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    So in the previous part they said write the equation for cracking which is this

    C40H82 --> C16H34 + C24H48


    Considering only the bonds broken and the bonds formed during the reaction, use the Data Booklet to calculate the enthalpy change for the reaction you wrote above

    ................................ .............
    How would we do this. I don't get what the markscheme does . The answer is +180
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    (Original post by Revision 99)
    So in the previous part they said write the equation for cracking which is this

    C40H82 --> C16H34 + C24H48


    Considering only the bonds broken and the bonds formed during the reaction, use the Data Booklet to calculate the enthalpy change for the reaction you wrote above

    ................................ .............
    How would we do this. I don't get what the markscheme does . The answer is +180
    Calculate the energy gone into breaking the bonds on the left hand side, then calculate the energy gone into making the bonds on the right hand side, and then find the difference? (This is what i remember from GCSE but don’t quote me )
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    Yeah, find the energy of those bonds broken on the left and then the energy of the bonds formed on the right. Subtract using the formula; energy required to break bonds-energy required to form bonds.
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    (Original post by Revision 99)
    So in the previous part they said write the equation for cracking which is this

    C40H82 --> C16H34 + C24H48


    Considering only the bonds broken and the bonds formed during the reaction, use the Data Booklet to calculate the enthalpy change for the reaction you wrote above

    ................................ .............
    How would we do this. I don't get what the markscheme does . The answer is +180
    It may be useful to think of an analogy before answering the question. Imagine you had a building made of lego bricks (the bricks represents the atoms and the lego sticking together represents the bonds and the building represents the molecule). If you wanted to make another building ( =another molecule so the products) you would separate all the bricks from each other (= breaking all the bonds in the current moleule) and rearrging all the lego bricks in another way (forming new bonds) to form another building (= another molecules)

    In the same way. You want to decompose the 40 carbon alkane into its consituent atoms (breaking the old building into its consituent lego bricks) and rearrange them to form the products (forming new bonds).

    Consider bonds broken:
    for an alkane that has n carbon atoms, there are:
    n-1 C-C bonds
    2n+2 C-H bonds
    if you then substitute n=40 you will find number of each type of bonds needing to be broken.

    FIND THE SUM OF THIS ENTHALPY (THIS IS +VE AS BREAKING BONDS IS ENDOTHERMIC PROCESS)
    Now we have 40 isolated C atoms and 82 isolated hydrogen atoms

    Consider bonds formed:
    forming C16H34 another alkane, then using the above rules:
    n-1 C-C bonds formed means 15 C-C bonds being formed.
    2n+2 C-H bonds formed means 24 C-H bonds being formed.
    FIND THE SUM OF THESE ENTHALPY(THIS WILL BE -VE AS BOND FORMING IS EXOTHERMIC PROCESS)

    forming C16H48 the alkene, then there are for n carbons long alkene with 1 double bond)
    n-2 C-C bond (think about it)
    1 C=C bond
    2n C-H bond

    using n=16 you get
    14 C-C bonds being formed
    1 C=C bond being formed
    32 C-H bonds being formd FIND THE SUM OF THESE ENTHALPY (AGAIN +VE VALUE) AND ADD TO THE ENTHALPY OF C16H34

    final step:
    to find enthalpy change you do total enthalpy of products minus total enthalpy of reactant. This will get you +180 kJmol^-1
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    (Original post by dip0)
    It may be useful to think of an analogy before answering the question. Imagine you had a building made of lego bricks (the bricks represents the atoms and the lego sticking together represents the bonds and the building represents the molecule). If you wanted to make another building ( =another molecule so the products) you would separate all the bricks from each other (= breaking all the bonds in the current moleule) and rearrging all the lego bricks in another way (forming new bonds) to form another building (= another molecules)

    In the same way. You want to decompose the 40 carbon alkane into its consituent atoms (breaking the old building into its consituent lego bricks) and rearrange them to form the products (forming new bonds).

    Consider bonds broken:
    for an alkane that has n carbon atoms, there are:
    n-1 C-C bonds
    2n+2 C-H bonds
    if you then substitute n=40 you will find number of each type of bonds needing to be broken.

    FIND THE SUM OF THIS ENTHALPY (THIS IS +VE AS BREAKING BONDS IS ENDOTHERMIC PROCESS)
    Now we have 40 isolated C atoms and 82 isolated hydrogen atoms

    Consider bonds formed:
    forming C16H34 another alkane, then using the above rules:
    n-1 C-C bonds formed means 15 C-C bonds being formed.
    2n+2 C-H bonds formed means 24 C-H bonds being formed.
    FIND THE SUM OF THESE ENTHALPY(THIS WILL BE -VE AS BOND FORMING IS EXOTHERMIC PROCESS)

    forming C16H48 the alkene, then there are for n carbons long alkene with 1 double bond)
    n-2 C-C bond (think about it)
    1 C=C bond
    2n C-H bond

    using n=16 you get
    14 C-C bonds being formed
    1 C=C bond being formed
    32 C-H bonds being formd FIND THE SUM OF THESE ENTHALPY (AGAIN +VE VALUE) AND ADD TO THE ENTHALPY OF C16H34

    final step:
    to find enthalpy change you do total enthalpy of products minus total enthalpy of reactant. This will get you +180 kJmol^-1

    Thank you so much!! Would never have understood the concept without this thorough explanation. Means a lot !!
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    (Original post by Revision99)
    Thank you so much!! Would never have understood the concept without this thorough explanation. Means a lot !!
    no need to mention.

    note: Whenever you are given bond enthalpies, you use this procedure to break the compound into its constituent toms (unless you can find a shortcut where you dont have to separate all the atoms) then use the equation enthalpy of product - enthalpy of reactant.
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    (Original post by dip0)
    no need to mention.

    note: Whenever you are given bond enthalpies, you use this procedure to break the compound into its constituent toms (unless you can find a shortcut where you dont have to separate all the atoms) then use the equation enthalpy of product - enthalpy of reactant.
    In essence when you are finding the total enthalpy of reactants and the total enthalpy of products you can imagine plotting these values on a free energy diagram, then the enthalpy change is products - reactants.
 
 
 
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