THERMOCHEMISTRY

Chapter 6 Thermochemistry

Definitions #1 Energy: The capacity to do work or produce heat Potential Energy: Energy due to position or composition Kinetic Energy: Energy due to the motion of the object

Radiant energy comes from the sun and is earth’s primary energy source Some types Energy Radiant energy comes from the sun and is earth’s primary energy source Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy available by virtue of an object’s position Which are Potential? Kinetic? 6.1

Definitions #2 Law of Conservation of Energy: Energy can neither be created nor destroyed, but can be converted between forms The First Law of Thermodynamics: The total energy content of the universe is constant

First law of thermodynamics DEsystem + DEsurroundings = 0 or DEsystem = -DEsurroundings C3H8 + 5O2 3CO2 + 4H2O Exothermic chemical reaction! Chemical energy lost by combustion = Energy gained by the surroundings system surroundings 6.3

State Functions depend ONLY on the present state of the system ENERGY IS A STATE FUNCTION A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane! WORK IS NOT A STATE FUNCTION WHY NOT???

, pressure, volume, temperature Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature DE = Efinal - Einitial DP = Pfinal - Pinitial DV = Vfinal - Vinitial DT = Tfinal - Tinitial Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. 6.3

E = q + w E = change in internal energy of a system q = heat flowing into or out of the system -q if energy is leaving to the surroundings +q if energy is entering from the surroundings w = work done by, or on, the system -w if work is done by the system on the surroundings +w if work is done on the system by the surroundings

Work, Pressure, and Volume Expansion Compression +V (increase) -V (decrease) -w results +w results Esystem decreases Esystem increases Work has been done by the system on the surroundings Work has been done on the system by the surroundings

Work is not a state function! Work Done On the System w = F x d w = -P DV DV > 0 -PDV < 0 wsys < 0 P x V = x d3 = F x d = w F d2 Work is not a state function! Dw = wfinal - winitial initial final 6.3

W = -0 atm x 3.8 L = 0 L•atm = 0 joules A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands from a vacuum (0.0 atm) to a constant pressure of 3.7 atm? w = -P DV (a) DV = 5.4 L – 1.6 L = 3.8 L P = 0 atm W = -0 atm x 3.8 L = 0 L•atm = 0 joules (b) DV = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm w = -3.7 atm x 3.8 L = -14.1 L•atm w = -14.1 L•atm x 101.3 J 1L•atm = -1430 J 6.3

Enthalpy and the First Law of Thermodynamics DE = q + w q = DH and w = -PDV At constant pressure: DE = DH - PDV DH = DE + PDV 6.4

DH = H (products) – H (reactants) Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0 6.4

Energy Change in Chemical Processes Endothermic: Reactions in which energy flows into the system as the reaction proceeds. + qsystem - qsurroundings Exothermic: Reactions in which energy flows out of the system as the reaction proceeds. - qsystem + qsurroundings

2H2 (g) + O2 (g) 2H2O (l) + energy Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) 2H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + H2O (s) H2O (l) energy + 2HgO (s) 2Hg (l) + O2 (g) 6.2

Endothermic Reactions

Exothermic Reactions

Review so far… Energy is…. 1st law of thermodynamics Potential verses Kinetic Energy State Function Enthalpy (q, heat, …) Work is… E = q + w and w = -P DV Endothermic Exothermic

Calorimetry The amount of heat absorbed or released during a physical or chemical change can be measured… …usually by the change in temperature of a known quantity of water 1 calorie is the heat required to raise the temperature of 1 gram of water by 1 C 1 BTU is the heat required to raise the temperature of 1 pound of water by 1 F

Heat (q) absorbed or released: The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = m x s Heat (q) absorbed or released: q = m x s x Dt q = C x Dt Dt = tfinal - tinitial 6.5

1 joule = 4.184 calories The Joule The unit of heat used in modern thermochemistry is the Joule 1 joule = 4.184 calories

A Bomb Calorimeter (commonly used by Oak Ridge HS)

A Cheaper Calorimeter (commonly used by WSHS)

Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Substance Specific Heat (J/g·K) Water (liquid) 4.184 Ethanol (liquid) 2.44 Water (solid) 2.06 Water (vapor) 1.87 Aluminum (solid) 0.897 Carbon (graphite,solid) 0.709 Iron (solid) 0.449 Copper (solid) 0.385 Mercury (liquid) 0.140 Lead (solid) 0.129 Gold (solid)

Calculations Involving Specific Heat OR or  q = Cp * m * T s = Specific Heat Capacity q = Heat lost or gained T = Temperature change m = mass in grams

Chemistry in Action: Fuel Values of Foods and Other Substances C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J Substance DHcombustion (kJ/g) Apple -2 Beef -8 Beer -1.5 Gasoline -34

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero. DH0 (O2) = 0 f DH0 (C, graphite) = 0 f DH0 (O3) = 142 kJ/mol f DH0 (C, diamond) = 1.90 kJ/mol f 6.6

Hess’s Law “In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

Hess’s Law

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ 2H2 + O2  2 H2O -571.66 kJ Step #3: Multiply reaction #3 by 2

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ 2H2 + O2  2 H2O -571.66 kJ CH4 + 2O2  CO2 + 2H2O -890.36 kJ Step #4: Sum up reaction and H

Calculation of Heat of Reaction Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O Hrxn =  Hf(products) -   Hf(reactants)     Substance Hf   CH4 -74.80 kJ O2 0 kJ CO2 -393.50 kJ H2O -285.83 kJ Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ] Hrxn = -890.36 kJ

The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD DH0 rxn dDH0 (D) f cDH0 (C) = [ + ] - bDH0 (B) aDH0 (A) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) 6.6

C (graphite) + 1/2O2 (g) CO (g) CO (g) + 1/2O2 (g) CO2 (g) C (graphite) + O2 (g) CO2 (g) 6.6

Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ rxn S(rhombic) + O2 (g) SO2 (g) DH0 = -296.1 kJ rxn CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS2 C(graphite) + 2S(rhombic) CS2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ 2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -296.1x2 kJ rxn + CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJ rxn C(graphite) + 2S(rhombic) CS2 (l) DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 6.6

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - DH0 rxn 6DH0 (H2O) f 12DH0 (CO2) = [ + ] - 2DH0 (C6H6) DH0 rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ -5946 kJ 2 mol = - 2973 kJ/mol C6H6 6.6

Chemistry in Action: Bombardier Beetle Defense C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2H2O (l) DH0 = ? C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g) DH0 = 177 kJ/mol H2O2 (aq) H2O (l) + ½O2 (g) DH0 = -94.6 kJ/mol H2 (g) + ½ O2 (g) H2O (l) DH0 = -286 kJ/mol DH0 = 177 - 94.6 – 286 = -204 kJ/mol Exothermic!

DHsoln = Hsoln - Hcomponents The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? 6.7

The Solution Process for NaCl DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol 6.7