Energy and Chemical Reactions

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Transcript Energy and Chemical Reactions

Energy and Chemical Reactions

Original by L. Scheffler 1

5.1 Exothermic and endothermic reactions

5.1.1

5.1.2

5.1.3

5.1.4

Assessment statement Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction ( ). State that combustion and neutralization are exothermic processes.

Apply the relationship between temperature change, enthalpy change and the classification of a reaction as endothermic or exothermic.

Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of the enthalpy change for the reaction.

Teacher’s notes Standard enthalpy change is the heat energy transferred under standard conditions — pressure 101.3 kPa, temperature 298 K. Only ∆H can be measured, not H for the initial or final state of a system.

5.2 Calculation of enthalpy changes

5.2.1

5.2.2

5.2.3

5.2.4

5.3 Hess’s law

5.3.1

5.4 Bond enthalpies

5.4.1

5.4.2

Assessment statement Teacher’s notes Calculate the heat energy change when the temperature of a pure substance is changed.

Students should be able to calculate the heat energy change for a substance given the mass, specific heat capacity and temperature change using q = mcΔT.

Design suitable experimental procedures for measuring the heat energy changes of reactions.

Students should consider reactions in aqueous solution and combustion reactions.

Use of the bomb calorimeter and calibration of calorimeters will not be assessed.

Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water.

Evaluate the results of experiments to determine enthalpy changes.

Students should be aware of the assumptions made and errors due to heat loss.

Assessment statement Teacher’s notes Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.

Students should be able to use simple enthalpy cycles and enthalpy level diagrams and to manipulate equations. Students will not be required to state Hess’s law.

Assessment statement Define the term average bond enthalpy.

Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic.

Teacher’s notes

Heat and Temperature

Heat is energy that is transferred from one object to another due to a difference in temperature Temperature is a measure of the average kinetic energy of a body Heat is always transferred from objects at a higher temperature to those at a lower temperature

Factors Affecting Heat Quantities

The amount of heat contained by an object depends primarily on three factors:

–

The mass of material

–

The temperature

–

The kind of material and its ability to absorb or retain heat.

Heat Quantities

The heat required to raise the temperature of 1.00 g of water 1 o C is known as a calorie The SI unit for heat is the joule . It is based on the mechanical energy requirements. 1.00 calorie = 4.184 Joules The energy required to raise 1 pound of water of 1 o F is called a British Thermal Unit or BTU The BTU is widely used in the USA to compute energy capacities of heating and air conditioning equipment

Calorimetry

Calorimetry involves the measurement of heat changes that occur in chemical processes or reactions. The heat change that occurs when a substance absorbs or releases energy is really a function of three quantities:

– The mass – The temperature change – The heat capacity of the material 7

Heat Capacity and Specific Heat

The ability of a substance to absorb or retain heat varies widely.

The heat capacity depends on the nature of the material.

The specific heat of a material is the amount of heat required to raise the temperature of 1 gram of a substance 1 o C (or Kelvin)

Specific Heat values for Some Common Substances

Substance Water (liquid) Water (steam) Water (ice) Copper Aluminum Ethanol Lead C J g -1 K -1 4.184 2.080 2.050 0.385 0.897 2.44 0.127 C J mol -1 K -1 75.327 37.47 38.09 24.47 24.2 112 26.4

Heat Exchange

When two systems are put in contact with each other, there will be a net exchange of energy between them unless they are at thermal equilibrium, i.e. at the same temperature.

Heat will flow from the substance at the higher temperature to that at a lower temperature 10

Heat Changes

The heat equation may be stated as D

Q = m C

T

where: D Q = Change in heat m = mass in grams C = specific heat in J g -1 o C -1 D T = Temperature change 11

Temperature Changes

Measuring the temperature change in a calorimetry experiment can be difficult since the system is losing heat to the surroundings even as it is generating heat.

