Respuesta :
Considering the combustion of acetylene, the standard enthalpy of formation of acetylene is 227.0 kJ/mol.
Let's consider the following thermochemical equation.
C₂H₂(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g) ΔH°rxn = −1255.8 kJ
Given the standard enthalpies of formation (ΔH°f), and consider that the enthalpy of formation of simple substances in the standard state is zero, we can calculate the standard enthalpy of the reaction (ΔH°rxn) using the following expression.
[tex]\Delta H \° _{rxn} = \Sigma \Delta H \° _{f} (p) \times n_p - \Sigma \Delta H \° _{f} (r) \times n_r[/tex]
where,
- n: moles
- r: reactants
- p: products
We will use this equation to calculate the standard enthalpy of formation of acetylene.
[tex]\Delta H \° _{rxn} = \Delta H \° _f(CO_2(g)) \times 2mol + \Delta H \° _f(H_2O(g)) \times 1mol - \Delta H \° _f(C_2H_2(g)) \times 1mol - \Delta H \° _f(O_2(g)) \times 5/2mol\\\\\Delta H \° _f(C_2H_2(g)) \times 1mol = \Delta H \° _f(CO_2(g)) \times 2mol + \Delta H \° _f(H_2O(g)) \times 1mol - \Delta H \° _{rxn} - \Delta H \° _f(O_2(g)) \times 5/2mol\\\\\Delta H \° _f(C_2H_2(g)) \times 1mol = (-393.5kJ/mol) \times 2mol + (-241.8kJ/mol) \times 1mol - (-1255.8 kJ)\\\Delta H \° _f(C_2H_2(g)) = 227.0 kJ/mol[/tex]
Considering the combustion of acetylene, the standard enthalpy of formation of acetylene is 227.0 kJ/mol.
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The standard enthalpy of formation, [tex]\Delta H_{f} ^{o}[/tex], of C₂H₂ is 227 kJ/mol
From the question,
We are to determine the [tex]\Delta H_{f} ^{o}[/tex] of C₂H₂
From the given chemical equation,
C₂H₂(g) + ⁵/₂O₂(g) → 2CO₂(g) + H₂O(g) [tex]\Delta H_{rxn}^{o}[/tex] = −1255.8 kJ
The standard enthalpy of a reaction, [tex]\Delta H_{rxn}^{o}[/tex], is given by the formula
[tex]\Delta H_{rxn}^{o} = \sum \Delta H_{f}^{o}(products) - \sum \Delta H_{f}^{o}(reactants)[/tex]
From the question
[tex]\Delta H_{f} ^{o}[/tex] of CO₂ = -393.5 kJ/mol
[tex]\Delta H_{f} ^{o}[/tex] of H₂O = -241.8 kJ/mol
∴ [tex]\Delta H_{rxn}^{o}[/tex] = (2× [tex]\Delta H_{f} ^{o}[/tex] of CO₂ + [tex]\Delta H_{f} ^{o}[/tex] of H₂O) - ([tex]\Delta H_{f} ^{o}[/tex] of C₂H₂ + ⁵/₂×[tex]\Delta H_{f} ^{o}[/tex] of O₂)
Putting the parameters into the equation, we get
-1255.8 = (2× -393.5 + -241.8) - ([tex]\Delta H_{f} ^{o}[/tex] of C₂H₂ + ⁵/₂× 0)
(NOTE: [tex]\Delta H_{f} ^{o}[/tex] of O₂ = 0 kJ/mol)
-1255.8 = (-787 -241.8) - ([tex]\Delta H_{f} ^{o}[/tex] of C₂H₂ + 0)
-1255.8 = -1028.8 - ([tex]\Delta H_{f} ^{o}[/tex] of C₂H₂)
∴ [tex]\Delta H_{f} ^{o}[/tex] of C₂H₂ = -1028.8 + 1255.8
[tex]\Delta H_{f} ^{o}[/tex] of C₂H₂ = 227 kJ/mol
Hence, the standard enthalpy of formation, [tex]\Delta H_{f} ^{o}[/tex], of C₂H₂ is 227 kJ/mol
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