(c) The combustion of one mole of methanol takes place at 298 K and1 atm. After combustion CO2 (g) and H20 (1) are produced and726.0 KJ of heat is librated. Calculate standard enthalpy offormation of methanol. The standard enthalpies of formation ofCO2 (g) and H20 (1) are – 393.0 KJ mol-1 and -286.0 KJ mol-1respectively.OR
Answers
Answer:
Answer: Standard enthalpy of formation of benzene = 129.9 KJ/mol
Explanation: Enthalpy of the reaction is equal to the total sum of the standard enthalpies of the formation of products minus the total sum of the standard enthalpies of formation of reactants. It is represented by \Delta H_{reaction}ΔH
reaction
Mathematically,
\Delta H_{rxn}=\Delta H_{products}-\Delta H_{reactants}ΔH
rxn
=ΔH
products
−ΔH
reactants
For the combustion of benzene:
2C_6H_6(g)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)2C
6
H
6
(g)+15O
2
(g)→12CO
2
(g)+6H
2
O(l)
The standard enthalpy of the elements present in their standard state is always zero.
\Delta H_{rxn}=[6(\Delta H_{H_2O})+12(\Delta H_{CO_2})]-[2(\Delta H_{C_6H_6})+15(\Delta H_{O_2})]ΔH
rxn
=[6(ΔH
H
2
O
)+12(ΔH
CO
2
)]−[2(ΔH
C
6
H
6
)+15(ΔH
O
2
)]
Putting values in above equation, we get:
3267.0 kJ=[6(285.83})+12(-393.5)]-[2(\Delta H_{C_6H_6})+15(0)]
\Delta H_f^0{C_6H_6 =129.9kJ/mol[/tex]