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TYPES OF CHEMICAL REACTIONS: ENDOTHERMIC AND EXOTHERMIC REACTIONS
An endothermic reaction is one during which heat is absorbed from the surrounding. Most decomposition processes are endothermic reaction. Dissolution of ammonium chloride, ammonium tetraoxosulphate(vi) etc are endothermic process. In endothermic reaction, the enthalpy of product(s) is greater than that of reactant(s). change in enthalpy is equal to the sum of heat of product minus the sum of heat of reactant i.e △H = – . Where is sum of heat of products and is sum of heat of reactants. Enthalpy change in this regard is positive. i.e △H = +ve
Examples of Endothermic reaction
- N2(g) + O2(g) → 2NO(g) ∆H= +180.6KJmol -1
Exothermic reaction is one during which heat is released to the surrounding. Examples of exothermic reaction is combustion reaction. Dissolution of H2SO4, NaOH, KOH, neutralization and AlCl3 are exothermic process. In exothermic reaction, the enthalpy of product(s) is less than that of reactant(s). change in enthalpy is equal to the sum of heat of product minus the sum of heat of reactant i.e △H = – . Where is sum of heat of products and is sum of heat of reactants. Enthalpy change in this regard is negative. i.e △H = -ve
- Example of exothermic reaction.
C(s) + O2(g) → CO2(g)
∆H = – 408KJmol –1
Note: The negative and the positive signs indicated in the above figures indicate exothermic and endothermic reactions respectively
Heat of formation: Heat of formation of a compound is the heat change that occurs when one mole of a compound is formed from its elements in their standard states. The standard condition is 25Oc and 1 atm. The heat of formation of water is -286kJmol-1. The equation is given below: H2(g) + ½O2(g0 → H2O(l) ΔHf = -286kJmol-1
Heat of combustion: This is the heat evolved when one mole of a substance is completely burnt in oxygen under state condition. The heat of combustion of carbon is -393kJmol-1
Example:
Calculate the heat of reaction in the equation below given that the heat of combustion of carbon and hydrogen are -393kJmol-1 and -286kJmol-1 respectively and heat of formation of butane is -125kJmol-1.
C4H10(l) + O2(g) → 4CO2 + 5H2O(l)
Solution
△H = –
= [-393+(-286)] – (-125+0)
= -2877kJmol-1
NOTE: The enthalpy of the formation of an element is zero
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