Reaction Energy
Section 1 - Thermochemistry Virtually every chemical reaction is accompanied by a change in energy. Chemical reactions usually absorb or release energy as heat. You learned in Chapter 12 that heat is also absorbed or released in physical changes, such as melting a solid or condensing a vapor. Thermochemistry is the study of the transfers of energy as heat that accompany chemical reactions and physical changes.
Heat and Temperature The heat absorbed and released as heat in a chemical or physical change is measured in a calorimeter Known quantities of reactants are sealed in container which is submerged in a known quantity of water The energy given off (or absorbed) during reaction is equal to the energy absorbed (or given off) by the known quantity of water Determined from temperature change of water
The direction of energy transfer is determined by the temperature differences between the objects within a system The energy is transferred as heat from the hotter brass bar to the cooler water This energy transfer will continue until the bar and the water reach the same temperature
Heat cannot be measured directly, but temperature can Temperature - a measure of the average kinetic energy of the particles in a sample of matter
Ability to measure temperature is based on heat transfer Amount of energy transferred as heat is measured in joules Joule - SI unit of heat as well as all other forms of energy Derived from units for force and length
Heat Heat - energy transferred between samples of matter because of a difference in their temperatures Energy transferred as heat always moves from higher to lower temperature (think diffusion!)
Specific Heat Quantity of energy transferred depends on nature and mass of material changing temperature, and size of temperature change 1 g Fe heated to 100.0 C and cooled to 50.0 C transfers 22.5 J of energy
Specific heat - amount of energy required to raise the temperature of one gram of substance by 1 C or 1K Either C or K can be used because the divisions on both scales are equal
Specific heat measured under constant pressure conditions (cp) cp is the specific heat at given pressure q is the energy lost or gained m is the mass of the sample ΔT is difference between initial and final temperatures
Sample Problem A 4.0 g sample of glass was heated from 274 K to 314 K, a temperature increase of 40 K, and was found to have absorbed 32 J of energy as heat. a. What is the specific heat of this type of glass? b. How much energy will the same glass sample gain when it is heated from 314 K to 344 K?
1. Analyze Given: m = 4.0 g ΔT = 40 K q = 32 J
2. Plan a. The specific heat, cp, of the glass is calculated using the equation given for specific heat. b. The rearranged specific heat equation is used to find the energy gained when the glass was heated.
3. Compute a. b.
Practice Problems 1. Determine the specific heat of a material if a 35 g sample absorbed 48 J as it was heated from 293 K to 313 K. 0.069 J/(g K) 2. If 980 kj of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be? 329 K
Enthalpy of Reaction Energy absorbed as heat during reaction represented by H H symbol for quantity called enthalpy Enthalpy change amount of energy absorbed by system as heat during a process at constant pressure
ΔH Energy absorbed/released as heat during chemical reaction represented by ΔH H is symbol for a quantity called enthalpy - the heat content of a system at constant pressure There is no way to measure enthalpy directly Only changes in enthalpy can be measured
The Greek letter delta (Δ) stands for change in ΔH literally means change in enthalpy Enthalpy change - amount of energy absorbed or lost by a system as heat during a process at constant pressure Always difference between enthalpies of the products and reactants ΔH = H products - H rectants
Enthalpy of reaction - the quantity of energy released or absorbed as heat during a chemical reaction (Difference between the stored energy of reactants and products) If mixture of H 2 and O 2 is set on fire, water will form and energy is released explosively Energy released comes from reactants as they form the products
Because energy is released, it is exothermic Energy of product (water) must be less than the energy of the reactants before it was set on fire 2H 2 (g) + O 2 (g) > 2H 2 O(g) This equation does not tell you that energy is given off as heat
Experiments done show that 483.6kJ of energy given off when 2 mol of H2O(g) are formed at 298.15K (STP) Change the equation to show energy given off 2H2(g) + O2(g) > 2H2O(g) + 483.6kJ This equation is called a thermochemical equation, because it includes the quantity of energy released or absorbed as heat during the reaction
Thermochemical equation Must always understand the coefficients are numbers of moles and NOT numbers of molecules Quantity of energy depends on the amounts of reactants and products Quantity of energy released during formation of water is proportional to amount of water formed 4H 2 (g) + 2O 2 (g) 4H 2 O(g) + 967.2kJ H 2 (g) + ½O 2 (g) H 2 O(g) + 241.8kJ
If you reverse the equation, energy must be ADDED to water to break it down to H 2 and O 2 2H 2 O(g) + 483.6kJ 2H 2 (g) + O 2 Because energy needs to be added (is absorbed) this reaction is endothermic Amount of energy absorbed by water to form H 2 and O 2 is equal to the energy released when water is formed
Physical states of reactants and products must ALWAYS be included in thermochemical reactions because they influence the overall amount of energy exchanged Example: H 2 O(l) H 2 O(g) H 2 O(s) H 2 O(g)
Thermochemical equations usually written by assigning the value of ΔH instead of writing the energy as a reactant or product 2H 2 (g) + O 2 (g) 2H 2 O(g) ΔH = -483.6kJ ΔH is negative number because energy is given off during the reaction If reaction is endothermic, as the reverse reaction is, ΔH becomes positive 2H 2 O(g) 2H 2 (g) + O 2 (g) ΔH = +483.6kJ
When writing thermochemical equations (TCEs), keep the following in mind: The coefficients in a balanced TCE represent the numbers of moles, not the number of molecules. This allows us to write fractions as coefficients when necessary. The physical state of the product or reactant involved in a reaction is an important factor and MUST be included in the TCE. The change in energy represented by a TCE is directly proportional to the number of moles of substances undergoing a change.