By plotting a graph of time v temperature it is possible to extrapolate back to what the maximum temperature would have been had the system not been losing heat to the surroundings.

A time v temperature graph 12

Heat Transfer Problem 1

Calculate the heat that would be required an aluminum cooking pan whose mass is 400 grams, from 20 o C to 200 o C. The specific heat of aluminum is 0.902 J g -1 o C -1 .

Solution

Q = mC

T = (400 g) (0.902 J g -1 o C -1 )(200 o C – 20 o C

)

= 64,944 J

Heat Transfer Problem 2

What is the final temperature when 50. grams of water at 20 o C is added to 80. grams water at 60 o C? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g -1 o C -1

Solution:

Q (Cold) =

Q ( hot ) mC

T= mC

Let T = final temperature (50 g) (4.184 J g -1 o C -1 )(T- 20 o C) = (80 g) (4.184 J g -1 o C -1 )(60 o C- T) (50 g)(T- 20 o C) = (80 g)(60 o C- T) 50T -1000 = 4800 – 80T 130T =5800 T = 44.6 o C 14

Phase Changes & Heat

Energy is required to change the phase of a substance The amount of heat necessary to melt a substance is called the Heat of fusion ( D H fus ).The heat of fusion is expressed in terms of 1 mole or 1 gram It takes 6.00 kJ of energy to melt 1 mole (18 grams) of ice into liquid water. This is equivalent to about 335 J per gram

The amount of heat necessary to boil a substance is called the Heat of vaporization (

H vap ) It may be expressed in terms of 1 mole or 1 gram

takes 40.6 kJ of energy to boil away 1 mole (18 grams) of water. This is equivalent to about 2240 J per gram.

Phase Changes & Heat

AB, CD, E(on up) – Temp Change BC – Phase Change #1; DE – Phase Change #2 16

Molar Heat Data for Some Common Substances

Substance Mercury, Hg

Q fus 2.29kJ/mol

Q vap 59.1kJ/mol Ethanol, C 2 H 5 OH Water, H 2 O Ammonia, NH 3 Helium, He Acetone Methanol, CH 3 OH 5.02kJ/mol 6.00kJ/mol 5.65kJ/mol 0.02kJ/mol 5.72kJ/mol 3.16kJ/mol 38.6kJ/mol 40.6kJ/mol 23.4kJ/mol 0.08kJ/mol 29.1kJ/mol 35.3kJ/mol

Heat Transfer Problem 3

How much energy must be lost for 50.0 g of liquid wax at 85.0˚C to cool to room temperature at 25.0˚C?

(C solid wax = 2.18 J/g˚C, m.p. of wax = 62.0 ˚C, C liquid wax =2.31 J/g˚C; MM = 352.7 g/mol,

D Q

total

= (50g)(2.31J g -1 ˚C -1 D

H fusion =70,500 J/mol)

)(62˚C-85˚C) D Q = mC liquid wax D T + (50g/352.7gmol

-1 )(-70,500J mol -1 ) + (50g)(2.18J g -1 ˚C -1 )(25˚C-62˚C) + n( D Q fusion ) + mC solid wax D T D

total

= (-2656.5 J) + (-9994.3 J)+ (-4033 J)

total

= -16,683.8 J

D Q total = D Q total= = D Q liquid wax + D Q solidification + D Q solid wax mC liquid wax D T +n( D Q fusion ) + mC solid wax D T 18

Heat Transfer Problem 4

Steam at 175 ° C that occupies a volume of 32.75 dm to end as liquid water at 20 o C?

