Energetics

Energy = force x distance (Joules)

Exothermic and Endothermic reactions

In chemical reactions, we need energy usually in the form of heat. Energy is absorbed to break the bonds of the reactants and energy is given out when new bonds are formed in the products.

Exothermic reactions Endothermic reactions

Heat is the energy transferred between objects that are at different temperatures.

The amount of heat transferred depends on the amount of the substance.

◦Energy is measured in units called joules (J).

Heat and Temperature

Temperature is a measure of “hotness” of a substance and represent the average kinetic energy of the particles in a substance.

It does not depend on the amount of the substance.

Do both beakers contain the same amount of heat?

Energy Changes in Chemical Reactions All chemical reactions are accompanied by some

form of energy change Exothermic Energy is given out Endothermic Energy is absorbed

Activity : observing exothermic and endothermic reactions

Enthalpy (H) is the heat content that is stored in a chemical system.

We measure the change in enthalpy ∆H i.e. the amount of heat released or absorbed when a chemical reaction occurs at constant pressure, measured in

kilojoules per mole (kJmol-1).

∆H = H(products) – H(reactants)

Enthalpy (H)and Enthapy change(ΔH)

Enthalpy

Enthalpy Level Diagram -Exothermic Change For exothermic reactions, the reactants have more

energy than the products, and the enthalpy change, ∆H = H(products) - H(reactants)

∆H is negative since H(products) < H(reactants)

There is an enthalpy decrease and heat is released to the surroundings

Self-heating cans◦ CaO (s) + H₂O (l) Ca(OH)₂ (aq)

Combustion reactions

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

neutralization (acid + base)◦ NaOH(aq) + HCl(aq) NaCl(aq) + H₂O(l)

Respiration◦C₆H₁₂O₆ (aq) + 6O₂ (g) 6CO₂ (g) + 6H₂O (l)

Examples of Exothermic Reactions

Enthalpy Level Diagram -Endothermic Change For endothermic reactions, the reactants have less

energy than the products, and the enthalpy change, ∆H = H(products) - H(reactants)

∆H is positive since H(products) < H(reactants)

There is an enthalpy increase and heat is absorbed from the surroundings

Enthalpy

Self-cooling beer can◦ H ₂O (l) H₂O (g)

Thermal decomposition CaCO₃ (s) CaO (s) + CO ₂ (g)

Photosynthesis 6CO₂ (g) + 6H₂O (l) C₆H₁₂O₆ (aq) + 6O₂ (g)

Examples of Endothermic Reactions

Amount of heat required to raise the temperature of a unit mass of a substance by 1 degree or 1 kelvin.

Uint : Jg-1 0C-1

Specific heat capacity

The specific heat capacity of alminium is 0.90 Jg-1 0C-1 . If 0.90J of energy is put into 1g of aluminium, the temperaturewill be raised by 10C.

Calculating heat absorbed and released

q = c × m × ΔTq = heat absorbed or releasedc = specific heat capacity of substancem = mass of substance in gramsΔT = change in temperature in Celsius

Heat given off by a process is measured through the temperture change in another substance (usually water).

Due to the law of conservation of energy, any energy given off in a process must be absorbed by something else, we assume that

the energy given out will be absorbed by the water and cause a temperature change.

calculate the heat through the equation Q = mcΔT

Calorimetry

How much heat is required to increase the temperature of 20 grams of nickel (specific heat capacity 440Jkg-1 0C-1) from 500C to 700C?

Example

The standard enthalpy change of combustion for a substance is the heat released when 1 mole of a pure substance is completely burnt in excess oxygen under standard conditions.

Example,CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔHƟ

c=-698 kJmol-1

The heat given out is used to heat another

substance,e.g. water with a known specific heat capacity.

The experiment set-up can be used to determine the enthalpy change when 1 mole of a liquid is burnt.

Enthalpy change of combustion reactions

Example : refer to page 185

Problems with calorimetry Loss of heat to the surroundings (exothermic reaction);

absorption of heat from the surroundings (endothermic reaction). This can be reduced by insulating the calorimeter.

Using incorrect specific heat capacity in the calculation of heat change. If copper can is used, the s.h.c. of copper must be accounted for.

Others include – e.g incomplete combustion. Some of the ethanol could be used to produce CO & soot & water (less heat is given out)

Use bomb calorimeter – heavily insulated & substance is ignited electronically with good supply of oxygen

If 1g of methanol is burned to heat 100g of water, raising its temperature by 42K, calculate the enthalpy change when 1 mole of methanol is burnt.

