Chemical Energetics.pdf

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1 CHEMICAL ENERGETICS The total chemical energy of a substance is called its enthalpy (heat content). It is an indication of the energetic stability and not kinetic stability. Chemical energy consists of kinetic energy and potential energy. The potential energy, due to electrostatic force of attraction between the particles, is usually the main component of the total chemical energy. E.g. A + B C + D Enthalpy change: ΔHreaction = ∑ () − ∑ () = HC + HD – (HA + HB) The activation energy of a reaction (Ea) is the minimum energy which the reacting particles must possess in order to overcome the energy barrier before becoming products. Temperature = 298K Pressure = 1 atm Concentration of solution = 1.0 mol dm -3 Elements in their standard states under standard conditions are assigned zero enthalpy (H=0). A substance in its most stable physical state at 298 K and 1 atm is in its standard states. *C(s) refers to the graphite allotrope since graphite is energetically more stable than diamond. When a substance exists as allotropes, state the allotrope referred to. The enthalpy change when molar quantities of reactants as specified by the chemical equation react to form products at 1 atm and 298K (standard conditions). The enthalpy change when 1 mole of a substance is formed from its constituent elements in their standard states at 298K and 1 atm. The energy released when 1 mole of a substance is completely burnt in excess oxygen at 298K and 1 atm. ENTHALPY & ENTHALPY CHANGES STANDARD CONDITIONS & STANDARD STATES ACTIVATION ENERGY STANDARD ENTHAPHY CHANGE OF REACTION STANDARD ENTHAPHY CHANGE OF FORMATION STANDARD ENTHAPHY CHANGE OF COMBUSTION

Transcript of Chemical Energetics.pdf

Page 1: Chemical Energetics.pdf

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CHEMICAL ENERGETICS

The total chemical energy of a substance is called its enthalpy (heat content).

It is an indication of the energetic stability and not kinetic stability.

Chemical energy consists of kinetic energy and potential energy. The potential energy, due to

electrostatic force of attraction between the particles, is usually the main component of the

total chemical energy.

E.g. A + B → C + D

Enthalpy change: ΔHreaction = ∑ 𝐻 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − ∑ 𝐻 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)

= HC + HD – (HA + HB)

The activation energy of a reaction (Ea) is the minimum energy which the reacting particles must

possess in order to overcome the energy barrier before becoming products.

Temperature = 298K

Pressure = 1 atm

Concentration of solution = 1.0 mol dm-3

Elements in their standard states under standard conditions are assigned zero enthalpy (H=0).

A substance in its most stable physical state at 298 K and 1 atm is in its standard states.

*C(s) refers to the graphite allotrope since graphite is energetically more stable than diamond.

When a substance exists as allotropes, state the allotrope referred to.

The enthalpy change when molar quantities of reactants as specified by the chemical equation react

to form products at 1 atm and 298K (standard conditions).

The enthalpy change when 1 mole of a substance is formed from its constituent elements in their

standard states at 298K and 1 atm.

The energy released when 1 mole of a substance is completely burnt in excess oxygen at 298K and

1 atm.

ENTHALPY & ENTHALPY CHANGES

STANDARD CONDITIONS & STANDARD STATES

ACTIVATION ENERGY

STANDARD ENTHAPHY CHANGE OF REACTION

STANDARD ENTHAPHY CHANGE OF FORMATION

STANDARD ENTHAPHY CHANGE OF COMBUSTION

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C(s) + 1

2O2 (g) → CO (g) ΔH is not ΔHc

o(C) but ΔHfo(CO), since CO is produced via incomplete

combustion of carbon.

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)

Note water is in liquid state and not gaseous state because water exists as liquid under

standard conditions.

Always exothermic

The more exothermic the reaction, the higher the energy value of the fuel.

The heat evolved when 1 mole of water is formed in the neutralization reaction between an acid

and a base, at 298K and 1 atm.

Exothermic since it involves the attraction of H+ and OH- ions to form an O – H bond

Enthalpy change of neutralization of a strong acid and a strong base is almost the same for

all strong acids and bases (-57.0 kJ mol-1).