Enthalpy of Formation Formation of water from H 2 and O 2 is a composition reaction TC data are often recorded as the heats of composition (formation) reactions
To make comparisons easy, heats of formation given for standard states of reactants and products Whatever state the reactants/products are at STP (standard temperature and pressure, 298.15K, 0atm) So, standard state of water is liquid
ΔH To indicate that a value represents measurements on substances in their standard states, a sign is added to enthalpy symbol, giving ΔH for the standard heat of a reaction Adding subscript f further indicates a standard heat of formation (ΔH f) Appendix Table A-14 shows heat of formation for synthesis of one mole of the compound listed
Stability and Heat of Formation If large amount of energy released when compound formed, the compound has high negative heat of formation (negative ΔH is exothermic, heat released)
Elements in standard state defined as having ΔH f = 0 The ΔH f of CO 2 is -393.5 kj/mol gas produced So, CO 2 is more stable than the elements that form it
Positive Enthalpies of Formation Compounds with positive ΔH f are slightly unstable and will suddenly decompose to their elements if the conditions are right Ex. HI (hydrogen iodide) is a colorless gas that somewhat decomposes at room temperature ΔH f = +26.5kJ/mol
As it decomposes, violet iodine vapor becomes visible
Compounds with very high positive heat of formation sometimes very unstable and may react or decompose violently Ex. Ethyne (acetylene), C 2 H 2 (ΔH f = +226.7kJ/mol) reacts violently with oxygen and must be stored in cylinders as a solution in acetone Mercury fulminate, HgC 2 N 2 O 2, has ΔH f = +270kJ/mol and its instability makes it useful as a detonator for explosives
Enthalpy of Combustion Combustion (burning) reactions make significant amount of energy as light and heat when substance combined with oxygen Heat of combustion energy that is released as heat by the complete combustion of one mole of a substance
Defined in terms of one mole of reactant, whereas heat of formation is defined in terms of one mole of product ΔH c refers specifically to heat of combustion Appendix A-5 gives list of heats of combustion
CO 2 and H 2 O are products of complete combustion of organic compounds containing ONLY H and C, or H and C and O (hydrocarbons)
Example Propane is a major component of fuel used for outdoor gas grills (barbeques) It reacts with O2 in the air and produces CO 2 and H 2 O and energy (light and heat) Complete combustion of one mole of propane, C 3 H 8 is described by the following TCE C3H8(g) + O2(g) CO2(g) + H2O(l) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) ΔHc = -2219.2kJ/mol
Calculating Enthalpies of Reaction TCEs can be rearranged and added to give enthalpy changes for reactions not included in data tables Basis for calculating heats of reaction is known as Hess s law the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process.
Energy difference between reactants and products is independent of the route taken to get from one to the other Measured heats of reaction can be combined to calculate heats of reaction that are difficult/impossible to actually measure
Example: methane gas (CH4) Let s calculate the heat of formation for the formation of methane gas, CH4, from its elements, H2 and solid C (graphite) at 25 C (298.15K) C(s) + H2(g) CH4(g) ΔH f =?
