Energy & Enthalpy Thermodynamics (rev. 0910) Chapter 5 Energy & Enthalpy Thermodynamics (rev. 0910)

Definition Thermodynamics- is the study of energy transformations.

Chemical Reactions Chemical reactions involve not just the conversion of reactants into products, but also involve an energy change in the form of heat—heat released as the result of a reaction, or heat absorbed as a reaction proceeds. Energy changes accompany all chemical reactions and are due to rearranging of chemical bonding.

Making Bonds Addition of energy is always a requirement for the breaking of bonds but the breaking of bonds in and of itself does not release energy. Energy release occurs when new bonds are formed.

Bond Energy If more energy is released when new bonds form than was required to break existing bonds, then the difference will result in an overall release of energy. If, on the other hand, more energy is required to break existing bonds than is released when new bonds form, the difference will result overall in energy being absorbed.

Overall Reaction Energy Whether or not an overall reaction releases or requires energy depends upon the final balance between the breaking and forming of chemical bonds.

Energy is... the ability to do work or produce heat. conserved. made of heat and work. a state function. (Energy is a property that is determined by specifying the condition or “state” (e.g., temperature, pressure, etc.) of a system or substance.) independent of the path, or how you get from point A to B.

Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. 6.7

Energy While the total internal energy of a system (E) cannot be determined, changes in internal energy (E) can be determined. The change in internal energy will be the amount of energy exchanged between a system and its surroundings during a physical or chemical change. Δ E = E final - E initial

Definitions Work is a force acting over a distance. Heat is energy transferred between objects because of temperature difference. (Heat is not a property of a system or substance and is not a state function. Heat is a process—the transfer of energy from a warm to a cold object.)

System vs Surroundings The Universe is divided into two halves. system and the surroundings. In a chemistry setting, a system includes all substances undergoing a physical or chemical change. The surroundings would include everything else that is not part of the system.

Heat Most commonly, energy is exchanged between a system and its surroundings in the form of heat. Heat will be transferred between objects at different temperatures. Thermochemistry is the study of thermal energy changes.

Definition of Heat Heat energy (or just heat) is a form of energy which transfers among particles in a substance (or system) by means of kinetic energy of those particle. In other words, under kinetic theory, the heat is transfered by particles bouncing into each other. http://physics.about.com/od/glossary/g/heat.htm

Definition of Temperature Temperature is a measurement of the average kinetic energy of the molecules in an object or system and can be measured with a thermometer or a calorimeter. It is a means of determining the internal energy contained within the system.

Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. Temperature = Thermal Energy 400C 900C greater thermal energy 6.2

Heat vs Temperature Note that temperature is different from heat, though the two concepts are linked. Temperature is a measure of the internal energy of the system, while heat is a measure of how energy is transferred from one system (or body) to another. The greater the heat absorbed by a material, the more rapidly the atoms within the material begin to move, and thus the greater the rise in temperature. http://physics.about.com/od/glossary/g/temperature.htm

Exo vs Endo Exothermic reactions release energy to the surroundings. Endothermic reactions absorb energy from the surroundings.

Heat Potential energy

Heat Potential energy

Three Parts Every energy measurement has three parts: A unit ( Joules of calories). A number. a sign to tell direction. negative - exothermic positive- endothermic

Surroundings System Energy DE <0

Surroundings System Energy DE >0

Same rules for heat and work Heat given off is “negative”. Heat absorbed is “positive”. Work done by the system on the surroundings is negative. Work done on the system by the surroundings is positive.

First Law of Thermodynamics The energy of the universe is constant. It is also called the: Law of conservation of energy. q = heat w = work In a chemical system, the energy exchanged between a system and its surroundings can be accounted for by heat (q) and work (w). DE = q + w Take the system’s point of view to decide signs.

Thermodynamics DE = q + w DE is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -PDV when a gas expands against a constant external pressure 6.7

Enthalpy and the First Law of Thermodynamics DE = q + w At constant pressure, q = DH and w = -PDV DE = DH - PDV DH = DE + PDV 6.7

Conservation of Energy Energy exchanged between a system and its surroundings can be considered to off set one another. The same amount of energy leaving a system will enter the surroundings (or vice versa), so the total amount of energy remains constant.