Explanation: Strong acids and strong bases ionize completely in dilute aqueous solution.

Reaction between them is effectively the reaction between aqueous H+ and OH- ions.

Enthalpy change of neutralization involving a weak acid or a weak base is slightly less

exothermic than -57.0 kJ mol-1.

Explanation: Weak acids/bases do not ionize completely in dilute aqueous solutions. During

neutralization, energy is absorbed to ionize the unionized weak acid/base. Ionization,

which involves bond breaking, is an endothermic process. Less energy is released and the

resulting ΔHoneutralisation is less exothermic.

ΔHoatom for an element is the energy required when 1 mole of gaseous atoms is formed from the

element at 298K and 1 atm.

E.g. Na (s) → Na (g) ΔHoatom = +107 kJ mol-1

1

2Br2 (g) → Br (g) ΔHo

atom = +112 kJ mol-1

Noble gases have enthalpy change of atomization of 0.

ΔHoatom for a compound is the energy required to convert 1 mole of the compound into gaseous

atoms at 298K and 1 atm.

ΔHoatom is always positive as atomization involves breaking bonds.

Depending on the type of substance, atomization is associated with different processes.

E.g. Na (s) → Na (g) ΔHoatom = ΔHo

fusion + ΔHovapourization

Br2 (l) → 2 × ΔHoatom = ΔHo

vapourization + B.E. (Br – Br)

CH4 (g) → C (g) + 4H (g)

STANDARD ENTHALPY CHANGE OF NEUTRALIZATION

STANDARD ENTHALPY CHANGE OF ATOMIZATION

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Bond dissociation energy is the energy required to to break 1 mole of a particular covalent bond in a

specific molecule in the gaseous state.

Always positive

It is a measure of the strength of covalent bonds.

In molecule like methane, the successive bond dissociation energies of each of the C – H

bonds are different.

Bond dissociation energy of the same type of bond in different molecules may differ.

Bond energy is the average energy required to break 1 mole of a covalent bond in the gaseous

state.

Note that when bond energy is used, all reactants and products are in their gaseous states.

The energy evolved when 1 mole of solid ionic compound is formed from its constituent gaseous

ions.

Always negative

It is a measure of the strength of ionic bonds and the stability of the ionic compounds.

There is always a difference between theoretical lattice energy and experimental lattice

energy, as one refers to the value found from experimental results using the Born-Haber

cycle while the latter refers to the value calculated based on a model which assumes that

the compound is completely ionic.

A small difference shows that the ionic model is a good one for the compound while a large

difference shows that there is covalent character in the ionic bond.

The first ionization energy is the energy required to remove 1 mole of electrons from 1 mole of

gaseous atoms to form 1 mole of singly charged gaseous cations.

Always positive

The first electron affinity is the enthalpy change when 1 mole of electrons is added to 1 mole of

gaseous atoms to form 1 mole of singly charged gaseous anions.

BOND DISSOCIATION ENERGY

BOND ENERGY

LATTICE ENERGY

IONIZATION ENERGY

ELECTRON AFFINITY

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1st EA is usually negative due to the attraction between the incoming electron and the atom.

(electron is added to a neutral atom and there is attraction between them)

2nd and subsequent EA are always positive because energy is required to overcome the

electrostatic repulsion between the incoming electron and the anion. (electron is added to a

negatively charged anion and repulsion needs to be overcome)

The energy evolved when 1 mole of free gaseous ions is dissolved in an infinite volume of water at

298K and 1 atm.

ΔHohydration is always negative (for both cation and anion) as heat energy is evolved in

forming ion-dipole interactions between the ions and the polar water molecules.

Magnitude of ΔHohydration depends on its charge density:

The enthalpy change when 1 mole of solute is completely dissolved in an infinite volume of solvent

at 298K and 1 atm.

ΔHosolution can be either positive or negative.

If it is highly positive, the salt is likely to be insoluble.

If it is negative, the salt is likely to be soluble.

