Entropy, Enthalpy, and Gibb’s Free Energy ENERGY OF REACTIONS Entropy, Enthalpy, and Gibb’s Free Energy

ENERGY SO FAR Review: Energy is the ability to do work or produce heat There is potential and kinetic energy Energy can neither be created or destroyed Every compound needs energy to increase temperature or to change from one state of matter to another

ENERGY SO FAR Units of energy: Joule (J) Calorie (cal) 1 cal = 4.184J 1kJ = 1000J 1kcal = 1000cal

REVIEW 255 cal = _____________ J 4200 J = ______________cal Convert the following: 255 cal = _____________ J 4200 J = ______________cal 55kcal = ______________ J 6325 J = ______________ kJ 3.85 kJ = _______________kcal

OTHER ASPECTS OF ENERGY Energy is also an important component of chemical reactions Example: 4Fe(s) + 3O2(g)  2Fe2O3(s) + 1625kJ In this example, you combine iron and oxygen to produce iron (III) oxide. This reaction also produces 1625kJ of energy/heat.

ENERGY AND REACTIONS To better explain the energy changes in reactions, chemists have come up with: Enthalpy (H): the heat content of substances under constant pressure For a chemical reaction, we describe the change in enthalpy. This is called: Enthalpy (or heat) of reaction (δHrxn): the change in heat or energy in a chemical reaction

HOW DO WE MEASURE δHrxn δHrxn = Hproducts - Hreactants To measure the heat produced or used by a reaction, scientists again use calorimeters. (REVIEW: What is a calorimeter?) To calculate, we have the following formula: δHrxn = Hproducts - Hreactants

PREVIOUS EXAMPLE Previously, we looked at the following: 4Fe(s) + 3O2(g)  2Fe2O3(s) + 1625kJ According to this equation, we produced (or lost) 1625kJ of energy to the environment (you feel this as hot) - EXOTHERMIC Therefore Hreactants > Hproducts

WHAT THIS MEANS For the reaction: Whenever Hreactants > Hproducts, the convention is to give a negative sign to the δHrxn Therefore, exothermic reactions are always a negative “-” δHrxn On the right side of the arrow

ENDOTHERMIC Since we know that exothermic reactions have a negative heat of reaction: WHAT IS THE SIGN FOR AN ENDOTHERMIC REACTION?

WHAT THIS MEANS Endothermic reactions are always a positive “+” δHrxn On the left side of the arrow You have to add energy to the reactant side to equal the products Therefore we always have to add energy to an endothermic reaction: Example: 27kJ + NH4NO3 (s)  NH4+(aq) + NO3-(aq)

CALCULATING HEATS OF REACTION Chemists have measured different heats of reaction for combustion reactions. In each case, they do this under a condition of standard pressure and temperature. They call these heats of combustion: δHcomb

What is the general format for a combustion reaction? REVIEW What is the general format for a combustion reaction?

COMBUSTION REACTION Molecule + O2  CO2 + H2O Therefore, these are values for the combustion of 1 MOLE of the molecule Example: We will look at the combustion of 1 mole of glucose (sugar) C6H12O6 + 6O2  6CO2 + 6H2O + 2808kJ δHcomb = -2808kJ/mole

DIMENSIONAL ANALYSIS How much energy is produced when you combust 2.5 moles of glucose 2.5 moles | -2808kJ = -7020kJ | 1mole

CALCULATING HEAT OF COMBUSTION Example: You start with 3.55x103 g of glucose (C6H12O6). How much energy is released when glucose goes through a complete combustion reaction? STEPS: Convert to moles Use the δHcomb for glucose to calculate energy δHcomb = -2808kJ/mole

ANSWER δHcomb = -2808kJ/mole Molar mass of glucose: 180g/mole 3.55x103 g | 1 mole | -2808kJ | 180g | 1 mole -5.54x104 kJ

TRY THESE The heat of combustion for octane (C8H18) is -5471kJ/mole. If you start with 1550g of octane, how much energy is released? Sucrose (C12H22O11), or table sugar, has a heat of combustion of -5644kJ/mole. If you add 2.5g of sucrose to your cereal, how much energy will be added to your cereal?

ANSWERS -7.44X104 kJ -41 kJ

TRY THIS You have a cup filled with 125mL of glucose (C6H12O6). If the density of glucose is 1.54g/mL, how many moles of glucose do you have? If the heat of combustion for glucose is -2805 kJ/mole, what is the heat produced from the cup of glucose?

ANSWER 1.07 moles glucose -3.00 x103 kJ

TRY THIS A combustion reaction with octane (C8H18) releases a total of -5.55x104 kJ of energy. If the heat of combustion for octane is -250kJ/mole, how many grams of octane did you start with?