Note: Specific heat capacity of water is 4.18 Jg-1 0C-1

Example

Practice questions page 187 #1-4

Enthalpy change of neutralisation (ΔHn)

The standard enthalpy change of neutraisation is the enthalpy change that takes place when 1 mole of H+ is completely neutralised by an alkali under standard conditions.

Example,NaOH(g) + HCl(g) NaCl(g) + H2O(l) ΔHƟ=-57 kJmol-1

The enthalpy change of neutralisation of a strong acid and a strong alkali is almost the same as they undergo complete ionisationof ions in water.

Reaction between strong acid and strong base involves H+(aq) + OH-(aq) H2O(l) ΔHƟ=-57 kJmol-1

For sulfuric acid, the enthalpy of neutralisation equation is ½ H2SO4(aq) + KOH(aq) ½K2SO4(aq) + H2O(l) ΔHƟ=-57 kJmol-1

Enthalpy change in solutions

Example : refer to page 188

Enthalpy change of neutralisation (ΔHn) The standard enthalpy change of neutraisation is the enthalpy change

that takes place when 1 mole of H+ is completely neutralised by an alkali under standard conditions.

Example,NaOH(g) + HCl(g) NaCl(g) + H2O(l) ΔHƟ=-57 kJmol-1

The enthalpy change of neutralisation of a strong acid and a strong alkali is almost the same as they undergo complete ionisationof ions in water.

Enthalpy change of solution (ΔHsol) The enthalpy change when 1 mol of solute is dissolved in excess solvent to form a

solution of ‘infnite dilution’ under standard conditions. NH4 NO3(s) in excess water NH4

+ (aq)+ NO 3

-(aq)

Enthalpy change in solutions

Example : refer to page 188

For neutralisation between a weak acid, a weak base or both, the enthalpy of neutraisation will be smaller than -57 kJmol-1 (less exothermic)

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)

ΔHƟ=-55.2 kJmol-1

Some of the energy released is used to ionise the acid.

200.0cm3 of 0.150 M HCl is mixed with 100.0cm3 of 0.350 M NaOH. The temperature rose by 1.360C. If both solutions were originally at the same temp, calculate the enthalpy change of neutralisation.

Assume that the density of the solution is 1 gcm-3 and the specific heat capacity is 4.18J Jg-1 0C-1.

Example

-56.8kJmol-1

The experimental change of neutralisation is -56.8 kJmol-1 The accepted literature value is -57.2 kJmol-1

Possible errors

(1) Heat loss to the environment. (2) Assumptions that (a) the denisty of NaOH and HCl solutions are the same as

water.(b) the specific heat capacity of the mixture are the same as

that of water

When 3 g of sodium carbonate are added to 50 cm3 of 1.0 M HCl, the temperature rises from 22.0 °C to 28.5°C. Calculate the enthalpy change for the reaction. Assume that the density of the solution is 1 gcm-3 and the specific heat capacity is

4.18J Jg-1 0C-1.

Example

Example : refer to page 189 dissolving ammonium chloride

The experimental change of solution is +13.8 kJmol-1 The accepted literature value is 15.2 kJmol-1

Possible errors (page 189)

(1) Absorption of heat from the environment. (2) Assumptions that the specific heat capacity of the solution

is the same as that of water(3) The mass of ammonium chloride is not taken into

consideration when working out the heat energy released.

100.0 cm3 of 0.100 mol dm-3 copper II sulphate solution is placed in a styrofoam cup. 1.30 g of powdered zinc is added and a single replacement reaction occurs. The temperature of the solution over time is shown in the graph below. Determine the enthalpy value for this reaction.

First step

Make sure you understand the graph.

Extrapolate to determine the change in temperature.

The extrapolation is necessary to compensate for heat loss while the reaction is occurring. Why would powdered zinc be used?

Example

Determine the limiting reactant

Calculate Q

Calculate the enthalpy for the reaction.

100.0 cm3 of 0.100 mol dm-3 copper II sulphate solution is placed in a styrofoam cup. 1.30 g of powdered zinc is added and a single replacement reaction occurs. The temperature of the solution over time is shown in the graph below. Determine the enthalpy value for this reaction.

The following measurements are taken:

Mass of cold water (g) Temperature rise of the water (0C) The loss of mass of the fuel (g)

We know that it takes 4.18J of energy to raise the temperature of 1g of water by 10C. This is called the specific heat capacity of water, c, and has a value of 4.18Jg-1K-1.

Hence, energy transferred can be calculated using: Energy transfer = mcΔT (joules)

If one mole of the fuel has a mass of M grams, then: Enthalpy transfer = m x 4.18 x T x M/y where y is mass loss of fuel.