Metric Units The SI (Metric System) unit for all forms of energy is the joule (J).

Heat and Work DE = q + w - q is exothermic -q = -∆H +q is endothermic -w is done “by” the system +w is done “on” the system Note: ∆H stands for enthalpy which is the heat of reaction

Practice Problem A gas absorbs 28.5 J of heat and then performs 15.2 J of work. The change in internal energy of the gas is: (a) 13.3 J (b) - 13.3 J (c) 43.7 J (d) - 43.7 J (e) none of the above

Answer (b) E = q + w 28.5 J - 15.2 J = + 13.3 J

Practice Problem Which of the following statements correctly describes the signs of q and w for the following exothermic process at 1 atmosphere pressure and 370 Kelvin? H2O(g) → H2O(l) (a) q and w are both negative (b) q is positive and w is negative (c) q is negative and w is positive (d) q and w are both positive (e) q and w are both zero

Answer (c). An exothermic indicates q is negative and the gas is condensing to a liquid so it is exerting less pressure on its surroundings indicating w is positive.

What is work? Work is a force acting over a distance. w= F x Dd P = F/ area d = V/area w= (P x area) x D (V/area)= PDV Work can be calculated by multiplying pressure by the change in volume at constant pressure. Use units of liter•atm or L•atm

Pressure and Volume Work Work refers to a force that moves an object over a distance. Only pressure/volume work (i.e., the expansion/contraction of a gas) is of significance in chemical systems and only when there is an increase or decrease in the amount of gas present.

Work needs a sign If the volume of a gas increases, the system has done work on the surroundings. work is negative w = - PDV Expanding work is negative. Contracting, surroundings do work on the system W is positive. 1 L•atm = 101.3 J

Example When, in a chemical reaction, there are more moles of product gas compared to reactant gas, the system can be thought of as performing work on its surroundings (making w < 0) because it is “pushing back,” or moving back the atmosphere to make room for the expanding gas. When the reverse is true, w > 0.

Compressing and Expanding Gases Compressing gas Work on the system is positive Work is going into the system Expanding gas Work on the surroundings is negative Work is leaving the system

Clarification Info If the reaction is performed in a rigid container, there may be a change in pressure, but if there is no change in volume, the atmosphere outside the container didn’t “move” and without movement, no work is done by or on the system. If there is no change in volume (V= 0), then no work is done by or on the system (w= 0) and the change in internal energy will be entirely be due to the heat involved ( ΔE = q).

Examples What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure? If 2.36 J of heat are absorbed by the gas above. what is the change in energy? How much heat would it take to change the gas without changing the internal energy of the gas?

Enthalpy The symbol for Enthalpy is H H = E + PV (that’s the definition) DH = DE + PDV (at constant pressure)

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 6.3

DH = DE + PDV Using DH = DE + PDV the heat at constant pressure qp can be calculated from: DE = qp + w (if w = - PDV then…) DE = qp – PDV (now rearrange) qp = DE + P DV = DH

Examples of Enthapy Changes KOH(s) → K(aq) + OH-1 (aq) ΔHsolution = - 57.8 kJ mol1 C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔHcombustion = -2221kJ/mol H2O(s) → H2O(l) ΔHfusion = 6.0 kJ/mol Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(s) ΔHreaction - 852 kJ/mol Ca(s) + O2(g) H2(g) → Ca(OH)2(s) ΔHformation - 986 kJ/mol1

3 Methods There are a variety of methods for calculating overall enthalpy changes that you should be familiar with. The three most common are the: the use of Heats of Formation Hess’s Law the use of Bond Energies

Heat of Reaction To compare heats of reaction for different reactions, it is necessary to know the temperatures at which heats of reaction are measured and the physical states of the reactants and products. Look in the Appendix of the textbook to find Standard Enthapy tables.

Standard Enthalpy of Formation Measurements have been made and tables constructed of Standard Enthalpies of Formation with reactants in their “standard states”. Use the symbol DHºf Standard state is the most stable physical state of reactants at: 1 atmosphere pressure specified temperature—usually 25 °C 1 M solutions For solids which exist in more than one allotropic form, a specific allotrope must be specified.

