SureDenEnthalpy Change or Heat Change of Reaction
Enthalpy change or Heat change of Reaction
The heat of reaction is defined as heat evolved or absorbed when a certain number of moles of the reactant as expressed by balanced chemical equation for that reaction are completely converted into the product.
For e.g. 2H2 (g) + O2 (g) → 2H2O Δ H = -468.8 KJ ( exothermic reaction)
Factors on which heat or enthalpy of a reaction depends
(1) Physical state of Reactant or Product-
H2(g) + ½ O2(g) à H2O(g) Δ8H = - 248.8 KJ/mol
H2(g) + ½ O2(g) à H2O(L) Δ8H = 285.9 KJ/mol
H2O (L) → H2O (g) Δr H = 37.1 KJ/mol
(2). Allotropic form of the element-
Some heat energy is also involved when one allotropic form of an element changes into another
S (Rhombic) + O2 (g) → SO2 Δ H = 297.4 KJ
S (monoclinic) + O2 (g) → SO2 Δ H = 299.9 KJ
S (Rhombic) → S (Monoclinic) = 2.5 KJ
(3) Temperature- Enthalpy of a reaction also depend upon the temperature at which the reaction occur. Because variation in heat capacity with temperature. Therefore, the enthalpy of reaction are generally expressed at 298 K.
(4). Reaction being carried out at constant Volume or Constant pressure- When a chemical reaction is carried out at constant volume, the heat change is called heat of reaction at constant volume & is equal to change in internal energy. While most of the reaction carried out at constant pressure, then heat of reaction is equal to enthalpy change. This is because heat capacity of different substances is different at constant volume & constant pressure.
Enthalpy of Formation (Δ HF°)
Enthalpy of formation of substance is defined as the amount heat change that occurs when one mole of the substance is formed in its standard state from its constituent elements. (Also in their respective standard state)
For example- The enthalpy of formation of water is
H2(g) + ½ O2(g) à H2O(L)
Δ HF° = -285.8 KJ mol-1
Enthalpy of complicated reaction
Δ Hr° = ∑ Enthalpy of product - ∑enthalpy of reactant
Numerical- Calculate the standard Enthalpy Change & standard internal energy change for the reaction.
OF2 (g) + H2O (g) → O2 + 2HF
when a standard enthalpy of reaction of OF2 is + 23 KJ mol
H2O (g) = -241.8 KJ mol-1
HF (g) = -268.6 KJ mol-1
O2 (g) = 0 3 – 2 = 1
Δ Hr° = [0 + (2 x -268.6)] – [23 + (-241.8)]
[-537.2] – [-218.8]
-537.6 + 218.8
-318.4 KJ
ΔE - ΔH - Δng RT
ΔE = -318.4 – 1 x (8.314/1000) x 298
= - 318.4 – (2477/1000)
= -318.4 – 2.477
= -320.8 KJ
Note- By convention, standard enthalpy of formation of an element in reference state are taken as 0.
Enthalpy for different types of Reaction
- Enthalpy of Combustion (ΔHC°) → The enthalpy change that occur when 1 mole of a substance is completely burned in excess of oxygen or air at any given temperature is known as enthalpy of combustion of that substance.
Example- C(s) + O2 (g) → CO2 (g) + ΔHC° = -393.5 KJ mol-1
- CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (L) ΔHC° = -890.35 KJ
- Enthalpy or Bond enthalpy → Bond enthalpy is the amount of energy required to break 1 mole bonds of a particulate type between two atoms in a gaseous state of a substance.
For example- H – H (g) → H (g) + H (g)
ΔHC = 435 KJ mol-1
But in poly atomic molecule, the bond enthalpy is the average enthalpy of different bonds
→ CH3 (g) + H (g) ΔH° = 427 KJ
CH2 – H → CH2 + H (g) ΔH° = 439 KJ
HC – H → CH + H (g) ΔH° = 452 KJ
CH → C (s) + H (g) ΔH° =347 KJ
We take the arrange of all the energies
((427 + 439 + 452 + 347) / 4) = 416 KJ mol-1
Application of Bond enthalpy
- The bond enthalpy can be used for calculating the enthalpy of reaction, enthalpy of formation of compound,
Formula used is:- If bond enthalpy is given
∆Hr = ∑ Bond enthalpy of Reactant - ∑B.E. of product reactant