Some salts with positive ΔHosolution are soluble e.g. NaCl, KCl, NH4NO3 due to a positive

entropy change.

The dissolution of an ionic solid MX(s) can be divided into 2 processes:

1. Breaking up the solid ionic lattice to form isolated gaseous ions. The process is endothermic and

the enthalpy change is –LE.

2. Hydration of the gaseous ions. Process is exothermic (forming ion-dipole interactions between ion

and water).

M+ (g) + X- (g) + aq → M+ (aq) + X- (aq) ΔH = ΔHohydration (M+) + ΔHo

hydration (X-)

Overall: MX(s) → M+ (aq) + X- (aq)

STANDARD ENTHALPY CHANGE OF HYDRATION

|𝚫𝐇(𝐡𝐲𝐝𝐫𝐚𝐭𝐢𝐨𝐧)| ∝ 𝒒

𝒓

STANDARD ENTHALPY CHANGE OF SOLUTION

RELATIONSHIP BETWEEN LATTICE ENERGY, ENTHALPY CHANGE OF HYDRATION & ENTHALPY

CHANGE OF SOLUTION

ΔHosolution = sum of ΔHo

hydration - L.E.

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The enthalpy change ΔH of a reaction is determined only by the initial and final states of the system

and is independent of the reaction pathway taken.

Calculation of Enthalpy Change of Reaction

1. Construct a balanced chemical equation for the reaction. Label it as equation (1).

2. Inspect the given enthalpy changes of the reactions and write down the chemical equations

for these reactions in such a way that

i. Reactants appear on the left

ii. Products appear on the right

3. Add the equations by canceling out the same substances.

4. Enthalpy change of reaction is equal to the algebraic sum of the given enthalpy changes of

the reactions that have been added.

Example:

Find the standard enthalpy change of formation of CO if the standard enthalpy changes of the

combustion of C and CO are – 393 kJ mol-1 and – 283 kJ mol-1 respectively.

Equation (1): C (s) + 1

2O2 (g) → CO (g) ΔHf

o (CO) = ?

Given reactions:

Equation (2): C (s) + O2 (g) → CO2 (g) ΔHco (C) = - 393 kJ mol-1

Equation (3): CO (g) + 1

2O2 (g) → CO2 (g) ΔHc

o (CO) = - 283 kJ mol-1

Reversing equation (3), CO2 (g) → CO (g) + 1

2O2 (g) -ΔHc

o (CO) = + 283 kJ mol-1

Equation (1) = Equation (2) + Reverse of equation (3)

Hence, ΔHfo (CO) = - 393 + 283 = -110 kJ mol-1

Determine Enthalpy Change Using Standard Enthalpy Change of Formation & Standard Enthalpy

Change of Combustion

HESS’S LAW

CALCULATIONS OF ENTHALPY CHANGES

ΔHoreaction = ∑ 𝒎 ΔHf

o (products) -∑ 𝒏 ΔHfo (reactants)

ΔHoreaction = ∑ 𝒏 ΔHc

o (reactants) -∑ 𝒎 ΔHco (products)

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A spontaneous process is one that in the absence of any barrier such as activation energy,

take place naturally in the direction stated. The change occurs without a need for

continuous input of energy from outside of the system.

It is irreversible as it cannot be brought back to its original state again under the same

condition.

Sign of ΔH by itself does not predict if a reaction is spontaneous.

Entropy

Entropy ‘S’ is a measure of the randomness/disorder of a system, reflection in the number of ways

that the energy of a system can be distributed through the motion of its particles. SI units: J mol-1K-1

For a spontaneous change, the total entropy must increase. Processes with positive entropy

change can occur spontaneously even though they may be endothermic.

Change in

temperature

As temperature increases, the average kinetic energy of the particles and the

range of the energies increase. There are more ways to disperse the energy

among the particles and entropy increases.

Change in phase During melting, heat is absorbed so kinetic energy increases. The order in the

solid is destroyed and the particles move more freely and are further away

from each other. There are more ways the particles and their energies can be

distributed. Entropy increases.