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 6.5

6.5

ΔH°formation It is important to recognize that the ΔH°formation (abbreviated as ΔH°f) is really just the heat of reaction for a chemical change involving the formation of a compound from its elements in their standard states.

Standard Enthalpies of Formation The standard heat of formation is the amount of heat needed to form 1 mole of a compound from its elements in their standard states. See the table in the Appendix Remember: For an element the value is 0

Equation Practice You need to be able to write the equation correctly before solving the problem. Try… What is the equation for the formation of NO2 ? Try writing the equation.

Practice Answer ½N2 (g) + O2 (g) ® NO2 (g) You must make one mole to meet the definition.

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.) 6.5

Since we can manipulate the equations We can use heats of formation to figure out the heat of reaction. Lets do it with this equation. C2H5OH +3O2(g) ® 2CO2 + 3H2O which leads us to this rule.

Hess’s Law Definition: When a reaction may be expressed as the algebraic sum of other reactions, the enthalpy change of the reaction is the algebraic sum of the enthalpy changes for the combined reactions.

Hess’ Law Enthalpy is a state function. It is independent of the path. We can add equations to come up with the desired final product, and add the DH values. Two rules to remember: If the reaction is reversed the sign of DH is changed If the reaction is multiplied, so is DH

Enthalpy As enthalpy is an extensive property, the magnitude of an enthalpy change for a chemical reaction depends upon the quantity of material that reacts. This means: if the amount of reacting material in an exothermic reaction is doubled, twice the quantity of heat energy will be released.

Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) DH = 6.01 kJ If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = -6.01 kJ If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) 2H2O (l) DH = 2 x 6.01 = 12.0 kJ 6.3

Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) DH = 6.01 kJ H2O (l) H2O (g) DH = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ 1 mol P4 123.9 g P4 x 3013 kJ 1 mol P4 x 266 g P4 = 6470 kJ 6.3

For the oxidation of sulfur dioxide gas SO2(g) + ½O2(g) → SO3(g) ΔH° = - 99 kJ/mol Doubling the reaction results in: 2SO2(g) + O2(g) → 2SO3(g) ΔH° = - 198 kJ/mol Notice that if you double the reaction, you must double the ΔH° value.

Sign Change of ΔH 2SO2(g) + O2(g) → 2SO3(g) ΔH° = - 198 kJ/mol If the reaction is written as an endothermic reaction: 2SO3(g) → 2SO2(g) + O2(g) ΔH° = + 198 kJ/mol

Tips for Hess’s Law Problems It is always a good idea to begin by looking for species that appear as reactants and products in the overall reaction. This will provide a clue as to whether a reaction needs to be reversed or not. Second, consider the coefficients of species that appear in the overall reaction. This will help determine whether a reaction needs to be multiplied before the overall summation.

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 6.5

Example C(s) + O2(g) → CO2(g) ΔH°f = - 394 kJ/mol 2H2(g) + O2(g) → 2H2O(l) ΔH°f = - 572 kJ/mol CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔH°f = + 891 kJ/mol ----------------------------------------------------- C(s) + 2H2(g) → CH4(g) ΔH°f = - 75 kJ/mol

Example C(s) + O2(g) → CO2(g) ΔH°f = - 394 kJ/mol 2H2(g) + O2(g) → 2H2O(l) ΔH°f = - 572 kJ/mol CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔH°f = + 891 kJ/mol ----------------------------------------------------- C(s) + 2H2(g) → CH4(g) ΔH°f = - 75 kJ/mol

Hess’s Law Example Given DHº= +77.9kJ DHº= +495 kJ DHº= +435.9kJ Calculate DHº for this reaction

Example Given calculate DHº for this reaction DHº= -1300. kJ

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 6.5

Problems to Try Try 12-3 Practice Problems

O2 NO2 -112 kJ H (kJ) 180 kJ NO2 68 kJ N2 2O2

Collegeboard. (2007-2008). Professional Development workshop materials: Special focus thermochemistry. http://apcentral.collegeboard.com/apc/public/repository/5886-3_Chemistry_pp.ii-88.pdf