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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 OTHER ASPECTS OF ENERGY Energy is also an important component of chemical reactions Example: 4Fe(s) In + 3O2(g) 2Fe2O3(s) + 1625kJ 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 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 You would have to add 1625kJ of energy to the products to equal the reactants WHAT THIS MEANS For the reaction: The reactants have 1625kJ more than the products – you have to add energy to the product side δHrxn = Hreactants - Hproducts δHrxn = -1625kJ Therefore, exothermic reactions are always a negative “-” δHrxn 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 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) δHrxn = Hproducts - Hreactants δHrxn = +27kJ 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 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 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 ANSWER Molar mass of glucose: 180g/mole δHcomb = -2808kJ/mole 3.55x103 g | 1 mole | -2808kJ | 180g | 1 mole -5.54x104 kJ TRY THESE 1. 2. 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 4 1.-7.44X10 2.-41 kJ 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 -3.00 x10 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 4 2.53X10 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 ; 2SO3 2SO2 + O2 ; How δH = -297kJ δH = 198kJ 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 ; 2SO2 + O2 2SO3 ; NOTE: δH = -297kJ δH =-198kJ 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; 2SO2 + O2 2SO3 ; δH = -594kJ δH =-198kJ 2S + 3O2 2SO3 Step 3: you add the two reactions together to get the final product 2S + 2O2 2SO2; 2SO2 + O2 2SO3 ; δH = -594kJ δH = -198kJ__ 2S + 3O2 2SO3; δH =-792kJ__ NOTE: If a compound is on opposite sides of a reaction, they cancel out SUMMARY 1. 2. 3. 4. 5. 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; 2H2 + O2 2H2O; H2 + O2 H2O2; ΔH = ? δH = -572kJ δH = -188kJ 2H2O2 2H2O + O2 Step 1: We need to reverse the second reaction 2H2 + O2 2H2O; H2O2 H2 + O2; Step δH = -572kJ δH = 188kJ 2: You need 2 moles of H2O2 2H2 + O2 2H2O; 2H2O2 2H2 + 2O2; δH = -572kJ δH = 376kJ 2H2O2 2H2O + O2 Step 3: Combine the reactions 2H2 + O2 2H2O; 2H2O2 2H2 + 2O2; δH = -572kJ 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 1. 2. 3. Ca + 2C CaC2; dHrxn = -62.8kJ CO2 C + O2; dHrxn = -393kJ 2CaCO3 + 2CO2 2CaC2 + 5 O2; dHrxn = +1538kJ OTHER PRACTICE 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 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 = Hreactants – Hproducts We have a similar equation for entropy δSrxn = Sproducts - Sreactants If the entropy increases during a reaction, then Sproducts > Sreactants and δSrxn is positive If the entropy decreases during a reaction, then Sproducts < Sreactants and δSrxn 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: 1. 2. 3. 4. CF(g) + F2(g) CF3 (g) NH3 (g) NH3 (aq) C10H8 (s) C10H8 (l) H2O (l) H2 (g) + O2 (g) ANSWERS 1. 2. 3. 4. 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 TRY3THIS CH4 + NH HCN + 3 H2 N2 C H2 + 3 H2 2 NH3; + 2 H2 CH4; ∆H = -91.8 kJ ∆H = -74.9 kJ + 2 C + N2 2 HCN; ∆H = +270.3 kJ TRY THIS N2H4 + H2 → 2NH3 N2H4 + CH4O CH2O + N2 + 3H2 kJ N2 + 3H2 → 2NH3 CH4O → CH2O + H2 ΔH = -37 ΔH = -46 kJ ΔH = -65 kJ TRY THIS 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 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 In 1878, 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) GIBB’S FREE ENERGY Gibbs Free Energy Equation: δGrxn δG = δHrxn – TδSrxn (kJ): represents free energy δH (kJ): represents change in enthalpy T (K): temperature in kelvins δS (J/K): represents change in entropy 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) If negative (-): if the Gibb’s free energy is negative, the reaction is spontaneous EXAMPLE N2(g) + 3H2(g) 2NH3(g) 1. 2. 3. 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 1. 2. 3. 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 A Chemical reaction has the following information: 1. 2. δHrxn = 145 kJ δSrxn = 322 J/K T = 109°C What is the Gibb’s free energy? Is this reaction spontaneous? ANSWER 1. 2. 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 δHrxn = 145 kJ δSrxn = 322 J/K T = 109°C 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 WHEN ARE REACTIONS SPONTANEOUS? Reactions δHrxn < 0; δSrxn > 0 δHrxn < 0; δSrxn < 0 (spontaneous at lower temperatures only) δHrxn > 0; δSrxn > 0 (spontaneous at higher temperatures only Reactions are spontaneous when: are not spontaneous when: δHrxn > 0; δSrxn < 0