We can use the combustion reactions of the elements, and of methane C(s) + O 2 (g) CO 2 (g) ΔH c = -393.5kJ/mol H 2 (g) + 1/2O 2 (g) H 2 O(l) ΔH c = -285.8kJ/mol CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) ΔH c = -890.8kJ/mol
Principles for combining TCEs If a reaction is reversed, the sign of ΔH is also reversed. Multiply the coefficients of the known equations so that when added together they give the desired TCE
Must reverse the combustion reaction for CH4, as we want it as a product, not as a reactant CO 2 (g) + 2H 2 O(l). CH 4 (g) + 2O 2 (g) ΔH = +890.8 kj/mol Now 2 formula units of water are used as reactant, will need 2 formula units of water as product
In combustion reaction for H2 as it is written, it only makes one formula unit of water Must multiply coefficients and the value of ΔH by 2 to get the desired quantity of water H 2 (g) + ½ O 2 (g) H 2 O(l) ΔHc = -285.8kJ/mol 2H 2 (g) + O 2 (g) 2H 2 O(l) ΔH = 2(-285.8kJ/mol)
Now we are ready to add the three equations together using Hess s law to give heat of formation for methane and the balanced equation C(s) + O 2 (g) CO 2 (g) ΔH c = -393.5 kj/mol 2H 2 (g) + O 2 (g) 2H 2 O(l) ΔH c = 2(-285.9 kj/mol) CO 2 (g) + 2H 2 O(l) CH 4 (g) + 2O 2 (g) ΔH = +890.8 kj/mol C(s) + 2H2(g) CH4(g) ΔH f = -74.3 kj/mol
Sample Problem Calculate the heat of reaction for the combustion of nitrogen monoxide gas, NO, to form nitrogen dioxide gas, NO 2, as given in the following thermochemical equation. NO(g) + ½ O 2 (g) NO 2 (g) Use the heat-of-formation data in Appendix Table A-14 (page 902). Solve by combining the known thermochemical equations. Verify the result by using the general equation for finding heats of reaction from heats of formation.
1. Analyze Given: ½ N 2 (g) + ½ O 2 (g) NO(g) ΔH f = +90.29 kj/mol ½ N 2 (g) + O 2 (g) NO 2 (g) ΔH f = +33.2 kj/mol Unknown: ΔH for NO(g) + ½ O2(g) NO2(g)
2. Plan The ΔH requested can be found by adding the ΔHs of the component reactions as specified in Hess s law. The desired equation has NO(g) and ½ O 2 (g) as reactants and NO 2 (g) as the product. ½ N 2 (g) + ½ O 2 (g) NO(g) ΔH f = +90.29 kj/mol ½ N 2 (g) + O 2 (g) NO 2 (g) ΔH f = +33.2 kj/mol NO(g) + O 2 (g) NO 2 (g)
We need an equation with NO as a reactant. Reversing the first reaction for the formation of NO from its elements and the sign of ΔH yields the following thermochemical equation. NO(g) ½ N 2 (g) + ½ O 2 (g) ΔH = 90.29 kj/mol The other equation should have NO 2 as a product, so we can retain the second equation for the formation of NO 2 from its elements as it stands. ½ N 2 (g) + O 2 (g) NO 2 (g) ΔH f = +33.2 kj/mol
3. Compute NO(g) ½ N 2 (g) + ½ O 2 (g) ΔH = 90.29 kj/mol ½ N 2 (g) + O 2 (g) NO 2 (g) ΔH f = +33.2 kj/mol NO(g) + ½ O 2 (g) NO 2 (g) ΔH = 57.1 kj/mol * Note the cancellation of the ½ N 2 (g) and the partial cancellation of the O 2 (g)
Practice Problem 1 Calculate the heat of reaction for the combustion of methane gas, CH 4, to form CO 2 (g) + H 2 O(l). 890.2 kj/mol
Practice Problem 2 Carbon occurs in two distinct forms. It can be the soft, black material found in pencils and lock lubricants, called graphite, or it can be the hard, brilliant gem we know as diamond. Calculate ΔH for the conversion of graphite to diamond for the following reaction. Cgraphite(s) Cdiamond(s) The combustion reactions you will need follow. C graphite (s) + O 2 (g) CO 2 (g) ΔH c = 394 kj/mol C diamond (s) + O 2 (g) CO 2 (g) ΔH c = 396 kj/mol 2 kj/mol
Determining Enthalpy of Formation When C is burned in a limited supply of O 2, CO is formed C is first oxidized to CO 2 Then part of CO 2 is reduced with C to give some CO Because the two reactions happen at the same time, it is impossible to directly measure the heat of formation of CO(g) from C(s) and O 2 (g) C(s) + ½ O 2 (g) CO(g) ΔH f =?
