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    Values of enthalpy changes of combustion can be used to calculate enthalpy change of formation.
    The enthalpy change for the reaction in equation 2.1 is the enthalpy change of formation of propane.

    3C + 4H2---------------------------> C3H8

    The table below shows the enthalpy changes of combustion of carbon, hydrogen and propane.


    carbon -394
    hydrogen -286
    propane -2219

    Use these data to calculate the enthalpy change of formation









    I keep getting the wrong answer to this and don't know why.
    Could somebody do it with working out please
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    products - reactants

    2219 - ((3 x 394) + (286 x 4))

    Work that out
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    (Original post by Notsocleverstudent)
    products - reactants

    2219 - ((3 x 394) + (286 x 4))

    Work that out
    why is Hydrogen times 4 and not times 8

    also, why is it 2219 and not -2219

    thanks
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    (Original post by SteveCrain)
    why is Hydrogen times 4 and not times 8

    also, why is it 2219 and not -2219

    thanks
    Because hydrogen by itself is diatomic, and so formula of hydrogen is H2. And in that equation, you have 4 lots of H2.

    And I'm not sure where the - comes from in 2219? :confused: Unless it was stated in the question.
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    (Original post by Notsocleverstudent)
    Because hydrogen by itself is diatomic, and so formula of hydrogen is H2. And in that equation, you have 4 lots of H2.

    And I'm not sure where the - comes from in 2219? :confused: Unless it was stated in the question.
    kk; the -2219 is the bond enthalpy for combustion of propane. You got the right answer, though I still don't understand why you chuck the minus
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    (Original post by SteveCrain)
    bump
    You are going from the reactants (C and H2) to the products (C3H8) VIA the combustion products (CO2 and H2O)

    3C + 4H2 ---- step 1 ---> 3CO2 + 4H2O ---- step 2 ----> C3H8


    The second step then involves the reverse of the combustion enthalpy of propane, so the value is 2219 instead of -2219...
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    (Original post by charco)
    You are going from the reactants (C and H2) to the products (C3H8) VIA the combustion products (CO2 and H2O)

    3C + 4H2 ---- step 1 ---> 3CO2 + 4H2O ---- step 2 ----> C3H8


    The second step then involves the reverse of the combustion enthalpy of propane, so the value is 2219 instead of -2219...
    How would I know the enthalpy changes for the combustion products? I still don't see why it would be + instead of -. Can you explain it again?
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    (Original post by SteveCrain)
    kk; the -2219 is the bond enthalpy for combustion of propane. You got the right answer, though I still don't understand why you chuck the minus
    All combustion reactions are exothermic so AH is negative so you always stick in a minus at the end regardless.
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    (Original post by SteveCrain)
    How would I know the enthalpy changes for the combustion products? I still don't see why it would be + instead of -. Can you explain it again?
    You are given the values in the question.

    You are told that:

    C3H8 ------- combustion -----> 3CO2 + 4H2O ................. has an enthalpy change of -2219kJ

    Well then according to the law of conservation of energy:

    3CO2 + 4H2O --- reverse reaction ---> C3H8 ................ has an enthalpy change of +2219kJ

    (note that I've left out the oxygen for simplicity)

    In your Hess's law calculation the second step corresponds to the reverse of the enthalpy of combustion of propane, so you change the sign to positive.
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    (Original post by SteveCrain)
    kk; the -2219 is the bond enthalpy for combustion of propane. You got the right answer, though I still don't understand why you chuck the minus
    The question asks for Enthalpy of Formation

    Is the answer + or - 107?
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    (Original post by Notsocleverstudent)
    The question asks for Enthalpy of Formation

    Is the answer + or - 107?
    minus
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    (Original post by Notsocleverstudent)
    The question asks for Enthalpy of Formation

    Is the answer + or - 107?
    The sum you posted earlier is quite correct:

    2219 - ((3 x 394) + (286 x 4))

    The enthalpy change of the first step = ((3 x -394) + (-286 x 4)) = -2326kJ

    The enthalpy change for the second step = +2219kJ

    Hence the enthalpy change from reactants to products = -2326 + 2219 = -107kJ
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    (Original post by charco)
    The sum you posted earlier is quite correct:

    2219 - ((3 x 394) + (286 x 4))

    The enthalpy change of the first step = ((3 x -394) + (-286 x 4)) = -2326kJ

    The enthalpy change for the second step = +2219kJ

    Hence the enthalpy change from reactants to products = -2326 + 2219 = -107kJ
    You put the bolded values as negative

    And when it comes to calculating enthalpy change, it always either Reactants - Products or Products - Reactants

    But you added the 2 values :confused:
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    (Original post by SteveCrain)
    minus
    Nvm, just found out why.

