17 Questions around this concept.
Given C(graphite)+O2(g) → CO2(g) ;
rH0=−393.5 kJ mol−1
H2(g)+ O2(g) → H2O(l) ;
rH0=−285.8 kJ mol−1
CO2(g)+2H2O(l) → CH4(g)+2O2(g) ;
rH0=+890.3 kJ mol−1
Based on the above thermochemical equations, the value of rH0 at 298 K for the reaction
C(graphite)+2H2(g) → CH4(g) will be :
When 9 in ice melts at 00 C, x KJ heat is required, then enthalpy of fusion of ice will be
Heat of Formation
The amount of heat evolved or absorbed or change in enthalpy when 1 mole of a substance is obtained from its constituents or free elements.
$\Delta \mathrm{H}^{\circ}=\sum \mathrm{H}^{\circ} \mathrm{p}-\sum \mathrm{H}^{\circ} \mathrm{r}$
Once the value of Ho at 25o C for any species has been assigned the value of Ho at other temperatures Can be found out by using Kirchoff's equation as follows.
$\int_{298}^{\mathrm{T}} \mathrm{dH}^{\circ}=\int_{298}^{\mathrm{T}} \mathrm{CpdT}$
$\mathrm{H}_{\mathrm{T}}^{\circ}-\mathrm{H}_{298}^{\circ}=\int_{298}^{\mathrm{T}} \mathrm{CpdT}$
1. Standard heat of formation of a free element is taken as zero.
For example, In carbon—graphite form is taken as standard state and in sulphur, the monoclinic form is the standard state.
2. The heat of formation may be +ve or -ve.
If $\Delta \mathrm{H}$ is -ve compound is exothermic.
If $\Delta \mathrm{H}$ is + ve compound is endothermic.
3. The stability of the exothermic compound is more than that of the endothermic compound hence, the greater the liberated energy greater is the stability of the compound.
Example, HF > HCI > HBr > HI
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