Enthalpy changes of combustion of fuels

Given that:Vol of water = 100 cm3

Temp rise = 34.50CMass of methanol burned = 0.75gSpecific heat capacity of water = 4.18 Jg-10C-1

Calculate the molar enthalpy change of the combustion of methanol.

Example

What is the big assumption made with this type of experiment?

States that If a reaction consists of a number of steps, the

overall enthalpy change is equal to the sum of enthalpy of individual steps.

the overall enthalpy change in a reaction is constant, not dependent on the pathway take.

Hess’s Law

measured under standard conditions: pressure of 1 atmosphere (1.013 x 105 Pa), temperature of 250C (298K) and concentration of 1 moldm-1.

e.g.N2(g) + 3H2(g) 2NH3(g) ΔHƟ = -92 kJmol-1

The enthalpy change of reaction is -92 kJmol-1

92 kJ of heat energy are given out when 1 mol of nitrogen reacs with 3 mols of hydrogen to form 2 mols of ammonia.

Standard enthalpy changes, ΔHƟ

Calculate the enthalpy change for the formation of sodium chloride solution from solid sodium hydroxide.

NaOH(aq)

NaOH(s) NaCl(s) + H2O(l)

+ HCl(aq)

1. Indirect path: NaOH(s) + (aq) NaOH(aq) ΔHƟ1=-43kJmol-1

2. NaOH(aq) + HCl (aq) NaCl(aq) + H2O(l) ΔHƟ1=-57kJmol-1

3. NaOH(s) + HCl (aq) NaCl(aq) + H2O(l).

Reaction in aqueous soln

ΔH2ΔH1

Indirect path

Direct path

+ HCl(aq)+ H2O(l)

Calculate the enthalpy change for the combustion of carbon monoxide to form carbon dioxide.

C(s) + O2(g) CO2(g) ΔHƟ =-394 kJmol-1

2C(s) + O2(g) 2CO(g) ΔHƟ = -222kJmol-1

2CO(s) + O2(g) 2CO2(g)

2CO(g) + O2(g) 2CO2(g)

2C(s)+O2(g)+O2(g)

Combustion reaction (using cycles)

ΔHƟ

ΔHƟ

= -394kJmol-1 ΔHƟ

= -222kJmol-1

ΔHƟ = -(-222)+2(-394) = -566kJmol-1

Combustion reaction (manipulating equations)

Example : refer to page 196 evaporation of water & 197 formation of ethanol from ethene

Calculate the enthalpy change for the thermal decomposition of calcium carbonate.

CaCO3(s) CaO(s) + CO2(g)CaCO3(s) +2HCl(aq) CaCl2(aq) + H2O(l) ΔHƟ

1=-17 kJmol-1

CaO(s) +2HCl(aq) CaCl2(aq) + H2O(l) ΔHƟ1=-195kJmol-1

CaCO3(s) CaO(s) +CO2(g)

CaCl2(aq) + H2O(l) +CO2(g)

*Example : Decomposition reaction

ΔH

-195kJmol-1

Direct path

Indirect path

+ 2HCl(aq)-17 kJmol-1

Calculate the enthalpy of hydration of anhydrous copper(II)sulfate change.

CuSO4(s) +5H2O(l) CuSO4.5H2O (s)

CuSO4(s) +5H2O(l) CuSO4.5H2O (s)

Cu2+(aq) + SO42-

(aq)

*Example : Enthalpy of hydration of an anhydrous salt.

ΔH

ΔH2ΔH1

Direct pathway

Indirect pathway

From the following data at 250C and 1 atmosphere pressure:

Eqn 1: 2CO2(g) 2CO(g) + O2(g) ΔHƟ=566 kJmol-1

Eqn 2: 3CO(g) + O3(g) 3CO2(g) ΔHƟ=-992 kJmol-1

Calculate the enthalpy change calculated for the conversion of oxygen to 1 mole of ozone,i.e. for the reaction O2(g) O3 (g)

*Example

3

2

Calculate the enthalpy change for the conversion of graphite to diamond under standard thermodynamic conditions. C (s,graphite) + O2(g) CO2 (g) ΔHƟ=-393 kJmol-1

C (s, diamond) + O2(g) CO2(g) ΔHƟ=-395 kJmol-1

*Example

Practice questions page 199 #7-9

Enthalpy changes can also be calculated directly from bond enthalpies.

The bond enthalpy is the amount of energy required to break one mole of a specified covalent bond in the gaseous state.

