Reaction Energy

Section 1 - ThermochemistryVirtually 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 TemperatureThe 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

HeatHeat - 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 HeatQuantity 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 ProblemA 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. AnalyzeGiven:

m = 4.0 g

ΔT = 40 K

q = 32 J

2. Plana. 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. Computea.

b.

Practice Problems1. 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 ReactionEnergy 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

ΔHEnergy 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 = Hproducts - Hrectants

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 H2 and O2 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

2H2(g) + O2(g) —> 2H2O(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

4H2(g) + 2O2(g) → 4H2O(g) + 967.2kJ

H2(g) + ½O2(g) → H2O(g) + 241.8kJ

If you reverse the equation, energy must be ADDED to water to break it down to H2 and O2

2H2O(g) + 483.6kJ → 2H2(g) + O2

Because energy needs to be added (is absorbed) this reaction is endothermic

Amount of energy absorbed by water to form H2 and O2 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:

H2O(l) → H2O(g)

H2O(s) → H2O(g)

Thermochemical equations usually written by assigning the value of ΔH instead of writing the energy as a reactant or product

2H2(g) + O2(g) → 2H2O(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

2H2O(g) → 2H2(g) + O2(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 FormationFormation of water from H2 and O2 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 CO2 is -393.5 kJ/mol gas produced

So, CO2 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), C2H2 (ΔH˚f = +226.7kJ/mol) reacts violently with oxygen and must be stored in cylinders as a solution in acetone

Mercury fulminate, HgC2N2O2, has ΔH˚f = +270kJ/mol and its instability makes it useful as a detonator for explosives

Enthalpy of CombustionCombustion (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

ΔHc refers specifically to heat of combustion

Appendix A-5 gives list of heats of combustion

CO2 and H2O are products of complete combustion of organic compounds containing ONLY H and C, or H and C and O (hydrocarbons)

ExamplePropane is a major component of fuel used for outdoor gas grills (barbeques)

It reacts with O2 in the air and produces CO2 and H2O and energy (light and heat)

Complete combustion of one mole of propane, C3H8 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) + O2(g) → CO2(g) ΔH˚c = -393.5kJ/mol

H2(g) + 1/2O2(g) → H2O(l) ΔH˚c = -285.8kJ/mol

CH4(g) + 2O2(g) → CO2(g) + 2H2O(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

CO2(g) + 2H2O(l) .→ CH4(g) + 2O2(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

H2(g) + ½ O2(g) → H2O(l) ΔHc = -285.8kJ/mol

2H2(g) + O2(g) → 2H2O(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) + O2(g) → CO2(g) ΔH˚c = -393.5 kJ/mol

2H2(g) + O2(g) → 2H2O(l) ΔH˚c = 2(-285.9 kJ/mol)

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔH˚ = +890.8 kJ/mol

C(s) + 2H2(g) → CH4(g) ΔH˚f = -74.3 kJ/mol

Sample ProblemCalculate the heat of reaction for the combustion of nitrogen monoxide gas, NO, to form nitrogen dioxide gas, NO2, as given in the following thermochemical equation.

NO(g) + ½ O2(g)→ NO2(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. AnalyzeGiven:

½ N2(g) + ½ O2(g) → NO(g) ΔH˚f = +90.29 kJ/mol

½ N2(g) + O2(g) → NO2(g) ΔH˚f = +33.2 kJ/mol

Unknown:

ΔH˚ for NO(g) + ½ O2(g) → NO2(g)

2. PlanThe 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 O2(g) as reactants and NO2(g) as the product.

½ N2(g) + ½ O2(g) → NO(g) ΔH˚f = +90.29 kJ/mol

½ N2(g) + O2(g) → NO2(g) ΔH˚f = +33.2 kJ/mol

NO(g) + O2(g) → NO2(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 N2(g) + ½ O2(g) ΔH˚ = −90.29 kJ/mol

The other equation should have NO2 as a product, so we can retain the second equation for the formation of NO2 from its elements as it stands.

½ N2(g) + O2(g) → NO2(g) ΔH˚f = +33.2 kJ/mol

NO(g) + ½ O2(g) → NO2(g) ΔH˚ = −57.1 kJ/mol

3. ComputeNO(g) → ½ N2(g) + ½ O2(g) ΔH˚ = −90.29 kJ/mol

½ N2(g) + O2(g) → NO2(g) ΔH˚f = +33.2 kJ/mol

* Note the cancellation of the ½ N2(g) and the partial cancellation of the O2(g)

Practice Problem 1Calculate the heat of reaction for the combustion of methane gas, CH4, to form CO2(g) + H2O(l).

