Respuesta :
Answer: The enthalpy change of the reaction comes out to be [tex]1.01\times 10^4kJ/mol[/tex]
Explanation:
The chemical reaction for the combustion of octane follows the equation:
[tex]2C_{8}H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(16\times \Delta H^o_f_{(CO_2(g))})+(18\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(C_8H_{18}(l))})+(25\times \Delta H^o_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(C_8H_{18}(l))}=-250.1kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(16\times (-393.5))+(18\times (-241.8))]-[(2\times (-250.1))+(25\times (0))]=10148.2kJ/mol=1.01\times 10^4kJ/mol[/tex]
Hence, the enthalpy change of the reaction comes out to be [tex]1.01\times 10^4kJ/mol[/tex]
The heat of reaction for the combustion of octane is -10148.2 kJ/mol.
We can obtain the heat of combustion from the balanced reaction equation and the heats of formation of the chemical species involved.
The equation of the reaction is;
2C8H18(l) + 25 O2(g) ------> 16CO2(g) + 18H2O(g)
Also, we have the information;
ΔH∘f C8H18(l) = −250.1 O2 kJ/mol
ΔH∘f CO2(g) = −393.5 kJ/mol
ΔH∘f H2O(g) = −241.8 kJ/mol
ΔH∘f O2(g) = 0 kJ/mol
Now;
ΔHrxn = ∑ΔH∘f (products) - ΔH∘f(reactants)
ΔHrxn = [(16(−393.5 kJ/mol) + 18(−241.8 kJ/mol)] - [2(−250.1 O2 kJ/mol) + 25(0 kJ/mol)]
ΔHrxn = -10148.2 kJ/mol
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