3 and a pressure of 2.60 atm. How much energy would it need to lose

Solution: n = PV/RT = (2.60 atm)(32.75 dm 3 ) (0.0821 dm 3 atm mol -1 K -1 )(448 K -1 ) = 2.315 mol

Q = (2.315 mol) (37.47 J mol -1 K -1 )(100 o C-175 o C) +(2.315 mol)(-40600 J mol -1 ) +(2.315 mol)(75.327 J mol -1 K -1 )(20 o C-100 o C)

Q = -6505.7J + -93989 J + -13950.6 J = -114445.3 J = -114.445 kJ

Chemical Reactions

In a chemical reaction Chemical bonds are broken Atoms are rearranged New chemical bonds are formed These processes always involve energy changes

Energy Changes

Breaking chemical bonds requires energy Forming new chemical bonds releases energy 21

Exothermic and Endothermic Processes

Exothermic processes release energy C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4H 2 O (g)

+ 2043 kJ

Endothermic processes absorb energy C(s) + H 2 O (g)

+113 kJ

 CO(g) + H 2 (g) 22

Energy Changes in endothermic and exothermic processes

In an endothermic reaction there is more energy required to break bonds than is released when bonds are formed. The opposite is true in an exothermic reaction.

Enthalpy Calculations

Enthalpy

Enthalpy is the heat absorbed or released during a chemical reaction where the only work done is the expansion of a gas at constant pressure

Enthalpy

Energy = Heat + Work Not all energy changes that occur as a result of chemical reactions are expressed as heat Work is a force applied over a distance.

Most energy changes resulting from chemical reactions are expressed in a special term known as enthalpy

Enthalpy

It is nearly impossible to set up a chemical reaction where there is no work performed.

The conditions for a chemical reaction are often set up so that work in minimized.

Enthalpy and heat are nearly equal under these conditions.

Enthalpy Changes

The change in enthalpy is designated by the symbol

–

H < 0 the process is exothermic.

–

H > 0 the process is endothermic.

–

Sometimes the symbol for enthalpy (

H) is used for heat (

–

In many cases where work is minimal, heat is a close approximation for enthalpy.

–

One must always remember that while they are closely related, heat and enthalpy are NOT identical

Energy and Enthalpy Changes

It is impractical to measure absolute amounts of energy or enthalpy.

Hence, we measure changes in enthalpy rather than total enthalpy Enthalpy is always measured relative to previous conditions.

Enthalpy is measured relative to the system.

Measuring Enthalpy

The amount of heat absorbed or released during a chemical reaction depends on the conditions under which the reaction is carried out including:

–

the temperature

–

the pressure

–

the physical state of the reactants and products

Standard Conditions

For most thermodynamic measurements standard conditions are established as

–

25 o C or 298 K

–

1.0 atmosphere of pressure

–

Note: this is a change from the gas laws where the standard temperature was 0 o C

Standard State

The pure form of a substance at standard conditions (25 o C and 1 atmosphere) is said to be in the standard state.

The most stable form of an element at standard conditions represents the standard state for that element.

Bond Enthalpies

One approach to determining an enthalpy change for a chemical reaction is to compute the difference in bond enthalpies between reactants and products The energy to required to break a covalent bond in the gaseous phase is called a bond enthalpy.

Bond enthalpy tables give the average energy to break a chemical bond. Actually there are slight variations depending on the environment in which the chemical bond is located

Bond Enthalpy Table

The average bond enthalpies for several types of chemical bonds are shown in the table below:

Bond Enthalpies

Bond enthalpies can be used to calculate the enthalpy change for a chemical reaction .

Energy is required to break chemical bonds . Therefore when a chemical bond is broken its enthalpy change carries a positive sign .

Energy is released when chemical bonds form . When a chemical bond is formed its enthalpy change is expressed as a negative value By combining the enthalpy required and the enthalpy released for the breaking and forming chemical bonds, one can calculate the enthalpy change for a chemical reaction

Bond Enthalpy Calculations

Example 1: Calculate the enthalpy change for the reaction N 2 + 3 H 2  2 NH 3

Bonds broken

1 N

N: = 945 3 H-H: 3(435) = 1305 Total = 2250 kJ

Bonds formed

2x3 = 6 N-H: 6 (390) = - 2340 kJ Net enthalpy change = + 2250 - 2340 =

- 90 kJ

Hess Law and Enthalpy Calculations

Standard Enthalpy Changes

The enthalpy change that occurs when the reactants are converted to products, both being in their standard states is known as the standard enthalpy change.

