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Thermochemistry

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It is the branch of chemistry which studies the energy changes in chemical reactions

Enthalpy Changes

figure A                                                 figure B

Endothermic reactions: (Figure A)

The reactants have LESS energy than the products, so they absorbed the energy from the surroundings, or we can say that there is a positive enthalpy change.

Exothermic reactions: (Figure B)

The  reactants have MORE energy than the products, so the reactants RELEASED energy to the surroundings, or we can say that  there is a negative enthalpy change.

Standard Enthalpy Change: ΔH^θ

The standard enthalpy change, that we represent “ΔH^θ“ is the one that occurs under standard conditions of pressure and temperature:

• Temperature: 298 K = 25 °C
• Pressure: 100 kPa (1 atm = 101 kPa)

IMPORTANT ENTHALPY CHANGES

Standard Enthalpy change of Combustion: ΔH^θ_c

The standard enthalpy change of combustion is the enthalpy change when 1 mole of a compound is burnt completely in oxygen under standard conditions (298K and 100kPa), all reactants and products being in their standard state.

CH₄(g) + O₂(g) → CO₂ (g)  +  2H₂O(g)     ΔHºc = -882 kJ mol–1

Standard Enthalpy change of Formation: ΔH^θ_f

The standard enthalpy change of formation is the enthalpy change when 1 mole of a compound is formed from its elements under standard conditions (298K and 100kPa), all reactants and products being in their standard state.

H₂(g) + ½O₂(g) → H₂O(l)      ΔHº_f = –285.8 kJ mol^–1

Standard Enthalpy change of reaction: ΔH^θ_r

The standard enthalpy change of reaction is the enthalpy change when the amounts of reactants (as they appear in the balanced equation) react to give products under standard conditions (298K and 100kPa), all reactants and products being in their standard state.

H₂(g) + F₂(g) = 2 HF     ΔHºr = -542 kJ

(notice that this is not the energy to form 1 mole but the amount of moles given in the stoichiometry balanced equation)

Standard Enthalpy change of Atomisation:ΔH^θ_at

Enthalpy change that occurs when 1 mole of gaseous atoms are formed from the element in its standard state.

Na(s) Na(g) ΔH = +107 kJ mol^-1

(notice that this energy will form ATOMS from molecules or lattices)

Standard Enthalpy change of Solution:ΔH^θ_sol

Enthalpy change that takes place when 1 mole of solute is dissolved in a solvent to form a dilute solution in standart conditions.

CaCl_2(s) Ca²⁺_(aq)+ 2Cl^–_(aq)ΔH_sol = -120 kJ mol^-1

At first sight, dissolution is a simple process, however it involves at least two stages.

• • 1 The solute bonds must be broken. (In this case, it is an IONIC COMPOUND so the energy required to brake the bonds is called LATTICE ENTHALPY)

CaCl₂(s)  Ca²⁺(g) + 2Cl^–(g)

• • 2 The separated solute particles must bond to the water (or solvent) molecules.

Ca²⁺(g) + 2Cl^–(g) + (aq) Ca²⁺(aq) + 2Cl^–(aq)

Some solution Enthalpies

Compound	ΔH(solution)/kJ mol-1	Compound	ΔH(solution)/kJ mol-1
Sodium hydroxide	-44.4	Ammonium chloride	14.6
Potassium hydroxide	-57.6	Ammonium nitrate	25.7
Sodium chloride	3.9	sulfuric acid	-96.2
Sodium bromide	-0.6	Nitric acid	-33.3
Sodium iodide	-7.5	Hydrogen fluoride	-61.5
Sodium fluoride	1.9	Hydrogen chloride	-74.8
Sodium chloride	3.9	Hydrogen bromide	-85.1
Sodium bromide	-0.6	Hydrogen iodide	-81.7
Sodium iodide	-7.5	Ethanoic acid	-1.5
Ref: CRC Handbook of chemistry and physics – Edition 44

Enthalpy of IonisationΔH^θ_i#

This is defined as the energy required to remove 1 mole of electrons from 1 mole of gaseous particles producing 1 mole of ions.

H(g)  H⁺(g) + e^–           ΔH^θ_i = -1312.0 kJ mol^-1

It can be stated as 1st ionisation energy, 2nd ionisation energy etc.

The 1st ionisation energy (ΔH^θ_i1)is the energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to produce 1 mole of singly charged positive ions.