We do know the heat of formation of CO 2 and the heat of combustion of CO C(s) + O 2 (g) CO 2 (g) ΔH f = -393.5 kj/mol CO(g) + ½ O 2 (g) CO 2 (g) ΔH c = -283.0 kj/mol We reverse the second equation because we need CO as a product CO 2 (g) CO(g) + ½ O 2 (g) ΔH c = +283.0 kj/mol
C(s) + O 2 (g) CO 2 (g) ΔH = 393.5 kj/mol CO 2 (g) CO(g) + ½ O 2 (g) ΔH = +283.0 kj/mol C(s) + ½ O 2 (g) CO(g) ΔH = 110.5 kj/mol
Sample Problem Calculate the heat of formation of pentane, C 5 H 12, using the information on heats of formation in Appendix Table A-14 and the information on heats of combustion in Appendix Table A-5. Solve by combining the known thermochemical equations.
1. Analyze Given: C(s) + O 2 (g) CO 2 (g) ΔH f = 393.5 kj/mol H 2 (g) + ½ O 2 (g) H 2 O(l) ` ΔH f = 285.8 kj/mol C 5 H 12 (g) + 8O 2 (g) 5CO 2 (g) + 6H 2 O(l) ΔH c = 3535.6 kj/mol Unknown: ΔH f for 5C(s) + 6H 2 (g) C 5 H 12 (g)
2. Plan Combine the given equations according to Hess s law We need C 5 H 12 as a product, so we reverse the equation for combustion of C 5 H 12 and the sign for ΔH c 5CO 2 (g) + 6H 2 O(l) C 5 H 12 (g) + 8O 2 (g) ΔH = +3535.6 kj/mol
Multiply the equation for formation of CO 2 by 5 to give 5C as a reactant. 5C(s) + 5O 2 (g) 5CO 2 (g) ΔH = 5( 393.5 kj/mol) Multiply the equation for formation of H 2 O by 6 to give 6H 2 as a reactant. 6H 2 (g) + 3O 2 (g) 6H 2 O(l) ΔH = 6( 285.8 kj/mol)
3. Compute 5C(s) + 5O 2 (g) 5CO 2 (g) ΔH = 5( 393.5 kj/mol) 6H 2 (g) + 3O 2 (g) 6H 2 O(l) ΔH = 6( 285.8 kj/mol) 5CO 2 (g) + 6H 2 O(l) C 5 H 12 (g) + 8O 2 (g) ΔH = +3535.6 kj/mol 5C(s) + 6H 2 (g) C 5 H 12 (g) ΔH f= 145.7 kj/mol
Practice Problem 1 Calculate the heat of formation of butane, C 4 H 10, using the balanced chemical equation and information in Appendix Table A-5 and Table A-14. Write out the solution according to Hess s law. 125.4 kj/mol
Practice Problem 2 Calculate the heat of combustion of 1 mol of nitrogen, N 2, to form NO 2 using the balanced chemical equation and Appendix Table A-14. (Hint: The heat of combustion of N 2 will be equal to the sum of the heats of formation of the combustion products of N 2 minus the heat of formation of N 2.) + 66.4 kj/mol
Practice Problem 3 Calculate the heat of formation for sulfur dioxide, SO 2, from its elements, sulfur and oxygen. Use the balanced chemical equation and the following information. S(s) + 3/2 O 2 (g) SO 3 (g) ΔH c= 395.2 kj/mol 2SO 2 (g) + O 2 (g) 2SO 3 (g) ΔH = 198.2 kj/mol 296.1 kj/mol
Section 2 Driving Force of Reactions The change in energy of a reaction system is one of two factors that allow chemists to predict whether a reaction will occur spontaneously and to explain how it occurs. The randomness of the particles in a system is the second factor affecting whether a reaction will occur spontaneously.
Enthalpy and Reaction Tendency Most chemical reactions in nature are exothermic Energy released Products have less energy than the reactants did Products more stable than reactants Trend in nature is for a reaction to continue in a direction that leads to a lower energy state
Might think that endothermic reactions cannot occur naturally because the products are at higher potential energy and less stable than reactants Expected to proceed only with assistance of outside influence (continued heating) Since they DO happen, MUST conclude that something other than enthalpy change must help determine whether a reaction will happen
Entropy and Reaction Tendency Naturally occurring endothermic process is melting Ice cube melts spontaneously at room temperature as energy transferred from warm air to ice Well-ordered arrangement of water molecules in ice is lost Less-ordered arrangement of liquid phase higher energy content water is formed A system can go from one state to another without an enthalpy change by becoming more disordered
Decomposition of ammonium nitrate is shown as follows 2NH4NO3(s) 2N2(g) + 4H2O(l) + O2(g) When ammonium nitrate, NH4NO3, decomposes, the entropy of the reaction system increases as one solid reactant becomes two gaseous products and one liquid product.