For diatomic molecule the bond enthalpy is defined as the enthalpy change for the process X-Y(g) X(g) + Y(g) [gaseous state]

Bond enthalpies (Bond energies)

Bond enthalpy can only be calculated for substances in the gaseous state.

Br2(l) 2Br(g) ΔHƟ= 224 kJmol-1

Bond Enthalpies

Br2(l) 2Br(g)

Br2(g)

Br-Br bond enthalpy

ΔH Ɵvap

enthalpy change of vaporisation

2 x ΔH Ɵat

atomisation

Energy must be supplied to break the van der Waals’ forces between the Bromine molecules and to break the Br-Br bonds. Endothermic process

Ave bond enthalpies are enthalpies calculated from a range of compounds,eg C-H bond enthalpy is based on the ave bond energies in CH4 , alkanes and other hydrocarbons.

Average bond enthalpies

Some average bond enthalpies

Bond Ave bond enthalpy, ΔHƟ (Kjmol-1)

Bond length (nm)

H-H 436 0.07

C-C 348 0.15

C-H 412 0.11

O-H 463 0.10

N-H 388 0.10

N-N 163 0.15

C=C 612 0.13

O=O 496 0.12

C Ξ C 837 0.12

NΞN 944

Refer to page 201

Bond breaking and FormingWhen a hydrocarbon e.g. methane (CH4)

burns, CH4 + O2 CO2 + H2O

What happens?

C

H

H

H

H

+O

O

O

O

C

HH

H

HO

OO

O

C

H

H H

HO

O

OO

ENERGYENERGY

Enthalpy Level (KJ)

Progress of Reaction

2 O=O

Bond Breaking

Bond Forming

4 C-H

4 H-O 2 C=O

CH4 + 2O2 CO2 + 2H2O

Bond breaking and Forming

CH4 + 2O2 CO2 + 2H2O

C

H

H

H

H

+O

O

O

O

C

H

H H

HO

O

OO

Why is this an exothermic reaction (produces heat)?

BondAve Bond Enthalpy (kJ/mol)

C-H 412

H-O 463

O=O 496

C=O 743

CH4 + 2O2 CO2 + 2H2O

Energy absorbed when bonds are broken = (4 x C-H + 2 x O=O)

Energy given out when bonds are formed = ( 2 x C=O + 4 x H-O)

= 4 x 412 + 2 x 496 = 2640 kJ/mol

CH

HH

H

+O

O

O

O

CH

H H

HO

O

OO

Break Form

= 2 x 803 + 4 x 464 = 3338 kJ/mol

Energy absorbed when bonds are broken (a) = 2640 kJ/mol

Energy released when bonds are formed (b)= 3338 kJ/mol

Enthalpy change, ΔH = ∑(bonds broken) - ∑(bonds made) = a + (-b)= 2640 – 3338= -698 kJ/mol

Why is this an exothermic reaction (produces heat)?

What can be said about the hydrogenation reaction of ethene?

H H C=C (g) + H-H (g) H-C-C-H (g) H H

Example

H H

H H

What can be said about the combustion of hydrazine in oxygen?

H H N-N (g) + O=O (g) NΞN (g) + 2 O (g) H H

Example

H H

ExampleCalculate the mean Cl-F bond enthalpy given that Cl2(g) + 3F2(g) 2ClF3(g) ΔHƟ= -164 kJmol-1

Bond enthalpy for Cl-Cl = 242 kJmol-1 and F-F = 158 kJmol-1

The bond enthalpy from an enthalpy change of reaction

Standard enthalpy change of atomisation is the enthalpy change when 1 mole of gaseous atoms is formed from the element under standard conditions.

Example C(s) C(g) Calculate the enthalpy change for the process 3 C(s) + 4H2(g) C3H8(g) ΔHƟ= -164 kJmol-1

Bond enthalpy for C-H = 412 kJmol-1 , H-H = 436 kJmol-1 and C-C = 348 kJmol-1

Using bond enthalpies & enthalpies of atomisation

(ΔH Ɵat )

ΔH Ɵat = 715 kJmol-1

Practice questions page 206 #10,12,13

The combustion of both C and CO to form CO2 can be measured easily but the combustion of C to CO cannot. This can be represented by the energy cycle.

ΔHx = -393 – (-283)

= - 110 kJmol-1

Hess’s Law - example

ΔHx

½O2(g)-393kJmol-1

C(s)+ ½O2(g) CO(g)

CO2(g)-283kJmol-1

½O2(g)

Calculate the standard enthalpy change when one mole of methane is formed from its elements in their standard states. The standard enthalpies of combustion of carbon, hydrogen and methane are -393, -286 and -890 kJmol-1 respectively.

Hess’s Law - example