It is designated as D H o .

H o reaction =

S D

H o products -

S D

H o reactants

Calculating Enthalpy from tables

The enthalpy of formation for compound is equal to the enthalpy change that occurs when a compound is formed from its elements The symbol for the bond enthalpy of formation is D H f Enthalpies of formation have been measured and tabulated for a large number of compounds 40

Enthalpies of Formation

Some enthalpies of formation for common compounds BaCO Ba(OH)

aO 3 CaCO 3 CaO 2 Ca(OH) 2 CaCl 2 -1219 -998 -554 -1207 -636 -987 -796 H 2 O (g) H 2 O (l) H 2 O 2 C 3 H 8 C 4 H 10 CO CO 2 -242 -286 -188 -104 -126 -110 -394 HCl (g) HCl (aq) NH 3 (g) NO NO 2 SO 2 Al 2 O 3 (s) -93 -167 -46 +90 +33.8

-297 -1670

See text: Brown, LeMay and Bursten, Chemistry the Central Science, 7 th edition pages 984-987 for additional values

Calculating Enthalpy from tables

Enthalpies of formation represent the enthalpy changes when compound forms from its elements The enthalpy of formation for a chemical reaction can be expressed as the difference between the enthalpy state of the products and that of the reactants

Hreaction =

S D

H o products –

S D

H o reactants

Sample Problem 1

Calcium carbonate reacts with hydrochloric acid according to the following equation: CaCO 3 (s) + 2HCl (aq)



CaCl 2 (aq) + H 2 O (l) + CO 2 (g) Calculate the enthalpy change for this reaction

D H o reaction = S D H o products – SD H o reactants D H o CaCO 3 D H o HCl (aq) D H o CaCl 2 D H o H 2 O (l) D H o CO 2 (g) -1207 -167 -796 -286 -394

Solution

SD H o products =(-796)+(-286)+(-394) = -1476 kJ SD H o reactants =( -1207 )+(2)(-167) = -1541 kJ D H o reaction

= -1476-(-1541) = +65 kJ

Sample Problem 2

Calculate the enthalpy change for the burning of 11 grams of propane C 3 H 8 (g) + 5 O 2

H o reaction =

S D

H (g) o



3 CO products – 2

(g) + 4 H H o reactants 2 O (g)

D H o C 3 H 8 D H o

O 2 (g)

D H o

H 2 O (g)

D H o CO 2 (g)

-104 0 -242 -394

Solution

SD H o products =(3)(-394)+(4)(-242) = -2150 kJ SD H o reactants =( -104 )+(5)(0) = -104 kJ D H o reaction

= -2150-(-104) = -2046 kJmol -1 Now 11 grams = 0.25 mole of propane (11 g/44 g mol -1 ) (0.25 mol )(-2046 kJ mol -1 ) = - 511.5 kJ

Some things to Remember

The enthalpy of formation table is stated in kJ mol -1 .

To find the sum of enthalpies of formation for reactants or products, multiply the number of moles of each substance by the enthalpy of formation for that substance.

Then find the difference: Products Reactants 45

Hess’ Law – Indirect Enthalpy Calculations by Rearranging Reactions

Hess’ Law provides a way to calculate enthalpy changes even when the reaction cannot be performed directly.

