Heat of Reaction | Enthalpy of Formation

Heat of reaction in term of xothermic and endothermic

Here we discussed the enthalpy of the reaction, enthalpy of formation, and enthalpy of combustion. The heat of reaction is evolved or absorbed during the chemical reaction. In a chemical reaction, old bonds of reactants are broken and new bonds of products are formed simultaneously. The heat is required to break the bonds of the reactants so this process is endothermic and heat is released during the bond formation so it is exothermic in nature. Let us a general chemical reaction:

Reactants Products

∆H = HP – HR

  • ∆H = Change in heat or enthalpy change
  • HP = Heat content of the product
  • HR = Heat content of the reactant

∆H may be positive or negative. In an exothermic reaction, energy is released so ∆H is negative. On the other hand, in an endothermic reaction, energy is absorbed so ∆H is positive.

∆H negative energy released exothermic HR>HP

∆H positive energy absorbed endothermic HR<HP

Graph of Heat of Reaction

Graph of Exothermic and Endothermic Reaction

Graphs of heat of reaction (exothermic and endothermic)

Let us a general example to understand the concept of endothermic and exothermic reactions:

A + B C + D + “x” Kcal ∆H = -“x” Kcal exothermic

A + B C + D – “x” Kcal ∆H = +”x” Kcal endothermic

we can also write as

A + B + “x” Kcal C + D ∆H = +”x” Kcal endothermic

Factors on which ∆H depends

1) The physical state of reactants and products

H2 (g) + 1/2 O2 (g) H2O (g) ∆H1

H2 (g) + 1/2 O2 (g) H2O (l) ∆H2

Where ∆H1≠ ∆H2

2) Allotropic form of reactants and products

C(diamond) + O2 CO2 (g) ∆H1

C(amorphous) + O2 CO2 (g) ∆H2

Where ∆H1≠ ∆H2

3) Solution in which reaction takes place (solvent)

4) Temperature and pressure (physical conditions)

Standard Heat of Reaction

The condition for standard enthalpy of reaction is at 298 K temperature, 1 atm pressure, and the concentration of an aqueous solution is 1 molar (1 mol/L). It is represented as ∆H°.

Let the standard enthalpy of some naturally occurring state substances is zero.

  • ∆H°Cl2 = 0
  • ∆H°O2 = 0
  • ∆H°H2 = 0
  • ∆H°Br2 = 0
  • ∆H°N2 = 0
  • ∆H°C (graphite) = 0

Carbon in graphite form is naturally occurring so the standard enthalpy of carbon in graphite form is zero but in diamond, the ∆H° is not zero.

  • ∆H°S8 (orthorhombic) = 0
  • ∆H°P4 (white) = 0

Enthalpy of Formation

The amount of heat absorbed or released when one mole of a compound is formed by its constituent elements in their free natural state.

H2 + 1/2 O2 H2O (g) ∆H = -241.82 kJ/mol

The enthalpy of the formation of water is the enthalpy of water. Let the reaction takes place at standard state. (1 atm, 25°C)

∆H = HP – HR

∆H°f = HH2O – (HH2 + HO2)

∆H°f = HH2O – (0)

∆H°f = HH2O

So enthalpy of any component is the enthalpy of the formation of that compound.

In the formation of CO2, at 1 atm and 25°C, when one mole of carbon reacts with one mole of oxygen to form one mole of carbon dioxide. The heat of the formation of CO2 is -393.5 kJ/mol.

C(s) + O2 CO2 ∆H = -393.5 kJ/mol

∆H = HP – HR

∆H°f = HCO2 – (HC + HO2)

∆H°f = HCO2 – (0)

∆H°f = HCO2

The enthalpy of formation of carbon dioxide is the enthalpy of carbon dioxide.

When 1 mole of H2 reacts with one mole of Cl2, 2 moles of HCl are produced. The ∆H for the reaction is -44 Kcal.

H2 + Cl2 2HCl ∆H = -44 kcal

Enthalpy is an extensive property. and heat of formation is defined for only one mole.

∆H°f (HCl) = -22 kcal

∆H = HP – HR

-44 = 2HHCl – (0)

-22 kcal = HHCl

The enthalpy of HCl is the enthalpy of formation of HCl.

Numericals

Calculate enthalpy change (∆H) for the given reaction:

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

The enthalpy of formation of CH4 is -74.5 KJ/mol, CO2 is -393.5 kJ/mol, and H2O is -285.8 KJ/mol.

∆H = HP – HR

∆H = (HCO2 + 2HH2O) – (HCH4 + 2HO2)

∆H = (-393.5 + 2×-285.5) – (-74.5 + 2×0)

∆H = -964.5 + 74.5

∆H = -890 KJ/mol

Calculate ∆H and∆U for the given reaction:

NH4NO3(s)⟶ N2​O(g)+2H2​O(l)

The heat of formation of NH4NO3(s), N2O, and H2O are −367.5 kJ, +81.46 kJ, and −285.78 kJ respectively at 1 atm and 25°C.

∆H = HP – HR

∆H = (HN2O + 2HH2O) – (HNH4NO3)

∆H = (+81.46 + 2×-285.5) – (-367.5)

∆H = -490.1 + 367.5

∆H = -122.6 KJ/mol

As

ΔH=ΔU + ΔngRT

ΔU=ΔH − ΔngRT

Δng = np-nR

Δng = 1 – 0 = 1

ΔU=-122.6 − (1) (8.314) (298) / 1000

ΔU=-122.6 − 2.47

ΔU= -125.07 KJ/mol

Enthalpy of Combustion

The amount of heat released when 1 mole of a compound is combusted completely in excess of air or oxygen. It is represented by ΔHc. ΔHc is always negative or exothermic in nature.

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ΔHc = -891 KJ/mol

The calorific value gives an idea about which fuel is better in terms of energy.

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