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    Would someone please help me with part C of this question?

    To help me with the part I need help with, the answer to part B is: -125

    (b) Given that the enthalpy of combustion to form carbon dioxide and steam is
    –2102 kJ mol–1 for propane and –1977 kJ mo1–1 for propene, determine the enthalpy change for the oxidation of 1 mol of propane to propene and steam
    C3H8(g) + ½ O2(g) ? C3H6(g) + H2O(g)
    using equations or a cycle to support your answer

    (c) State the number and type of bonds made and broken in the oxidation of propane to propene and steam. Use the mean bond enthalpies in the table above, together with your answer to part (b), to calculate the bond enthalpy of the O=O bond in the oxygen molecule.

    Bondsbroken..................... ................................ ................................ ........................

    Bondsformed..................... ................................ ................................ ........................

    Bond enthalpy of O=O ................................ ................................ .............................


    The working out the exam paper has is:

    (c) Bonds broken 2 × (C – C) + 8 × (C – H) + (O = O)
    or nett : 1 × (C – C) + 2 × (C – H) + (O = O) (1)
    Bonds formed 1 × (C – C) + 6 × (C – H) + 1 × (C = C) + 2 × (O – H)
    or nett : 1 × (C=C) + 2 × (O – H) (1)

    Bond enthalpy of O=O ?H = ?B(bonds broken) – ?B(bonds formed) (1)
    ? –125 = B(C – C) + 2B(C – H) + B(O = O) – B(C = C) – 2B(O – H)
    ? B(O = O) = 2(–125 –348 –2 × 413 + 612 + 2 × 463
    = +2 × 239 = 478 kJ mol–

    Can you please show me, how to get to the answer without using the net? I dont understand where the 2 comes from
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    (Original post by interact)
    Would someone please help me with part C of this question?

    To help me with the part I need help with, the answer to part B is: -125

    (b) Given that the enthalpy of combustion to form carbon dioxide and steam is
    –2102 kJ mol–1 for propane and –1977 kJ mo1–1 for propene, determine the enthalpy change for the oxidation of 1 mol of propane to propene and steam
    C3H8(g) + ½ O2(g) ? C3H6(g) + H2O(g)
    using equations or a cycle to support your answer

    (c) State the number and type of bonds made and broken in the oxidation of propane to propene and steam. Use the mean bond enthalpies in the table above, together with your answer to part (b), to calculate the bond enthalpy of the O=O bond in the oxygen molecule.

    Bondsbroken..................... ................................ ................................ ........................

    Bondsformed..................... ................................ ................................ ........................

    Bond enthalpy of O=O ................................ ................................ .............................


    The working out the exam paper has is:

    (c) Bonds broken 2 × (C – C) + 8 × (C – H) + (O = O)
    or nett : 1 × (C – C) + 2 × (C – H) + (O = O) (1)
    Bonds formed 1 × (C – C) + 6 × (C – H) + 1 × (C = C) + 2 × (O – H)
    or nett : 1 × (C=C) + 2 × (O – H) (1)

    Bond enthalpy of O=O ?H = ?B(bonds broken) – ?B(bonds formed) (1)
    ? –125 = B(C – C) + 2B(C – H) + B(O = O) – B(C = C) – 2B(O – H)
    ? B(O = O) = 2(–125 –348 –2 × 413 + 612 + 2 × 463
    = +2 × 239 = 478 kJ mol–

    Can you please show me, how to get to the answer without using the net? I dont understand where the 2 comes from
    It's just poorly written out.

    The enthalpy change of the equation is equal to the bond enthalpies of the reactants - bond enthalpies of the products

    BUT you will see that in the equation you have only 1/2 mole of oxygen bonds breaking.

    Hence when you rearrange to get the oxygen on its own you have to multiply by 2 to get the bond enthalpy of 1 mole of O=O bonds.

    bonds of 1/2(O2) = -125 - bonds(C3H8) + bonds(C3H6) + bonds(H2O)
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    (Original post by charco)
    It's just poorly written out.

    The enthalpy change of the equation is equal to the bond enthalpies of the reactants - bond enthalpies of the products

    BUT you will see that in the equation you have only 1/2 mole of oxygen bonds breaking.

    Hence when you rearrange to get the oxygen on its own you have to multiply by 2 to get the bond enthalpy of 1 mole of O=O bonds.

    bonds of 1/2(O2) = -125 - bonds(C3H8) + bonds(C3H6) + bonds(H2O)
    Thankyou. sorry but i'm just a little confused. should my answer be the bond enthalpy of 1 mole of O=O or 1/2 a mol of O=O
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    How do you rearrange the equation to get oxgen on its own?
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    (Original post by interact)
    How do you rearrange the equation to get oxgen on its own?
    Remember that the definition of bond enthalpy is the energy required to break 1 mole of bonds in a gaseous molecule.

    By Hess' law:

    bonds broken - bonds formed = enthalpy change of reaction

    bonds(C3H8) + bonds(1/2O2) - bonds(C3H6) - bonds(H2O) = ΔH(rxn)

    rearrange by taking terms to the other side and changing sign to leave bonds(1/2O2) on its own...

    bonds(1/2O2) = ΔH(rxn) - bonds(C3H8) + bonds(C3H6) + bonds(H2O)

    Now you work out the value of the right hand side...

    BUT this equals only 1/2 the bond enthalpy of oxygen ...

    SO you have to multiply by 2 to get the bond enthalpy of oxygen
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    Thankyou soooo much, i cannot thank you enough
 
 
 
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