house

Thermodynamic Laws 1st Law of Thermodynamics 2nd Law of Thermodynamics Energy cannot be created or destroyed 2nd Law of Thermodynamics The entropy of the universe is always increasing. Naturally favors a disordered state

When does a system become MORE disordered from a chemical reaction When does a system become MORE disordered from a chemical reaction? (ΔS > 0) Melting Vaporization More particles present in the products than the reactants 4C3H5N3O9 (l)  6N2 (g) + 12CO2 (g) + 10H2O (g) + O2 (g) Solution formation with liquids and solids Addition of heat

3rd Law of Thermodynamics The entropy (ΔS) of a perfect crystal is 0 at a temperature of absolute zero (0°K). No particle motion at all in crystal structure All motion stops

How do we determine if a chemical reaction is spontaneous? Change in entropy (ΔS) Gibbs Free Energy (ΔG)

Change in entropy (ΔS) For a chemical reaction to be spontaneous (ΔST > 0), there MUST be an increase in system’s entropy (Δssys> 0) and the reaction MUST be exothermic (Δssurr > 0). Exothermic reactions are favored, NOT endothermic reactions. Exothermic (ΔH < 0, ΔS > 0) Endothermic (ΔH > 0, ΔS < 0) ΔST = Δssys + Δssurr If ΔST > 0, then the chemical reaction is spontaneous

Example 1: Will entropy increase or decrease for the following? N2 (g) + 3H2 (g)  2NH3 (g) 2KClO3 (s)  2KCl (s) + 3O2 (g) CO(g) + H2O(g)  CO2 (g) + H2 (g) C12H22O11 (s)  C12H22O11 Decrease in entropy Increase in entropy Cannot tell

How do we calculate the entropy change (ΔS) in a chemical reaction? Same method as using the enthalpies of formation to calculate ΔH and use the same table. aA + bB  cC + dD ΔS° =[c (ΔS°C) + d(ΔS°D)] - [a (ΔS°A) + b (ΔS°B)]

Example 2: Calculate ΔS° for the following reaction at 25°C…. 4HCl(g) + O2 (g)  2Cl2 (g) + 2H2O (g) ΔS°= -128.8 J/K, entropy decreases

Free Energy and Equilibrium

ΔG = ΔG° + RTlnQ At equilibrium, ΔG = 0, so reaction quotient (Q) = equilibrium constant (K) At equilibrium ΔG° = - Rtlnk Enables the reaction’s equilibrium constant (K) to be calculated from the change in free energy (ΔG°)

What is the relationship between free energy(ΔG) and K? The magnitude of ΔG° indicates how far the chemical reaction in its standard state is from equilibrium. ΔG° = 0 , equilibrium ΔG°= large value, far from equilibrium ΔG° = small value, close to equilibrium The sign (+, - ) indicates which direction the reaction needs to shift to achieve equilibrium Positive (+) -------- shift to left, no reaction Negative (-) -------- shift to right, reaction goes to completion

Gibbs Free Energy

Change in Gibbs Free Energy (ΔG) ΔG = ΔH – TΔS Relates enthalpy and entropy to determine which has more importance in determining whether a reaction is spontaneous Combines energy transfer as heat (ΔH) and energy released to contribute to disorder (ΔS)

Change in Gibbs Free Energy (ΔG) ΔG = ΔH – TΔS ΔG < 0 , spontaneous reaction Energy available to do work ΔG > 0, nonspontaneous reaction Energy deficiency, no leftover energy and not enough energy for reaction

How can we apply the Gibbs equation to determine spontaneity of reaction? ΔG = ΔH – TΔS ΔH ΔS ΔG Result - + Spontaneous (all temperatures) Nonspontaneous (low temperatures)

Two Paths to Calculating ΔG ΔG = ΔH – TΔS Determine ΔH. What methods can we use? Determine ΔS. Then calculate ΔG

Two Paths to Calculating ΔG 2) Use Standard Free Energy of Formation (ΔGf °) values to determine ΔG Standard Free Energy of Formation (ΔGf °) --- ΔG° for the formation of 1 mole of a chemical compound in its standard state. ΔGf ° for element formation in their most stable state = 0. aA + bB  cC + dD ΔG° =[c (ΔGf°)C + d(ΔGf°)D] - [a (ΔGf°)A + b (ΔGf°)B ]

Example 2: Find ΔG for a chemical reaction given ΔH = -218 kJ and ΔS = -765 J/K at 32°C. B) At what temperature does this reaction become spontaneous? Assume only temperature changes.

Example 3: Calculate ΔG°rxn under standard conditions for the following reaction using ΔGf° values. Fe2O3 (s) + 2Al(s)  2Fe(s) + Al2O3 (s)

A spontaneous reaction is NOT necessarily fast!!!! Reaction rate involves kinetics ! !