Na(g)  Na⁺(g) + 1 e-         ΔH^θ_i1 = +496 kJ mol^-1

Remember : The value is quoted per mole of species and The particles must be gaseous

The 2nd ionisation energy (ΔH^θ_i2)is the energy required to remove 1 mole of electrons from 1 mole of gaseous singly charged atoms to produce 1 mole of doubled charged positive ions.

Na⁺(g) Na²⁺(g) + 1 e-ΔH^θ_i2 = +4562 kJ mol^-1

We can observe that the value for getting the second electron from the sodium ion is significantly larger than the first ionization for sodium. This happens because the first electron in sodium is in an outer level of energy and the Na+ is stable so it requires a much larger energy to get that second electron.

Predicting the position of an element based on its ionization energy values

We can infer in which group of the periodic table an element would be by knowing its successive ionization energy values.

Example: In the graph below you observe that the values for the first four ionization energies change in a constant rate but between the 4th and 5th there is a very big jump. That is telling us that we are in an inner level so the element has 4 electrons it its last level of energy and so it belongs in group IV.

Bond Dissociation Enthalpy:ΔHθ_bond

Energy required to break and separate one mole of bond so that the gaseous atoms exert no force on each other.

Bond dissociation is always an ENDOTHERMIC REACTION since we need to GIVE energy to break bonds.

Bond FORMING is always an EXOTHERMIC REACTION since the compounds are more stable than the single atoms.

Mean Values         .

H-H  436       H-F  562       N-N  163

C-C  346       H-Cl  431       N=N  409

C=C  611      H-Br  366      N=N  944

C=C  837      H-I  299       P-P  172

C-O  360       H-N  388      F-F  158

C=O  743      H-O  463      Cl-Cl  242

C-H  413       H-S  338      Br-Br  193

C-N  305       H-Si  318      I-I  151

C-F  484       P-H  322       S-S  264

C-Cl  338       O-O  146       Si-Si  176

Thermochemistry Calculations

1) CALORIMETRY

Calorimetry is the technique we use in the lab to measure the change in the temperature produced by energy released or absorbed in a chemical reaction. For this technique, a calorimeter is used. There are different types of calorimeters. The most common is the one in the picture below which consist of two polystyrene (styrofoam) cups, a cover, a thermometer and a stirrer. 

The principle of the calorimeter is that the known solutions will react to form products and heat will be absorbed or released. The water contained in the calorimeter (in the solutions) will be the recipient for that change in temperature, so

• If the reaction is exothermic, the reaction will release energy TO the water and the water temperature will INCREASE 
• If the reaction is endothermic, the reaction will absorb the energy FROM the water temperature will DECREASE 

Calorimetry relies on the following facts:

a) 1 gram of water needs 4.18 J (1 cal)  of energy to increase the temperature in 1 °C, so knowing the mass of water and the change in the temperature, we can calculate the amount of energy change. we can represent this change with the following formula: 

 Where  

• • Q represents the total energy absorbed or released BY THE WATER
  • m is the mass of water in the calorimeter *
  • Δt is the change in temperature we measured (always final – initial)

(most of the time, at this level in chemistry,  we assume that the density of the solution is 1 g ml^-1 so the mass of the solution will numerically equal to the volume of the solution.

B)Since we are in a closed system, the change in energy inside the calorimeter must add up to zero (all energy released by the reactants must be absorbed by the water and the calorimeter and vice versa. We do not calculate the energy absorbed or release by the calorimeter so the equations can be written as follows:

 Q_water + Q _reaction = 0      or        Q_water  = – Q _reaction

Enthalpy Changes

If the amount of energy is measured at constant pressure we call it ENTHALPY (H) and ENTHALPY CHANGE is represented as  “ΔH”

We can then express the formulas above as follows:

ΔH = m x c x Δt

and 

 ΔH_water + ΔH _reaction = 0      or        ΔH_water  = – ΔH _reaction

Examples of exercises:

A)Calculation of simple problems

Calculate the energy change, in joules, when 100 cm³ of water rises from 22 °C to 45 °C. (density of water = 1gcm^-1)

Q = m c Δt 

Q =100 g 4.18 J g^-1 °C^-1 (45-22)°C

Q = 9164 J

B) Calculating enthalpies of reactions

5g of ammonium nitrate (NH₄NO₃) was dissolved in 50cm³ of water. The temperature changed from 22^oC to 14^oC.

ΔH_water  = – ΔH _reaction

ΔH_water =50 g 4.18 J g^-1 °C^-1 (14-22)°C= – ΔH _reaction

ΔH_water = –1680 J = – ΔH _reaction

ΔH _reaction = + 1680 J .