In vaporizing, heat is absorbed so the kinetic energy increases. There is also a

large increase in volume and the particles move more freely and they are

further away from each other resulting in many more ways that the particles

and the energy can be distributed. Entropy increases.

Note: mention more way both the particles and their energies can be

distributed.

Change in number

of particles (for

gases only)

When there is an increase in the number of gas particles, there is a large

increase in entropy as the gases are the most disordered so the number of

ways that the particles and the energy can be distributed increase greatly.

If there is no change in the number of gas particles, entropy may increase or

decrease but will be relatively small numerically.

Mixing of particles When gases are mixed, each gas expands to occupy the whole container under

same constant pressure. The volume available for each gas is increased. There

are more ways the particles and their energies can be distributed. Entropy

ENTROPY CHANGES

FACTORS CAUSING ENTROPY CHANGES

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increases.

However, when the gases are mixed at constant volume, the entropy does

not change.

When a gas expands (due to removal of external pressure), the volume

available increases so entropy increases as there are more ways that the

particles and the energy can be distributed. Liquid with similar polarities (e.g.

benzene and hexane) mix together spontaneously.

Dissolution of an

ionic solid

Entropy increases because the ions in the solid are free to move in solution.

Entropy also decreases because water molecules that were originally free to

move become restricted by the hydration of the ions.

The overall entropy change depends on which of the above factors is more

significant. For ions with a single charge, the overall entropy is usually

positive.

ΔG = ΔH - TΔS

ΔG < 0 The reaction is feasible and takes place spontaneously.

Reaction is exergonic.

ΔG = 0 The system is at equilibrium. There is no net reaction in the forward or

backward direction.

ΔG = 0 during melting and boiling.

ΔG > 0 The reaction is not feasible and cannot take place spontaneously.

It is spontaneous in the reverse direction.

Reaction is endergonic.

ΔG is dependent on temperature.

It is not incorrect to assume that ΔH and ΔS remain roughly constant regardless of

temperature. However, this assumption would be incorrect when there is a change in phase

of one of the reactants or products as the temperature is increases. In such cases, ΔH and ΔS

change significantly.

While Gibbs free energy can be used to determine the feasibility of a reaction, it does not

take into account the kinetics of the reaction.

Some reactions are thermodynamically favourable but kinetically not feasible.

GIBBS FREE ENERGY

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ΔH ΔS -T ΔS ΔG

At Low Temperature At High Temperature

A Temperature

independent

cases

<0 >0 <0 Always < 0

Spontaneous at temperatures

B >0 <0 >0 Always > 0

Non-spontaneous at temperatures

C Temperature

dependent

cases

>0 >0 <0 >0

Non-spontaneous

<0

Spontaneous

D <0 <0 >0 <0

Spontaneous

>0

Non-spontaneous

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Always take note of the limiting reagent. For enthalpy change of neutralization for example,

if NaOH is the limiting reagent, doubling the concentration of H+ ions (e.g. using sulfuric acid

of same concentration and volume instead of HCl) will not cause the temperature of the

solution to go higher. This is because the same amount of water is produced, since OH- ions

are the limiting reactant. [T5 Q2]

Enthalpy change of the reaction between aqueous HCl and solid NaOH is higher than that of

the reaction between aqueous HCl and aqueous NaOH. This is because enthalpy change of

solution of NaOH is exothermic. [T5 Q2]

Always use the term more/less endothermic/exothermic rather than “higher” and “lower”.

When explaining why some reactions occur (e.g. why N2 (g) reacts with O2 (g) to produce NO

in car engines), state that the high temperature (condition) provides sufficient energy to

overcome the activation energy. [T5 Q4]

Note that the equation of ΔG = ΔH – TΔS can be in the form of y = mx + c, where ΔG = y and

ΔH = c. Gradient obtained from a ΔG – temperature graph gives ΔS. [T5 Q12]

Take note of the standard conditions, 1. 298K, 2. 1 atm, 3. Concentration of solution is 1 mol

dm-3. This is important as some of these conditions have to be inferred as they are not

explicitly stated. [T5 Q12] [Class Test Q2]

ΔHoreaction = ∑ 𝒎 ΔHf

o (products) -∑ 𝒏 ΔHfo (reactants) is also applicable to entropy changes

of formation.

