5. ENTHALPIES FOR DIFFERENT TYPES OF REACTIONS

ENTHALPIES FOR DIFFERENT TYPES OF REACTIONS

• Standard Enthalpy of Formation
Enthalpy of formation is defined as the change in enthalpy in the formation of 1 mole of a substance from its constituting elements under standard conditions of temperature at 298K and 1 atm pressure.

ΔH_r^º (298.15 K) and ΔG_r^º (298.15 K)

ΔG_r^º (T)

ΔH_r^º (T) and K_p (298.15 K)

K_p (T₁) and K_p (T₂) where T₁, T₂ ≠ T

ΔH_r^º (T) and ΔG_r^º (298.15 K)

Enthalpy of Combustion: It is defined as the heat energy or change in enthalpy that accompanies the combustion of 1 mole of a substance in excess of air or oxygen.

CO_(g) + ½ O_2(g)  CO_2(g) ΔHº = -283 kJ

• Thermochemical Equation
A balanced chemical equation together with the value of Δ_rH and the physical state of reactants and products is known as thermochemical equation.

Thermochemical equations

•  A typical chemical equation is

   S + O₂  SO₂

• It is called a thermochemical equation when we add information about ΔH…

   S + O₂ SO₂     ΔH = -296.9 kJ

• If we change the equation, then the ΔH also changes…

  SO₂  S + O₂       ΔH = +296.9 kJ

• If the reaction is reversed the sign is reversed
• Also, if numbers in the equation change, so will the amount of energy produced/absorbed:

   2S + 2O₂  2SO₂     ΔH = -593.8 kJ

Conventions regarding thermochemical equations
1. The coefficients in a balanced thermochemical equation refer to the number of moles of reactants and products involved in the reaction.

If the coefficients of the chemical equation are multiplied by some factor, the enthalpy change must also be multiplied by the same factor (ΔH is an extensive property).

H_2(g) + ½ O_2(g)  H₂O_(l)  ΔH = -286 kJ

Multiply by a factor of two:

2H_2(g) +  O_2(g) 2H₂O_(l)  ΔH = 2 x (-286kJ) = -572kJ

Multiply by a factor of one-half:

½ H_2(g) + ¼ O_2(g)  ½ H₂O_(l)  ΔH = ½ x (-286kJ) = -134kJ

• Hess’s Law of Constant Heat Summation
The total amount of heat evolved or absorbed in a reaction is same whether the reaction takes place in one step or in number of steps.

The law stats that the change in enthalpy for a reaction is the same whether the reaction takes place in one or a series of step. The Hess’s law can also be states as the enthalpy change for a chemical reaction is the same regardless of the path by which the reaction occurs.

For example, consider following two paths for the preparation of methylene chloride

Path I:

CH_{4 (g)} + 2 Cl_2(g) CH₂Cl_2(g) + 2HCl_(g)   ΔH₁^º = - 202.3kJ

Path II:

CH_{4 (g)} + Cl_2(g)  CH₃Cl_2(g) + HCl_(g)   ΔH₂^º = - 98.3kJ

CH₃Cl_(g) + Cl_2(g)   CH₂Cl_2(g) + HCl_(g)  ΔH₃^º = - 104.0kJ

Adding two steps

CH_4(g) + 2Cl_2(g)  CH₂Cl_2(g) + 2HCl_(g)  ΔH^º = - 202.3kJ

Thus whether we follow path 1 or path 2 the enthalpy change of the reaction is same.

ΔH₁^º = ΔH₂^º + ΔH₃^º = - 202.3kJ

• Born-Haber Cycle
It is not possible to determine the Lattice enthalpy of ionic compound by direct experiment. Thus, it can be calculated by following steps. The diagrams which show these steps is known as Born-Haber Cycle.