(heat absorbed by dissolving 5 g of ammonium nitrate)

C) Calculation Molar enthalpies

For the example B, calculate the Molar Enthalpy change (in KJ/mol) for the same reaction

We need to calculate how many moles of NH₄NO₃ reacted and then make the proportion for 1 MOLE of substance:

M_r(NH₄NO₃) = 14 + (1 x 4) + 14 + (3 x 16) = 80, so 1 mol of NH₄NO₃ = 80g

It means that 5g absorbed 1680 J so 80 g will absorb:

5g  ——————  1680 J

80 g  —————-  x 

x = 1680 J x 80 g mol^-1 / 5 g

x = 26880 J/mol or 26.89 kJ/mol

Or, we calculate first the amount of moles that the 5 g represent:

n = m/Mr   ==>  n = 5g /80g  = n= 0.0625 moles

It means that 0.0625 moles absorbed 1680 J so 1 mol will absorb:

0.0625 moles  ——————  1680 J

1 mole  ——————  x

x = 1680J / 0.0625 mol

x = 26880 J/mol or 26.88 kJ/mol

D) Calculating enthalpies of Neutralization (YOU NEED TO KNOW THIS)

The reaction between sodium hydroxide and hydrochloric acid is  NaOH + HCl ==> NaCl + H₂O.  A student carried the following experiment:

He placed 50 cm³ HCl(aq) 2.0 mol dm^-3  in a styrofoam calorimeter. Placed the thermometer and read the temperature after 3 minutes. The initial temperature of the acid was 20^oC. Then, he added 50 cm³ NaOH(aq) 2.0 mol dm^-3 to the calorimeter.  After the addition of the base, the temperature rise to 33.5  ^oC. 

Calculate the enthalpy of neutralisation. Give your answer in kJ mol^-1 to 1 decimal place.

1) calculate the amount energy absorbed by the water in the solution

Note: the mass of water in the reaction is the sum of the masses of each solution, in this case, we have two solutions of equal volume: 50 cm³, so the mass of water will be 100 g. The temperature recorded is the change in temperature of the water, so we can express the equation as follows: 

ΔH_water  = m c Δt  

ΔH_water =100 g 4.18 J g^-1 °C^-1 (33.5-20)°C 

ΔH_water = 5643 J

(This is the amount of energy absorbed by the water, since the reaction produced an increment in temperature, so it is exothermic) 

2) calculate the amount energy released  by the reaction in the solution

ΔH_water  = – ΔH _reaction

ΔH _reaction = – 5643 J or -5.643 kJ

(This is the amount of energy  released by the reaction, since it is exothermic)

3) calculate the amount energy released per mol of reactants based on the amount of moles that reacted in the solution

#of moles of acid  =  V _acid X M _acid  = 0.050 dm³ _x 2.0 mol dm^-3 = 0.1 mol of acid

 #of moles of base  = V _base X M _base = 0.050 dm³ _x 2.0 mol dm^-3 = 0.1 mol of base

since the molar ratio is 1/1 for the reaction: 0.1 moles of water will be formed. 

 NaOH      +       HCl      ==>      NaCl    +     H₂O.

1 mol     +       1 mol     ==>    1 mol       +   1 mol    

we can say that the proportion is the same for the reaction we produced

0.1 mol      +   0.1 mol     ==>    0.1 mol     +  0.1 mol    

The energy released by the reaction corresponds to 0.1 mol of water formed. 

0.1 moles  —————— -5.643 kJ

1 mole  ——————  x = – 56.43 kJ

or 56.4 kJ (to one decimal place)

Enthalpy change of combustion of fuels

When measuring the energy produced from a compound, water is used to measure the amount of heat; this is because of its specific absorption of energy. The experiment  is very inefficient. This is because not all of the energy is absorbed by the water; some is absorbed by the air, while some is absorbed by the container holding the water. Also, the ethanol may not have entirely reacted. 

To work out the enthalpy change:

1)Calculate the heat given out by the fuel, this can be done by calculating the amount of energy absorbed by the water,

ΔH_water  = m c Δt 

and

ΔH_water  = –ΔH_alcohol

2)Calculate the number of moles of fuel that has been burnt. This can be done by working out the difference between the mass of fuel before and after the burning. 

n =Mass of alcohol burnt
          Mr of alcohol

• Calculate the energy given out by 1 mole of the substance through proportions

ΔH_c =    ΔH measured   (alcohol)
              Number of moles burnt

Example of exercises: 

A student carried the following experiment:

He placed 100 cm³ of water into the beaker. The initial temperature of the water was recorded as 18^oC. The spirit burner containing the ethanol (C₂H₅OH) had a mass 18.62g. After burning its mass was 17.14g. The maximum temperature of the water reached 89^oC.

