Answer:
Explanation:
Hello,
Considering the chemical reaction, the enthalpy of reaction is given by:
ΔH°rxn=ΔfHCO2+ΔfHH2O-ΔfHC8H18
(ΔfHO2=0)
Taking into account that the reaction produces energy, ΔH°rxn is negative. No, solving for ΔfHC8H18:
ΔfHC8H18=-ΔH°rxn+8*ΔfHCO2+9*ΔfHH2O
ΔfHC8H18=-(-5104.1 kJ/mol)+9*(-292.74kJ/mol)+8*(-393.5 kJ/mol)
ΔfHC8H18=-678.56 kJ/mol
Best regards.
A particular isomer of C₈H₁₈, whose combustion reaction produces 5104.1 kJ of heat per mole of C₈H₁₈(g) consumed, has a standard enthalpy of formation of -220.1 kJ/mol.
Let's consider the following thermochemical equation.
C₈H₁₈(g) + 25/2 O₂(g) ⟶ 8 CO₂(g) + 9 H₂O(g) ΔH°rxn = −5104.1 kJ/mol
We can calculate the standard enthalpy of the reaction (ΔH°rxn) using the following expression.
[tex]\Delta H\° _{rxn} = \Delta H_f\°(p) \times n_p - \Delta H_f\°(r) \times n_r[/tex]
where,
We will use this equation to calculate the standard enthalpy of formation of this isomer of C₈H₁₈.
[tex]\Delta H\° _{rxn} = \Delta H_f\°(CO_2(g)) \times 8mol + \Delta H_f\°(H_2O(g)) \times 9mol - \Delta H_f\°(C_8H_{18}(g)) \times 1mol - \Delta H_f\°(O_2(g)) \times 25/2mol\\\\\Delta H_f\°(C_8H_{18}(g)) \times 1mol = \Delta H_f\°(CO_2(g)) \times 8mol + \Delta H_f\°(H_2O(g)) \times 9mol - \Delta H\° _{rxn} - \Delta H_f\°(O_2(g)) \times 25/2mol\\\\\Delta H_f\°(C_8H_{18}(g)) \times 1mol =(-393.5 kJ/mol) \times 8mol + (-241.8kJ/mol) \times 9mol - (-5104.1kJ) - (0kJ/mol) \times 25/2mol\\[/tex]
[tex]\Delta H_f\°(C_8H_{18}(g)) = -220.1 kJ/mol[/tex]
A particular isomer of C₈H₁₈, whose combustion reaction produces 5104.1 kJ of heat per mole of C₈H₁₈(g) consumed, has a standard enthalpy of formation of -220.1 kJ/mol.
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