Heat of combustion & formation-

## Heat of combustion –

“It is defined as the change in internal energy or heat enthalpy when one mole of substance is completely oxidized in O2.”

For Ex-

2H2(g) + O2(g)  —> 2H2O(g)     ;   ΔH = -115 Kcal.

2 mole H2 gives energy = 115 Kcal

1 mole H2 gives energy = 115/2 = 57.5 Kcal

Heat of combustion of H2 is -57.5 Kcal.

For Ex-

C(s) + O2(g)  —> CO2(g)    ;    ΔH = -94.3 Kcal.

1 mole Carbon  gives energy = 94.3 Kcal

Heat of combustion of Carbon  is – 94.3 Kcal.

Heat of combustion is always exothermic .It means ΔH =negative

Exception –

N2(g) +1/2O2(g) —> N2O(g) ; ΔH = positive

N2(g)+O2 (g) —> 2NO(g)  ; ΔH = positive

## Heat of formation-

“It is defined as the change in internal energy or heat enthalpy when one mole of substance is formed from its initial components or elements”

For Ex –

C(s) + O2(g)  —> CO2(g)    ;    ΔH = -94.3 Kcal.

In this reaction 1 mole of CO2 is formed from its elements i-e Carbon & oxygen.

Therefore , heat of formation of CO2  ΔH’ = -94.3 Kcal.

For Ex-

2H2(g) + O2(g)  —> 2H2O(g)     ;   ΔH = -115 Kcal.

In this reaction 2 mole of H2O is formed from its elements i-e Hydrogen & oxygen.

Therefore , heat of formation of  H2O  ‘ ΔH’ = -115 / 2 = -57.5 Kcal.

heat of formation of  H2O  ‘ ΔH’ = -57.5 Kcal.

## Numerical-

### Solution –

C2H2(g) + 2.5O2(g) —> 2Co2(g) + H2O(g) ; ΔH =  -310.62 Kcal   ———eq1

C(graphite) +O2(g) —> CO2(g) ; ΔH = -94.05 Kcal ————- eq2

H2 (g) + 1/2 O2(g) —> H2O(g) ; ΔH = -68.32 Kcal  ————-eq3

2C(graphite) +H2(g) —> C2H2(g) ; ΔH = ? Kcal  ——– eq4

Multiply Eq 2  by 2 & add eq3,

2C(graphite) +2O2(g)  + H2 (g) + 1/2 O2(g)—> 2CO2(g) + H2O(g)

ΔH = (-94.05 x 2 ) + (-68.32) = -256.42 Kcal

2C(graphite) +2O2(g)  + H2 (g) + 1/2 O2(g)—> 2CO2(g) + H2O(g) ;  ΔH = -256.42 Kcal —– eq 5

Now to get eq 4, substract eq 1 from eq5 ,

2C(graphite) +2O2(g)  + H2 (g) + 1/2 O2(g) – C2H2(g) – 2.5O2(g)—> 2CO2(g)+H2O(g)- 2Co2(g) – H2O(g) ; ΔH = -256.42 – (-310.62 )

= + 54.2 Kcal

2C(graphite)+ H2 (g) —> C2H2(g) ; ΔH = +54.2 Kcal