Thermochemistry

Thermochemistry

Azim khan

Bond Enthalpy or Bond energy• Amount of heat require to break 1 mole of

particular covalent bonds of gaseous molecule forming free gaseous atoms or radicals at constant temperature and pressure.

Bond dissociation enthalpyAmount of energy require to break 1 mole of of a particular bond of a particular polyatomic gaseous molecules forming free gaseous atoms and radicals at constant temperature and pressure.STEP 1: CH4(g) → CH3(g) + H(g) ; Δbond H° = +427 KJmol-1

STEP 2: CH3(g) → CH2(g) + H(g) ; Δbond H° = +439 KJmol-1

STEP 3: CH2(g) → CH(g) + H(g) ; Δbond H° = +452 KJmol-1

STEP 4: CH(g) → C(g) + H(g) ; Δbond H° = +347 KJmol-1

Average bond enthalpy ΔC-HH° = ¼ (ΔaH°) = ¼ (1665 KJmol-1) = 416 KJmol-1

Hess’s law of constant heat summation

• The heat of a reaction or the enthalpy change in chemical reaction depend upon initial state of reactants and final state of products and independent of the path which the reaction is brought about.

• Heat of reaction is same whether it carried out in Single step or series of step.

• State function.

• Example- A → C ∆rH˚• Same conversion carried out in 2 steps as,Reaction Enthalpy change

A → B ∆rH˚1

B → C ∆rH˚2

Then by Hess’s law, ∆rH˚ = (∆rH˚)1 + (∆rH˚)2

Hess’s law of constant heat summation

Application of Hess's law

• Hess’s law used for,• To calculate heat of formation, combustion,

neutralization, ionization etc• To calculate the heat of reaction which may not

take place normally or directly.• To calculate heat of extremely slow or fast

reactions.• To calculate enthalpies of reactants and

product.

Spontaneous process

• Process that take place on its own or without the intervention of the external agency or influence.• Example-• Flow of energy from higher pressure

to low pressure or a flow of heat from higher temperature to lower temperature.

• One the spontaneous process starts, it proceeds without the continuous external help.

• Product is more stable then reactant.• Product should have less energy than the reactants.• It take place direction in which energy is low.• Generally exothermic but some is endothermic. Eg.

Melting of ice.• Process proceed till an equilibrium is reached.• May be fast or slow.• Decrease enthalpy increase entropy.

Spontaneous process

• Example-• All natural process.• Flow of water.• Flow of heat.

Spontaneous process

Non- spontaneous process

• Process which does not take place on its own but takes place only with the intervention of external agency or influence.• Not natural eg. Compression of gas.• Endothermic.• Increase enthalpy, decrease entropy.

Order in a system

• when the atom, molecules or ions are arranged in perfect order then the system said to be order.

• Decrease entropy is a measure of order in sytsem.

Disorder in a system

• Molecule or ion are free to move randomly, then the system is said to disorder.

• Entropy high.• With disorder, energy of the system increases.• Natural process.• Eg. Flow of gas from high pressure to low

pressure.

Entropy• Entropy change(∆S) of a system in a process

which is equal to the amount of heat transferred in a reversible manner(qrev) divided by the absolute temperature(T), at which the heat is absorbed.

• ∆S = = • 1 e.u. = 1 JK-1

• Entropy measure the disorder in the system.• Higher the disorder more is entropy

Entropy change for a phase change

• Phase change physical state of matter.• During phase change, both the phases

exist at equilibrium and temperature remain constant.• ∆S = Solid to liquid• ∆S = liquid to gas• ∆S = solid to gas

Different process accompanying entropy change

• Solid to liquid to gas entropy increases.• Solid or liquid dissolve in a solvent, entropy of the

substance increase.• When gas dissolve in solvent, entropy decreases.• When the gas mix, entropy increase.• Increase molecular complexity result in the increase in

entropy.• Spontaneous process, expansion of gas, disorder of

the system increase entropy.• Eg. When the Tr leave classroom entropy increase.

Entropy and spontaneity of the process

• Spontaneous process result in disorders, hence increase in entropy(∆S>0).

• Take place with Decrease in free energy.• There are some process which are opposite.• Eg neutrilization reaction. Entropy decreases.• H(aq) + OH(aq) → H2O(l) ∆S<0

2nd law of thermodynamic• Total entropy of the system and its

surrounding increases in a spontaneous process.

• All natural process are spontaneous, the entropy of the universe increases.

• ∆STotal = ∆Ssystem + ∆Ssurrounding > 0• ∆Suniverse = ∆Ssystem + ∆Ssurrounding > 0

Gibbs free energy

• G = H – TS• At constant T and P, change in free

energy G for the system is, • ∆G = ∆H – T∆S.• Unit- SI unit is J ot kJ CGS unit is cal or kcal.

Name of the reaction

∆H ∆S ∆G Spontaneity of reaction

1)Exothermic -ve -ve -ve at low T

Spontaneous at low T

2)Endothermic +ve +ve -ve at high T

Spontaneous at high T

3) Exothermic -ve +ve -ve at all T Spontaneous at all T

4)Endothermic +ve -ve + at all T Non- spontaneous at all T

Relation between ∆G˚ and K• The free energy of any substance at a T is

represented as,• G= G˚ + RT ln[substance]• Consider a reversible reaction• A + B = C + D• ∆G = ∑Gproduct - ∑Greactanct

• ∆G = ∆G˚ + RTlnk since at equilibrium, ∆G=0

• ∆G˚ = -2.303 RT log10 k

Temperature condition for equilibrium

• In every thermodynamic process, a system has a tendency to equilibrium in final state.

• Weather final state is spontaneous or non- spontaneous both are balanced.

A B

• Free energy change ∆G is,• ∆G = ∆H - T∆S• Since at equilibrium, (i.e no energy gain or loose)

∆G=0.• 0 = ∆H - T∆S• T∆S = ∆H or T = • From this condition, ∆H & ∆S value depend upon

Temperature. But T is independent.• Hence, value ∆H & ∆S, the temperature of

equilibrium can be predicted.

Temperature condition for equilibrium

Third law of thermodynamic

Important of 3rd law of thermodynamic

• It gives the starting place from which to measure entropy.

• Entropy of a substance determine at any temperature higher then 0K at any state.

• Stand molar entropy (S˚) can be measure at 25˚C and 1 atm.

• Spontaneity of the reaction and ∆S˚ can be calculated.

At T=25˚C and pressure 1atm.

Entropy change at different T and constant pressure

• ∆S = qrev/ T = • For infinitesimal change, dS = .• If Cp = molar heat capacity at const P, then• Cp = or dH = Cp.dT• dS = • If temperature change from T1 to T2, then the

entropy change ∆S will be.• ∆S = • ∆S = 2.303 Cp log

Entropy change at different T and constant pressure

Usefulness of Standard molar entropy

• All reactant and products in chemical reaction, standard entropy of the reaction can be calculated.

• From entropy change of a reaction, the entropy of a reactant or product can be calculated.

• Spontaneity of the process can be predicated.

Residual entropy of the substance

• According to 3rd law of thermodynamic. S=0, T=0.

• The entropy of the substance is greater then zero at T=0.

• CO2 freezes it easily disorder.• If a crystal contains some impurities,

(substance not perfectly ordered due to different orientation of molecule).