1) calculate the amount of energy absorbed by the water 

ΔH = m x c x Δt

ΔH_water =100 g 4.18 J g^-1 °C^-1 (89-18)°C

ΔH_water =29678 J

and

 ΔH_water =- ΔH _reaction

ΔH _reaction =  -29.678 kJ

2) Calculate the number of moles of fuel burnt

a)calculate the mass of alcohol burnt

 mass of alcohol burnt= final mass – initial mass = 18.62 g  – 17.14 g 

 mass of alcohol burnt= 1.48g

b) calculate the molar mass of the alcohol

M_r(C₂H₅OH) = (2 x 12) + (1 x 5) + 16 + 16 = 46, so 1 mole = 46g

 1 mole(C₂H₅OH) =46 g/mol

c) calculate the number of moles of alcohol burnt.

n = m/Mr 

n= 1.48 g / 46 g/mol

n= 0.0322 moles

3) calculate the energy released by mole of alcohol burnt

0.0322 moles   ——————    -29.687 kJ

1 mole   ——————-     = x 

x=  -29.687kJ / 0.0322 mol

x= 922.kj/mol

The data book value for the heat of combustion of ethanol is -1367 kJmol^-1, showing lots of heat loss in the experiment!

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2) CALCULATIONS WHEN ENTHALPY CHANGES ARE IMPOSSIBLE TO MEASURE DIRECTLY

Laplace Law

A.L. Lavoisier and P.S. Laplace gave this law in 1780 which states that “the enthalpy of a reaction is exactly equal but opposite in sign for the reverse reaction.”

For example, if ΔH is the enthalpy change in going from A to B then the enthalpy change for the process B to A would be -ΔH. Thus, the enthalpy of formation of a compound is numerically equal but opposite in sign to the enthalpy of decomposition of the compound.

SO₂(g) → S(s) + O₂(g)   ΔH = +296.9 kJ              S(s) + O₂(g) → SO₂(g)  ΔH = -296.9 kJ

 

Hess’s Law

Regardless which route a chemical change takes, the total energy change stays the same (as long as the initial and final conditions are the same).

One very useful form of Hess’s Law says that if you can get a chemical equation by adding up two or more separate chemical equations, then the heat of the reaction is equal to the sum of the heats of reactions involved.

example:

a)    A  +  B  ==>  C  +  D                                     ΔH = β

b)   C   +  J   ==>  K  +  B                                     ΔH = Ω

 

c)   D  +  M  ==>  J  +  R                                      ΔH = δ

ADDING a, b and c:               A +  B + C + J + D + M   ==>   C +D + K +  B + J + R              ΔH = β + Ω + δ

 

or                                                            A +  M   ==>  K  + R                                    ΔH = β + Ω + δ                                                 

for practice online click in the link below:

 

CHAPTER 5

Thermochemistry

ChemTours

• State Functions and Path Functions
• Internal Energy
• Pressure–Volume Work
• Heating Curves
• Calorimetry
• Hess’s Law

 

 

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A2 LEVEL

Electron affinity

The electron affinity of an element is the energy given off when a neutral atom in the gas phase gains an extra electron to form a negatively charged ion.

A fluorine atom in the gas phase, for example, gives off energy when it gains an electron to form a fluoride ion. Note that the molecule of Fluorine must be broken first into atoms. (Atomisation enthalpy)

F(g) + e^– F^–(g)             ΔH^θ_EA= -328.0 kJ/mol

Lattice Enthalpy

There are two ways of defining Lattice enthalpy

The lattice formation enthalpy refers to the energy RELEASED when 1 mole of solid crystal is formed from its scattered gaseous ions. (always exothermic)

Ca²⁺(g) + 2Cl^–(g)  CaCl₂(s)     ΔH^θ_lat= – 2195.2 kj/mol

The lattice dissociation enthalpy refers to the energy ABSORBED when 1 mole of solid crystal is broken into its scattered gaseous ions. (always endothermic)

CaCl₂(s)    Ca²⁺(g) + 2Cl^–(g) ΔH^θ_lat= + 2195.2 kj/mol