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Reaction Energy
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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.
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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
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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
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Heat cannot be measured directly, but temperature can
Temperature - a measure of the average kinetic energy of the particles in a sample of matter
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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
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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!)
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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
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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
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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
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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?
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1. AnalyzeGiven:
m = 4.0 g
ΔT = 40 K
q = 32 J
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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.
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3. Computea.
b.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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.
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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
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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
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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
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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)
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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
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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
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As it decomposes, violet iodine vapor becomes visible
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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
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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
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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
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CO2 and H2O are products of complete combustion of organic compounds containing ONLY H and C, or H and C and O (hydrocarbons)
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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
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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.
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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
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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 = ?
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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
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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
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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
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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)
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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
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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.
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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)
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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)
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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
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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)
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Practice Problem 1Calculate the heat of reaction for the combustion of methane gas, CH4, to form CO2(g) + H2O(l).
−890.2 kJ/mol
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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
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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 = ?
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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
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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
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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.
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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)
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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
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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)
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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
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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
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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
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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
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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.
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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˚
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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)
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Each variable can have positive or negative value
This leads to four possible combinations of terms
+ ΔH, + ΔS
+ ΔH, - ΔS
- ΔH, + ΔS
- ΔH, - ΔS
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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
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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)
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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
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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
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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
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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.
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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
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2. PlanΔS, ΔH, T → ΔG
The value of ΔG can be calculated according to the following equation.
ΔG˚ = ΔH˚ − TΔS˚
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3. Compute
ΔG˚ = 176 kJ/mol − 298 K [0.285 kJ/(mol•K)]
= 176 kJ/mol − 84.9 kJ/mol
= 91 kJ/mol
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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