−890.2 kJ/mol

Practice Problem 2Carbon 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.

Cgraphite(s) + O2(g) → CO2(g) ΔH˚c = −394 kJ/mol

Cdiamond(s) + O2(g) → CO2(g) ΔH˚c = −396 kJ/mol

2 kJ/mol

Determining Enthalpy of Formation

When C is burned in a limited supply of O2, CO is formed

C is first oxidized to CO2

Then part of CO2 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 O2(g)

C(s) + ½ O2(g) → CO(g) ΔH˚f = ?

We do know the heat of formation of CO2 and the heat of combustion of CO

C(s) + O2(g) → CO2(g) ΔH ˚f = -393.5 kJ/mol

CO(g) + ½ O2(g) → CO2(g) ΔH ˚c = -283.0 kJ/mol

We reverse the second equation because we need CO as a product

CO2(g) → CO(g) + ½ O2(g) ΔH ˚c = +283.0 kJ/mol

C(s) + O2(g) → CO2(g) ΔH˚ = −393.5 kJ/mol

CO2(g) → CO(g) + ½ O2(g) ΔH˚ = +283.0 kJ/mol

C(s) + ½ O2(g) → CO(g) ΔH˚ = −110.5 kJ/mol

Sample ProblemCalculate the heat of formation of pentane, C5H12, 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. AnalyzeGiven:

C(s) + O2(g) → CO2(g) ΔH˚f = −393.5 kJ/mol

H2(g) + ½ O2(g) → H2O(l) ` ΔH˚f = −285.8 kJ/mol

C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(l) ΔH˚c = −3535.6 kJ/mol

Unknown:

ΔH˚f for 5C(s) + 6H2(g) → C5H12(g)

2. PlanCombine the given equations according to Hess’s law

We need C5H12 as a product, so we reverse the equation for combustion of C5H12 and the sign for ΔH˚c

5CO2(g) + 6H2O(l) → C5H12(g) + 8O2(g) ΔH˚ = +3535.6 kJ/mol

Multiply the equation for formation of CO2 by 5 to give 5C as a reactant.

5C(s) + 5O2(g) → 5CO2(g) ΔH˚ = 5(−393.5 kJ/mol)

Multiply the equation for formation of H2O by 6 to give 6H2 as a reactant.

6H2(g) + 3O2(g) → 6H2O(l) ΔH˚ = 6(−285.8 kJ/mol)

3. Compute5C(s) + 5O2(g) → 5CO2(g) ΔH˚ = 5(−393.5 kJ/mol)

6H2(g) + 3O2(g) → 6H2O(l) ΔH˚ = 6(−285.8 kJ/mol)

5CO2(g) + 6H2O(l) → C5H12(g) + 8O2(g)

ΔH˚ = +3535.6 kJ/mol

5C(s) + 6H2(g) → C5H12(g) ΔH˚f= −145.7 kJ/mol

Practice Problem 1Calculate the heat of formation of butane, C4H10, 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 2Calculate the heat of combustion of 1 mol of nitrogen, N2, to form NO2 using the balanced chemical equation and Appendix Table A-14. (Hint: The heat of combustion of N2 will be equal to the sum of the heats of formation of the combustion products of N2 minus the heat of formation of N2.)

+ 66.4 kJ/mol

Practice Problem 3Calculate the heat of formation for sulfur dioxide, SO2, from its elements, sulfur and oxygen. Use the balanced chemical equation and the following information.

S(s) + 3/2 O2(g) → SO3(g) ΔH˚c= −395.2 kJ/mol

2SO2(g) + O2(g) → 2SO3(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.

2NH4NO3(s) → 2N2(g) + 4H2O(l) + O2(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

SolidGas

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 EnergyProcesses 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 entropy-change 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

C2H4(g) + H2(g) → C2H6(g)

The entropy in this reaction decreases because there is a decrease in moles of gas

C2H4(g) + H2(g) → C2H6(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 ProblemFor the reaction NH4Cl(s) → NH3(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. AnalyzeGiven:

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. PlanΔS, 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 ProblemFor 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