ANSWER 2.53X104 g

HEATS OF REACTION As we said earlier, most heats of reaction or heats of combustion are measured using a calorimeter. Sometimes (because some reactions are toxic or unstable), we cannot use a calorimeter and must find an alternative way to measure heats of reaction.

HEATS OF REACTION Hess’s law: If you add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction

EXAMPLE: FIND THE HEAT OF REACTION 2S + 3O2  2SO3 This reaction can occurs in 2 steps: S + O2  SO2 ; δH = -297kJ 2SO3  2SO2 + O2 ; δH = 198kJ How do these reactions go together to form the final reaction?

2S + 3O2  2SO3 Step 1: reverse the second reaction because SO3 is a product. Therefore, you have to change the sign for the heat of reaction S + O2  SO2 ; δH = -297kJ 2SO2 + O2  2SO3 ; δH =-198kJ NOTE: Now the second reaction is exothermic

2S + 3O2  2SO3 Step 2: there are 2 moles of S in the first reaction. This means we have to double everything in the first reaction (including the heat of reaction) 2(S + O2  SO2 ; δH = -297kJ) 2S + 2O2  2SO2; δH = -594kJ 2SO2 + O2  2SO3 ; δH =-198kJ

2S + 3O2  2SO3 Step 3: you add the two reactions together to get the final product 2S + 2O2  2SO2; δH = -594kJ 2SO2 + O2  2SO3 ; δH = -198kJ__ 2S + 3O2  2SO3; δH =-792kJ__ NOTE: If a compound is on opposite sides of a reaction, they cancel out

SUMMARY You need to have known chemical reactions that you can combine to form the final chemical reaction You need the δH for each reaction If you need to reverse a reaction, you must also change the sign of δH If you need to multiply a reaction by a number, you must also multiply the δH When all reactions are completed, you then add them together to get the final δH

TRY THE FOLLOWING Combine the two reactions to find the heat of reaction when hydrogen peroxide (H2O2) breaks apart to form water and oxygen) 2H2O2  2H2O + O2; ΔH = ? 2H2 + O2  2H2O; δH = -572kJ H2 + O2  H2O2; δH = -188kJ

2H2O2  2H2O + O2 Step 1: We need to reverse the second reaction 2H2 + O2  2H2O; δH = -572kJ H2O2  H2 + O2; δH = 188kJ Step 2: You need 2 moles of H2O2 2H2O2  2H2 + 2O2; δH = 376kJ

2H2O2  2H2O + O2 Step 3: Combine the reactions 2H2 + O2  2H2O; δH = -572kJ 2H2O2  2H2 + 2O2; δH = 376kJ 2H2O2  2H2O + O2 δH = -196kJ NOTE: When you cancel out reactants and products, 2H2 cancels out 2H2, but only 1O2 is cancelled out in the products

TRY THIS 2 Ca + 2C + 3O2  2CaCO3 Ca + 2C  CaC2; dHrxn = -62.8kJ CO2  C + O2; dHrxn = -393kJ 2CaCO3 + 2CO2  2CaC2 + 5 O2; dHrxn = +1538kJ

OTHER PRACTICE (DID THIS ONE) 4 NH3 + 5 O2  4 NO + 6 H2O N2 + O2  2 NO ∆H = -180.5 kJ N2 + 3 H2  2 NH3 ∆H = -91.8 kJ 2 H2 + O2  2 H2O ∆H = -483.6 kJ

CH4 + NH3  HCN + 3 H2 TRY THIS (DID THIS ONE) N2 + 3 H2  2 NH3; ∆H = -91.8 kJ C + 2 H2  CH4; ∆H = -74.9 kJ H2 + 2 C + N2  2 HCN; ∆H = +270.3 kJ

TRY THIS (DID THIS ONE) N2H4 + H2 → 2NH3 N2H4 + CH4O  CH2O + N2 + 3H2 ΔH = -37 kJ N2 + 3H2 → 2NH3 ΔH = -46 kJ CH4O → CH2O + H2 ΔH = -65 kJ

TRY THIS (DID THIS ONE) 2C + 2H2O  CH4 + CO2 C + H2O  CO + H2; ∆H = 131.3kJ CO + H2O  CO2 + H2; ∆H = -41.2kJ CH4 + H2O  3H2 + CO; ∆H = 206.1kJ

TRY THIS (DID THIS ONE) 5C + 6H2  C5H12 C + O2  CO2 ΔH = -393.5kJ 2H2 + O2  2H2O ΔH = -571.8kJ C5H12 +8O2  5CO2 + 6H2O ΔH = -3536kJ

PREDICTING SPONTANEOUS CHEMICAL REACTIONS Energy is an essential ingredient for a chemical reaction Not all chemical reactions happen spontaneously To determine if a reaction occurs spontaneously, we need to look at another concept: ENTROPY