2NH 4 NO 3 (s) 2N 2 (g) + 4H 2 O(l) + O 2 (g) On left are 2 mol of solid ammonium nitrate On right are 3 mol of gaseous particles and 4 mol liquid
Entropy (S) a measure of the degree of randomness of the particles, such as molecules, in a system To understand entropy, think of solids, liquids and gases Liquid Gas Solid
SOLIDS - particles are in fixed position in small area and only vibrate, do not move Can determine the location of the particles because degree of randomness is low (they are not random) and so entropy is low When solid melts, particles are close together but they can move about somewhat LIQUIDS - system is more random, more difficult to describe location of particles Entropy is higher
GASES - when liquid evaporates, particles move quickly and much father apart Locating individual particle even more difficult System much more random Entropy of gas higher than liquid Entropy of liquid higher than solid
At absolute zero (0K) random motion stops Entropy of pure crystalline solid is zero at absolute zero As energy added, randomness of molecular motion increases
Entropy change - the difference between the entropy of the products and the reactants Increase in entropy represented by positive value for ΔS Decrease is negative value for ΔS
Forming a solution almost always involves an increase in entropy There is increase in randomness True for mixing gases, dissolving liquid in liquid, and dissolving solid in liquid
Free Energy Processes in nature are driven in two directions Toward lowest enthalpy (transfer of energy) Toward highest entropy (randomness) When these two oppose each other, the dominant factor decides the direction of change
To predict which factor will dominate in a system, a function has been defined to relate the enthalpy and entropy factors at a given temperature Free energy, G combined enthalpy-entropy function of a system Measures both the enthalpy-change and entropychange movements Natural processes move in direction that lowers free energy in a system
Only the change in free energy can be measured Change in free energy defined in terms of changes in entropy and enthalpy Free-energy change at a constant temperature and pressure, ΔG of a system is defined as the difference between the change in enthalpy (ΔH) and the product of the Kelvin temperature (T) and the entropy change (ΔS), which is defined as TΔS ΔG = ΔH - TΔS
This expression is for substances in their standard states The product TΔS and the quantities ΔG and ΔH have the same units, usually kj/ mol The units of ΔS for use in this equation are usually kj/(mol K)
Each variable can have positive or negative value This leads to four possible combinations of terms + ΔH, + ΔS + ΔH, - ΔS - ΔH, + ΔS - ΔH, - ΔS
ΔG = ΔH - TΔS If ΔH is negative and ΔS is positive, then both terms on the right are negative Both factors contribute to process being spontaneous So ΔG will always be negative Reaction is definitely spontaneous
ΔG = ΔH - TΔS ΔH is positive (endothermic process) and ΔS is negative (decrease in randomness) Reaction as written will NOT happen (ΔG very high positive integer)
When enthalpy and entropy changes are operating in different directions, sometimes one predominates, other times the other dominates If ΔH is negative and ΔS is negative ΔH leads to spontaneous process BUT ΔS opposes it C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) The entropy in this reaction decreases because there is a decrease in moles of gas
C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) ΔS = -0.1207 kj/(mol K) (large decrease in entropy) ΔH = -136.9 kj/mol (highly exothermic) Reaction still happens because enthalpy change dominates ΔG = ΔH - TΔS = -136.9 kj/mol 298K[-0.1207 kj /(mol K) ] = -101.1 kj/mol
Compare with process of manufacturing syngas (mixture of CO and H2), starting point for making many chemicals, including methanol Reaction is endothermic (ΔH = +206.1 kj/mol) ΔS = +0.215 kj/(mol K) Resulting ΔG is positive at room temp This tells us reaction will not occur at room temp even though entropy change is favorable ΔG = ΔH - TΔS = +206.1 kj/mol 298K[-0.215 kj/(mol K)] = +142.0 kj/mol
Sample Problem For the reaction NH 4 Cl(s) NH 3 (g) + HCl(g), at 298.15K, ΔH = +176 kj/mol and ΔS = +0.285 kj/(mol K). Calculate ΔG, and tell whether this reaction can proceed in the forward direction at 298.15 K.
1. Analyze Given: ΔH = 176 kj/mol at 298.15 K ΔS = 0.285 kj/(mol K) at 298.15 K Unknown: ΔG at 298.15 K
2. Plan ΔS, ΔH, T ΔG The value of ΔG can be calculated according to the following equation. ΔG = ΔH TΔS
3. Compute ΔG = 176 kj/mol 298 K [0.285 kj/(mol K)] = 176 kj/mol 84.9 kj/mol = 91 kj/mol
Practice Problem For the vaporization reaction Br2(l) Br2(g), ΔH = 31.0 kj/mol and ΔS = 93.0 J/(mol K). At what temperature will this process be spontaneous? above 333 K