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22 October, 01:20

The enthalpy of combustion of acetylene C2H2 is described by C2H2 (g) + (5/2) O2 (g) →2CO2 (g) + H2O (l),  H rxn = - 1299 kJ/mol. Calculate the enthalpy of formation of acetylene, given the following enthalpies of formation  H f[CO2 (g) ] = - 393.5 kJ/mol  H f[H2O (l) ] = - 285.8 kJ/mo

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  1. 22 October, 01:28
    0
    The enthalpy of formation of acetylene is 226.2 kJ/mol

    Explanation:

    Step 1: Data given

    C2H2 (g) + (5/2) O2 (g) → 2CO2 (g) + H2O (l) Heat of Reaction (Rxn) = - 1299kJ/mol

    Standard formation [CO2 (g) ] = - 393.5 kJ/mol

    Standard formation [H2O (l) ] = - 285.8 kj/mol

    Step 2: The balanced equation

    The formation of acetylene is:

    2C (s) + H2 (g) → C2H2 (g)

    Step 3: Calculate the enthalpy of formation of acetylene

    It doesn't matter if the process will happen in 1 step or in more steps. What matters is the final result. This is Hess' law of heat summation.

    To have the reaction of the formation of acetylene we have to take:

    ⇒ the reverse equation of the combustion of acetylene

    2CO2 (g) + H2O (l) → C2H2 (g) + (5/2) O2 (g)

    ⇒ The equation of formation of CO2 (multiplied by 2)

    2C (s) + 2O2 (g) → 2CO2 (g)

    ⇒ the equation of formation of H2O

    H2 (g) + 1/2 O2 (g) → H2O (l)

    2CO2 (g) + H2O (l) + 2C (s) + 2O2 (g) + H2 (g) + 1/2 O2 (g → C2H2 (g) + (5/2) O2 (g) + 2CO2 (g) + H2O (l)

    Final reaction = 2C (s) + H2 (g) → C2H2 (g)

    Calculate the enthalpy of formation of acetylene =

    ΔHf = 1299 kJ/mol + (2*-393.5) kJ/mol + (-285.8) kJ/mol

    ΔHf = 226.2 kJ/mol

    The enthalpy of formation of acetylene is 226.2 kJ/mol
  2. 22 October, 01:33
    0
    226.2 kJ/mol

    Explanation:

    Let's consider the following thermochemical equation for the combustion of acetylene.

    C₂H₂ (g) + (5/2) O₂ (g) → 2 CO₂ (g) + H₂O (l), ΔH°rxn = - 1299 kJ/mol.

    We can also calculate the enthalpy of the reaction per mole of acetylene from the enthalpies of formation.

    ΔH°rxn = 2 mol * ΔH°f (CO₂ (g)) + 1 mol * ΔH°f (H₂O (l)) - 1 mol * ΔH°f (C₂H₂ (g)) - 1 mol * ΔH°f (O₂ (g))

    1 mol * ΔH°f (C₂H₂ (g)) = 2 mol * ΔH°f (CO₂ (g)) + 1 mol * ΔH°f (H₂O (l)) - ΔH°rxn - 1 mol * ΔH°f (O₂ (g))

    1 mol * ΔH°f (C₂H₂ (g)) = 2 mol * (-393.5 kJ/mol) + 1 mol * (-285.8 kJ/mol) - (-1299 kJ) - 1 mol * (0 kJ/mol)

    ΔH°f (C₂H₂ (g)) = 226.2 kJ/mol
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