ENTROPY Entropy (S): a measure of the number of possible ways that the energy of a system can be distributed In other words, entropy is the tendency for molecules to spread out as far as possible from each other Since molecules spreading out is dependent on temperature, the unit for entropy is (J/K) – Joules/Kelvin

KELVIN TEMPERATURE Before we go any further, we need to review Kelvins: Kelvin (K) A temperature scale based on absolute 0 (the coldest possible temperature) The Celsius temperature converts to Kelvin: K = °C + 273

THERMODYNAMICS Thermodynamics: The laws of thermodynamics, in principle, describe the specifics for the transport of heat and work. The only law we are interested in is the second law of thermodynamics

SECOND LAW OF THERMODYNAMICS Second Law of Thermodynamics: spontaneous processes always proceed in such a way that the entropy of the universe increases.

PREDICTING CHANGES IN ENTROPY Reminder: ΔHrxn is the heat of reaction (enthalpy) We have a similar symbol for entropy Δ Srxn = the entropy in a reaction If the entropy increases during a reaction, then Δ Srxn is positive If the entropy decreases during a reaction, then Δ Srxn is negative

PREDICTING CHANGES IN ENTROPY Changing states of matter: entropy changes when you go between solid, liquid or gas H2O(l)  H2O (g); δSrxn > 0 Dissolving a gas in a solid or liquid always decreases entropy When you increase the number of gas particles in a reaction, entropy tends to increase Zn(s) + HCl(aq)  ZnCl2(aq) + H2(g); δSrxn > 0

PREDICTING CHANGES IN ENTROPY With some exceptions, entropy increases when a solid dissolves in a liquid The solid tends to break apart Random motion of particles increases as the temperature increases As you increase temperature, the entropy increases

PREDICT THE FOLLOWING Try to determine whether entropy increases or decreases for the following reactions: CF(g) + F2(g)  CF3 (g) NH3 (g)  NH3 (aq) C10H8 (s)  C10H8 (l) H2O (l)  H2 (g) + O2 (g)

ANSWERS Entropy decreases: You go from 2 molecules of gas to 1 Entropy decreases: You are dissolving a gas into a liquid Entropy increases: You have gone from a solid to a liquid Entropy increases: You are going from no gas to 2 molecules of gas

PUTTING IT ALL TOGETHER Generally, exothermic reactions are spontaneous (you do not need to add energy to make them work) Generally, spontaneous reactions will increase in entropy Yet, there is a way to know for sure if a reaction will occur spontaneously or not

GIBB’S FREE ENERGY William Gibbs (an American) combined enthalpy and entropy to determine if a reaction was spontaneous He called his equation Gibbs Free Energy equation Free energy: the energy that is available to do work (δG) The energy that holds a molecule together is called BOND ENERGY Reactions are either making or breaking these bonds by releasing or adding energy

δGrxn = δHrxn – TδSrxn GIBB’S FREE ENERGY Gibbs Free Energy Equation: δG (kJ): represents free energy δH (kJ): represents change in enthalpy T (K): temperature in kelvins δS (J/K): represents change in entropy

EXAMPLE δHrxn = -91.8kJ δSrxn = -197J/K T = 25°C N2(g) + 3H2(g)  2NH3(g) δHrxn = -91.8kJ δSrxn = -197J/K T = 25°C Why does the entropy decrease in this reaction? What is the Gibb’s free energy? Is this reaction spontaneous?

STEP 1: WRITE WHAT YOU KNOW δG = ? δH = -91.8kJ T = 25°C +273 = 298K δS = -197J/K = -0.197kJ/K NOTE: Temperature must be in Kelvin and entropy is converted to kJ so that it matches the units for δG and δH.

STEP 2: PUT THE NUMBERS IN THE FORMULA δGrxn = δHrxn – TδSrxn δGrxn = -91.8kJ – (298K)(-0.197kJ/K) δGrxn = -91.8kJ – (-58.7kJ) δGrxn = -91.8kJ + 58.7kJ δGrxn = -33.1kJ

STEP 3: DETERMINE IF IT IS SPONTANEOUS The reason why the entropy of this reaction decreased is that you are going from 4 moles of gas to 2 moles of gas. The amount of gas is reduced and lowers the entropy. δGrxn = -33.1kJ Since the Gibb’s free energy is negative, this is a spontaneous reaction.

TRY THIS δHrxn = 145 kJ δSrxn = 322 J/K T = 109°C A Chemical reaction has the following information: δHrxn = 145 kJ δSrxn = 322 J/K T = 109°C What is the Gibb’s free energy? Is this reaction spontaneous?