E.g. ΔSoreaction = ∑ 𝒎 ΔSf

o (products) -∑ 𝒏 ΔSfo (reactants) [T5 Q13]

ZnCO3 (s) + 2 HCl (aq) → ZnCl2 (aq) + H2O (l) + CO2 (g)

Given that 192 J of heat was liberated from the reaction and that HCl is the limiting reagent

(0.01 mol), the enthalpy change is given by: 1

2 ΔH =

−192

0.0100 = -19200 J mol-1

ΔH = -19200 × 2 = -38.5 kJ mol-1

This is because −192 𝐽

0.0100 𝑚𝑜𝑙 gives the enthalpy change for 1 mol of HCl. However, 2 mol of HCl

were stated in the equation. Multiplication by 2 is therefore necessary. [Class Exercise 2 Q2]

1

2O2 (g) → O (g) is a process of atomization. Note that oxygen now exists as atoms rather

than molecules. The 2 are hence different and enthalpy change of atomization must be

factored into calculation. [Class Exercise 3 Q2]

When making calculations involving bond energy, besides noting the all reactants and

products must exist in gaseous state, also take note that all bonds of reactants and

products must be taken into account as it is common to miss out some of them. [Class

Exercise 3 Q1]

Note that ion-dipole interaction is different from permanent dipole – permanent dipole

interactions. Ions such as OH- carries a full negative charge is not a partial negative charge

and hence the delta minus sign should not be labelled on hydroxide ions. Also, take note

that it is the O that carries the negative charge and this has to be taken note of in drawing

the ion-dipole interactions. [Class Exercise Q2]

Take note of the terms “energy released” and “energy required” in the definition of some

enthalpy terms. [Class Test Q1]

MISCONCEPTIONS & REMARKS

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When drawing energy level diagrams for Born Haber cycle, note that ionization of cations

comes before ionization of anions. This is because cations must lose the electrons first and

they can be gained by the other atoms to become anions. [Class Test Q1]

Always label "energy/kJ mol-1" on y-axis and "0" energy level for elements in their standard

states in an energy level diagram. [Class Test Q1]

When calculating a specific enthalpy change from an energy level diagram, e.g. first E.A. of O,

make it the subject of the equation and conduct the calculation. [Class Test Q1]

When reasoning if a reaction is feasible, compare the absolute values of ΔH and TΔS E.g. For a feasible reaction, ΔG is negative only when |TΔS| < |ΔH|. [Class Test Q2]

ΔHfo (O2(g)) and ΔHf

o (Cl2(g)) are zero as they are elements in their standard states. Hence do not wrongly use bond energy for these enthalpy changes. [Class Test Q2]

Relationship of equation ΔHosolution = sum of ΔHo

hydration - L.E. When making a comparison in terms of solubility (e.g. down a group), compare the change in magnitude of LE and the magnitude of enthalpy change of hydration for cation and/or anions. From there it is possible to determine if enthalpy change of solution is becoming more/less positive. [TP5 Q8 (c)]

When making comparison between enthalpy changes involving calculation using bond energy (e.g. and enthalpy change of combustion), take note of the assumption made when using bond energies. 1. The bond energies are calculated average values. 2. Reactants and products are all in the gaseous state. [TP5 Q4]

Understanding energetics by comparing the enthalpy change of combustion/formation between reactants and products:

In order to attain a more negative enthalpy change of combustion, the products must have been placed at a higher energy level than the reactants to reach the same energy level after combustion (visualize through an energy level diagram).

Similarly, in order to attain a more negative enthalpy change of formation from the elements (at energy level ‘0’) the products must have been placed at a lower energy level than the reactants. [BT 2013 Paper 1 Q19]

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Alternative way of deducing entropy change: if a reaction takes place without external assistance, its Gibbs Free Energy is negative, if the enthalpy change is also given, entropy change can be determined. [BT 2014 Paper 2 Q2]