If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy change for the individual steps

Techniques

Equations may be multiplied, divided, or reversed and then added together to form a new equation If an equation is multiplied or divided the enthalpy of the reaction is multiplied or divided by the same factor If the direction an equation is reversed the sign of the enthalpy is the opposite as well When adding equations together the enthalpies are added together as well 47

Hess’ Law: Example 1

N 2 (g) + O 2 (g) 2 NO (g) + O 2  (g) 2 NO (g) D H 1  2 NO 2 (g) D H 2 = +181 kJ = -113 kJ Find the enthalpy change for N 2 (g) + 2 O 2 (g)  2 NO 2 (g) 48

Hess’ Law: Example 1

The required equation is really the sum of the two given equations Solution: N 2 (g) + O 2 (g) 2 NO (g) + O 2  (g) 2 NO (g) D H 1  2 NO 2 (g) D H 2 = +181 kJ = -113 kJ ------------------------------------------------------------ N 2 (g) +2O 2 (g)+ 2 NO (g)  N 2 (g) +2O 2 (g)  2 NO (g) + 2 NO + 2 NO 2 (g) 2 (g) D H = D H 1 + D H 2 = +181 kJ +(-113) = + 68 kJ 49

Hess Law: Example 2

From the following reactions and enthalpy changes: 2 SO 2 (g) + O 2 2 S (s) +3 O 2 (g)  (g)  2 SO 3 2 SO 3 (g) D H = -196 kJ (g) D H = -790 kJ Find the enthalpy change for the following reaction: S (s) + O 2 (g)  SO 2 (g) Solution: 2 SO 3 (g) 2 S (s) +3 O 2  2 SO 2 (g)  (g) + O 2 2 SO 3 (g) D H = +196 kJ (g) D H = -790 kJ

--------------------------------------------------------------------------------------------------------------

Reversing the order of the first equation reverses the sign of D H 50

Hess Law Example 2

From the following reactions and enthalpy changes: 2 SO 2 (g) + O 2 2 S (s) +3 O 2 (g)  (g)  2 SO 3 2 SO 3 (g) D H = -196 kJ (g) D H = -790 kJ Find the enthalpy change for the following reaction: S (s) + O 2 (g)  SO 2 (g) 2 SO 3 (g) 2 S (s) +3 O 2  2 SO 2 (g)  (g) + O 2 2 SO 3 (g) D H = +196 kJ (g) D H = -790 kJ

--------------------------------------------------------------------------------------------------------------

2 SO 3 (g) +2 S(s) +

3 O 2 (g)  2 SO 3 (g)+2 SO 2 (g) + O 2 D H = -594 (g) kJ 2 S(s) + 2 O 2 (g)  2 SO 2 (g) D H = -594 kJ 51

Hess Law: Example 2

From the following reactions and enthalpy changes: 2 SO 2 (g) + O 2 2 S (s) +3 O 2 (g)  (g)  2 SO 3 2 SO 3 (g) D H = -196 kJ (g) D H = -790 kJ Find the enthalpy change for the following reaction: S (s) + O 2 (g)  SO 2 (g) 2 SO 3 (g)  2 S (s) +3 O 2 2 SO (g)  2 (g) + O 2 2 SO 3 (g) D H = +196 kJ (g) D H = -790 kJ

--------------------------------------------------------------------------------------------------------------

2 SO 3 (

) +2 S(

) +

3 O 2 (

)  2 SO 3 (

)+2 SO 2 (g) + O 2 D H = -594 (g) kJ 2 S(s) + 2 O 2 (g)  2 SO 2 (g) D H = -594 kJ S(s) + O 2 (g)  SO 2 (g) D H = -297 kJ 52

Born Haber Cycle

Born-Haber Cycle

Born-Haber Cycles are energy cycles for the formation of certain ionic compounds The enthalpy of formation for an uncombined element is therefore = 0 Application of Hess’ Law A Born-Haber cycle can be used to calculate quantities that are difficult to measure directly such as lattice energies The formation of an ionic compound as a sequence of steps whose energies can be determined.