    Since Combustion data was given, you would have to do Reactants - Products. For formation, it's Products - Reactants.


    Initially, I did it for formation but seeing as they gave you combustion values, you have to do it using Reactants - Products.

    So, your calculation would look like this:

    ((3 x -394) + (-286 x 4)) - (-2219) = -107

    Hopefully, that'll make it a lot easier for you (y)
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    (Original post by Notsocleverstudent)
    You put the bolded values as negative

    And when it comes to calculating enthalpy change, it always either Reactants - Products or Products - Reactants

    But you added the 2 values :confused:
    You are learning the way to do the questions by rote. This is never a good idea.

    You would be better off understanding WHY.

    I have set out my posts to explain how Hess's law is applied to enthalpy problems. You are just using an alternative route from the reactants to the products.

    In every case you add each step. It's just that you may have to change the value of the step you are taking because the reverse has been defined.

    There are only a few ways to do this BUT if you learn the equations by ROTE you will easily make mistakes.

    Understand the REASON that you apply the equations and work from first principles - that you you can't make mistakes.

    Your reaction can go via the elements in their standard states, in which case you use formation enthalpies (can you see why?)



    Look at the diagram and you see that step 1 is the REVERSE of the formation enthalpy, so you are going in the opposite direction to deltaHf and have to change the sign.
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    (Original post by charco)
    You are learning the way to do the questions by rote. This is never a good idea.

    You would be better off understanding WHY.

    I have set out my posts to explain how Hess's law is applied to enthalpy problems. You are just using an alternative route from the reactants to the products.

    In every case you add each step. It's just that you may have to change the value of the step you are taking because the reverse has been defined.

    There are only a few ways to do this BUT if you learn the equations by ROTE you will easily make mistakes.

    Understand the REASON that you apply the equations and work from first principles - that you you can't make mistakes.

    Your reaction can go via the elements in their standard states, in which case you use formation enthalpies (can you see why?)



    Look at the diagram and you see that step 1 is the REVERSE of the formation enthalpy, so you are going in the opposite direction to deltaHf and have to change the sign.
    I would initially use Hess's law to achieve an answer, then use the 2 equations to assure that my answer's correct. Also in this circumstance, using the equations can not give you a wrong answer as long as you can analyse the question correctly. I clearly didn't first time, I got a right answer but I took all values as being positive, but mistakes help improve you either way.

    Also, mark schemes only require calculations and the correct sign. Nothing beyond that.

    But thank you for actually allowing me to understand the process.

    Edit: Oh and I feel the way I've been taught at sixth form comes into play aswell
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    (Original post by charco)
    You are given the values in the question.

    You are told that:

    C3H8 ------- combustion -----> 3CO2 + 4H2O ................. has an enthalpy change of -2219kJ

    Well then according to the law of conservation of energy:

    3CO2 + 4H2O --- reverse reaction ---> C3H8 ................ has an enthalpy change of +2219kJ

    (note that I've left out the oxygen for simplicity)

    In your Hess's law calculation the second step corresponds to the reverse of the enthalpy of combustion of propane, so you change the sign to positive.
    I know how to get the answer and that the the route from Reactants to Combustion products has the same enthalpy change as the route from reactants to products to combustion products, but I still don't understand why you disregard the minus for -2219.
    I also don't understand where the diagram comes into it.
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    (Original post by SteveCrain)
    I know how to get the answer and that the the route from Reactants to Combustion products has the same enthalpy change as the route from reactants to products to combustion products, but I still don't understand why you disregard the minus for -2219.
    I also don't understand where the diagram comes into it.
    I didn't disregard the minus...

    The enthalpy change for a combustion reaction is exothermic (ie. negative)...

    ... BUT if you perform the equal but opposite change, i.e. go from the combustion products to the compound then you are reversing the reaction, so you have to reverse the sign of the enthalpy.

    THE NEXT BIT IS VITAL TO UNDERSTANDING

    According to the law of conservation of energy the energy change going from A --> B is equal to the negative of the enthalpy change going form B --> A
    (If this were not the case we would be able to 'make' energy from nothing)

    OK got that?

    So Hess's law takes you from A to B via a third state, C

    You want to find:

    A --> B

    You know that you can go from A --> C ............. -x kJ
    and B --> C ............... -y kJ

    and you have the values for their enthalpy changes (shown as -x and -y)

    So your route is to go first from A to C. No problem, you have the value = -x kJ

    BUT now you are at C and you want to go to B.

    Well you have the energy change of B --> C

    ... soooo the value for C --> B MUST be the reverse of -y kJ, which = +y kJ

    Hence your sum becomes:

    Enthalpy (A to B) = -x + y

    The diagram I included just shows this idea on an energy level chart. You go up it's positive, you go down it's negative.
 
 
 
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