ANSWER δG = +22kJ Since the Gibb’s free energy is positive, this reaction is NOT spontaneous Challenge: At what temperature would this reaction become spontaneous?

ANSWER We need to find what temperature δG<0 δHrxn = 145 kJ δSrxn = 322 J/K T = 109°C We need to find what temperature δG<0 0kJ = 145kJ – T(0.322kJ/K) T(0.322kJ/K) = 145kJ T = 450K the reaction has to be at least 450K to be spontaneous

TRY THIS ΔH = 55kJ ΔS = 225J/K At what temperature does this reaction become spontaneous?

ANSWER REMEMBER: Set the ΔG = 0 0 = 55,000J – T(225J/K) T(225J/K) = 55,000J 225J/K 225J/K T = 244K

WHEN ARE REACTIONS SPONTANEOUS? Reactions are spontaneous when: δHrxn < 0; δSrxn > 0 δHrxn < 0; δSrxn < 0 (spontaneous at lower temperatures only) δHrxn > 0; δSrxn > 0 (spontaneous at higher temperatures only Reactions are not spontaneous when: δHrxn > 0; δSrxn < 0

SPONTANEOUS? Like entropy or enthalpy, Gibb’s free energy can be positive or negative If positive (+): if the Gibb’s free energy is positive, the reaction is not spontaneous (you would need to add energy to make the reaction work) - ENDERGONIC If negative (-): if the Gibb’s free energy is negative, the reaction is spontaneous -EXERGONIC

ENERGETICS The graphs below show how energy is used as a chemical reaction proceeds

ENERGETICS AND REACTIONS Even if spontaneous, reactions don’t just occur They need energy to start (Energy of activation) Many times this energy cost is very high and needs to be reduced to make the reaction work A catalyst is a substance that lowers the amount of energy needed to start a chemical reaction SEE GRAPH

CATALYST

HOW CATALYSTS WORK (END HERE) Action of catalysts When reactions occur it is because molecules are smashing into each other Yet, they are like puzzle pieces and have to hit each other at an exact angle – EFFECTIVE COLLISION This takes energy to move the molecules around enough to hit at this angle A catalyst is a substance that “holds” molecules together so that it is easier to put them together

REACTION RATES Just because a reaction is spontaneous, it doesn’t mean that you’ll be able to see anything happening If a reaction is too slow, you won’t be able to see change for a long time A measure of how fast a reaction goes is called the REACTION RATE There are several factors that affect how fast a reaction goes and WHICH DIRECTION it goes

HOW DO REACTIONS WORK REMEMBER: When we talked about reactions we mentioned that a reaction can go either direction: Fe +O2  Fe2O3 The reaction can either go to the left or the right Actually, it is continually going left AND right

HOW DO REACTIONS WORK Fe +O2  Fe2O3 In this reaction iron and oxygen are making iron (III) oxide, but iron (III) oxide is also breaking apart at the SAME TIME The speed that this happens and which things are made or broken apart depends on many things The fact that they are constantly doing this is called DYNAMIC EQUILIBRIUM or CHEMICAL EQUILIBRIUM

CHEMICAL/DNYAMIC EQUILIBRIUM Reactions go to a certain ending that gives you a certain concentration of product At that point you won’t get a higher concentration, but the chemicals are forming or breaking apart at a rate that makes sure that the concentration is the same The reaction is still proceeding, but the concentrations are stay the same

HOW DO YOU KNOW WHICH WAY A REACTION GOES Le Chatelier was a scientist that studies reaction rates and which direction reactions go The things he found are called Le Chatelier’s Principles

EFFECTS OF CONCENTRATION PCl5  PCl3 + Cl2 If you add more reactant (PCl5), the reaction shifts to the right If you add more product (PCl3 OR Cl2), the reaction shifts to the left SUMMARY: If you add a higher concentration to one side of the reaction it shifts the reaction to make more of the other side

PRESSURE/VOLUME N2(g) + H2(g)  NH3(g) If you increase the pressure OR decrease the volume, the reaction shifts to the side with the fewest molecules (makes NH3) If you decrease the pressure OR increase the volume, the reaction shifts to the side with the most molecules (N2 and H2)

4Fe(s) + 3O2(g)  2Fe2O3(s) + heat TEMPERATURE 4Fe(s) + 3O2(g)  2Fe2O3(s) + heat EXOTHERMIC REACTION: If the temperature is increased, the reaction will shift to the left (Fe and O2) If the temperature is decreased, the reaction will shift to the right (Fe2O3) THE OPPOSITE IS TRUE OF ENDOTHERMIC REACTIONS

EFFECTS OF CATLYST A catalyst will not shift the reaction either way It will just make sure that the reaction reaches equilibrium much faster Increased reaction rate