Some Definitions

The enthalpy of atomization is the enthalpy change that occurs when one mole of gaseous atoms is formed from the element in the standard state under standard conditions Example: ½ Cl 2 (g)



Cl (g)

H o at = 121 kJ mol -1 The electron affinity is the enthalpy change that occurs when an electron is added to an isolated atom in the gaseous state: O (g) + e O (g) + e-

 

O O 2 (g)

H o (g)

H o = -142 kJ mol = +844 kJ mol -1 -1 The lattice enthalpy is the enthalpy change that occurs from the conversion of an ionic compound in the ground state into its gaseous ions LiCl (g)



Li + (g) + Cl (g)

H o = +846 kJ mol -1

Born Haber Cycle Diagram

The stepwise energy changes for the formation of NaCl

Born Haber Cycle for NaCl

The formation of NaCl can be considered as a five step process 1.

Na (s) + 1/2 Cl 2 (g)



NaCl (s) The vaporization of sodium metal to form the gaseous element. 2.

The dissociation of chlorine gas to gaseous chlorine atoms is equal to one half of the bond energy for a Cl-Cl covalent bond The ionization of gaseous sodium atoms to Na(g)



Na + The ionization of chlorine atoms. (This quantity is the negative electron affinity for the element chlorine.) The lattice energy on the formation of sodium chloride from the gaseous ions

Born-Haber Cycle for NaCl

The stepwise energy changes for the formation of NaCl: The vaporization of sodium metal to form the gaseous element .

Na (s)



Na (g) ∆H° sublimation = + 109 kJ mol -1 The dissociation of chlorine gas to gaseous chlorine atoms is equal to one half of the bond energy for a Cl Cl covalent bond 1/2 Cl 2 (g)



Cl (g) ∆H° diss = + 122 kJ mol -1 The ionization of gaseous sodium atoms to: Na (g)



Na + (g) + e ∆H° ionization = + 496 kJ mol -1 The ionization of chlorine atoms. (This quantity is the negative electron affinity for the element chlorine.) Cl (g) + e -



Cl (g) ∆H° elect.affinity

= - 368 kJ mol -1 The lattice energy on the formation of sodium chloride from the gaseous ions Na + (g) + Cl (g)



NaCl (s) ∆H° lattice = - 770 kJ mol -1

Entropy

Entropy

is defined as a state of disorder or randomness.

In general the universe tends to move toward release of energy and greater entropy.

Entropy

The statistical interpretation of thermodynamics was pioneered by James Clerk Maxwell (1831 – 1879) and brought to fruition by the Austrian physicist Ludwig Boltzmann (1844 –1906).

Entropy

Spontaneous chemical processes often result in a final state is more

Disordered

Random

than the original.

The Spontaneity of a chemical process is related to a change in randomness.

Entropy is a

thermodynamic property related to the degree of randomness or disorder in a system.

Reaction of potassium metal with water. The products are more randomly distributed than the reactants

Entropy and Thermodynamics

According to the second law or thermodynamics the entropy of the universe is always increasing.

This is true because there are many more possibilities for disorder than for order.

Royal Flush Nothing hand

Entropy

Entropy is Disorder

Disorder in a system can take many forms. Each of the following represent an increase in disorder and therefore in entropy: 1.

Mixing different types of particles. i.e. dissolving salt in water.

A change is state where the distance between particles increases. Evaporation of water.

Increased movement of particles. Increase in temperature.

Increasing numbers of particles. Ex.

2 KClO 3



2 KCl + 3O 2

Entropy States

The greatest increase in entropy is usually found when there is an increase of particles in the gaseous state.

The symbol for the change in disorder or entropy is given by the symbol,

S .

The more disordered the more positive a system becomes the value for

S will be.

Systems that become more ordered negative

S values.

have

Entropy, S

The entropy of a substance depends on its state: S (gases) > S (liquids) > S (solids) H 2 S o (J/K -1 mol -1 ) O (liquid) H 2 O (gas) 69.95

188.8

Entropy and States of Matter

S˚(Br 2 liquid) < S˚(Br 2 gas) S˚(H 2 O solid) < S˚(H 2 O liquid)

Entropy, Phase & Temperature

S increases slightly with T S increases a large amount with phase changes

Entropy and Temperature

The Entropy of a substance increases with temperature.

