Respuesta :
Answer: The enthalpy of formation of accetylene is, 226.2 kJ/mol
Explanation:
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
The chemical equation for the combustion of acetylene follows:
(1) [tex]C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(l)[/tex]
[tex]\Delta H^o_{rxn}=-1299kJ/mol[/tex]
The formation of [tex]CO_2[/tex] will be,
(2) [tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex] [tex]\Delta H_f_{(CO_2)}=-393.5kJ/mol[/tex]
The formation of [tex]H_2O[/tex] will be,
(3) [tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex] [tex]\Delta H_f_{(H_2O)}=-285.8kJ/mol[/tex]
The formation of [tex]C_2H_2[/tex] will be,
(4) [tex]2C(s)+H_2(g)\rightarrow C_2H_2(g)[/tex] [tex]\Delta H_f_{(C_2H_2)}=?[/tex]
Now we are reversing equation 1, multiplying equation 2 by 2 and then adding equation 1, 2 and 3, we get:
Reaction (1) :
[tex]2CO_2(g)+H_2O(l)\rightarrow C_2H_2(g)+\frac{5}{2}O_2(g)[/tex]
[tex]\Delta H^o_{rxn}=1299kJ/mol[/tex]
Reaction (2) :
[tex]2C(s)+2O_2(g)\rightarrow 2CO_2(g)[/tex] [tex]\Delta H_f_{(CO_2)}=2\times -393.5kJ/mol=-787kJ/mol[/tex]
Reaction (3) :
[tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex] [tex]\Delta H_f_{(H_2O)}=-285.8kJ/mol[/tex]
[tex]\Delta H_f_{(C_2H_2)}=\Delta H^o_{rxn}+\Delta H_f_{(CO_2)}+\Delta H_f_{(H_2O)}[/tex]
[tex]\Delta H_f_{(C_2H_2)}=(1299kJ/mol)+(-787kJ/mol)+(-285.8kJ/mol)[/tex]
[tex]\Delta H_f_{(C_2H_2)}=226.2kJ/mol[/tex]
Therefore, the enthalpy of formation of accetylene is, 226.2 kJ/mol