Molecular motions of heptane, C 7 H 16 Molecular motions of heptane at different temps.

Entropy - Boltzman

Ludwig Boltzman S = k Ln W Entropy is proportional to the number of degrees of freedom or possible configurations in a system.

Standard Entropy Values

The standard entropy,

S o , of a substance is the entropy change per mole that occurs when heating a substance from 0 K to the standard temperature of 298 K.

Unlike enthalpy, absolute entropy changes can be measured.

Like enthalpy, entropy is a state function . The change in entropy is the difference between the products and the reactants

S o =

S o (products) -

S o (reactants)

Standard Entropy Values

Some standard enthalpy values The amount of entropy in a pure substance depends on the temperature, pressure, and the number of molecules in the substance. Values for the entropy of many substances at have been measured and tabulated. The standard entropy is also measured at 298 K.

Factors That Determine Entropy States

The greater the disorder or randomness in a system the larger the entropy. Some generalizations The entropy of a substance always increases as it changes from solid to liquid to gas and vice versa.

When a pure solid or liquid dissolves in a solvent, the entropy of the substance increases.

When gas molecules escape from a solvent, the entropy increases.

Entropy generally decreases with increasing molecular complexity

Gibbs Free Energy

Spontaneity

A chemical reaction is spontaneous if it results in the system moving form a less stable to a more stable state.

Decreases in enthalpy and increases in entropy move a system to greater stability.

The combination of the enthalpy factor and the entropy factor can be expressed as the Gibbs Free Energy .

Gibbs Free Energy

The standard free energy change is defined by this equation

G o =

H o – T

S o Where

H o

S o = the enthalpy change = the entropy change T = Kelvin temperature A chemical reaction is spontaneous if it results in a negative free energy change .

Gibbs Free Energy

Possible Combinations for free energy change:

G o =

H o – T

S o

H-T

S < 0 (-) < 0 (-) > 0 (+) Always (-)

Always Spontaneous Never Spontaneous Spontaneous at High Temperature Spontaneous at Low Temperature

> 0 (+) > 0 (+) < 0 (-) < 0 (-) > 0 (+) > 0 (+) > 0 (+)

< 0 (-)

< 0 (-) > 0 (+) < 0 (-) Always (+) (-) if T large (+) if T small (+) if T large

(-) if T small 78

Free Energy Problem 1

A certain chemical reaction is exothermic with a standard enthalpy of - 400 kJ mol -1 . The entropy change for this reaction is +44 J mol -1 K -1 . Calculate the free energy change for this reaction at 25 o C. Is the reaction spontaneous?

Solution

Convert the entropy value to kJ. 44 J mol -1 kJ mol -1 K -1

G = - 400 kJ mol -1 K – (298 K)(0.044 kJ mol -1 -1 = 0.044 K -1 )

G = - 400 kJ mol -1

G = - 413.1 kJ mol -1 – 13.1 kJ mol -1 . Since

G is negative the reaction is spontaneous.

Note. Because

H <0 and

S >0, this reaction is spontaneous at all temperatures.

Free Energy Problem 2

A certain chemical reaction is endothermic with a standard enthalpy of +300 kJ mol -1 . The entropy change for this reaction is +25 J mol -1 K -1 . Calculate the free energy change for this reaction at 25 o C. Is the reaction spontaneous?

Solution

Convert the entropy value to kJ. 25 J mol -1 kJ mol -1 K -1

G = + 300 kJ mol -1 K – (298 K)(0.025 kJ mol -1 -1 = 0.025 K -1 )

G = + 300 kJ mol -1

G – 7.45 kJ mol -1 = + 292.55 kJ mol -1 . Since

G is positive the reaction is non-spontaneous.

Note. Because

H >0 and

S >0, this reaction is non spontaneous at low temperatures. It the temperature were substantially increased it would become spontaneous.