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    Hi, . what does the symbol when you have enthalpy change of r? i think the r stands for reaction, but if you were given a value for this e.g. for CO, what does this mean? is it the formation of CO? Many thanks
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    Delta r H (or Delta H r as it used to be written by certain A level exam boards) refers to the enthalpy of reaction, but the reaction that it is referring to (or enough information to construct the balanced equation yourself) must be given.
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    (Original post by Pigster)
    Delta r H (or Delta H r as it used to be written by certain A level exam boards) refers to the enthalpy of reaction, but the reaction that it is referring to (or enough information to construct the balanced equation yourself) must be given.
    .This question is about the reaction given below. CO(g) + H2O(g) CO2(g) + H2(g) Enthalpy data for the reacting species are given in the table below. Substance CO(g) H2O(g) CO2(g) H2(g) ΔH / kJ mol−1 −110 −242 −394 0 The standard enthalpy change for this reaction of carbon monoxide and steam is A +42 kJ mol−1 B −42 kJ mol−1 C +262 kJ mol−1 D −262 kJ mol−1 (multiple choice). if you treat the r as f, you get the wrong sign. how do you do this
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    (Original post by Bertybassett)
    .This question is about the reaction given below. CO(g) + H2O(g) CO2(g) + H2(g) Enthalpy data for the reacting species are given in the table below. Substance CO(g) H2O(g) CO2(g) H2(g) ΔH / kJ mol−1 −110 −242 −394 0 The standard enthalpy change for this reaction of carbon monoxide and steam is A +42 kJ mol−1 B −42 kJ mol−1 C +262 kJ mol−1 D −262 kJ mol−1 (multiple choice). if you treat the r as f, you get the wrong sign. how do you do this
    Draw a hesses law cycle.
    The first red arrow is going against the direction of the first black arrow so you must change the sign from - to +.
    Also remember for the final answer write Delta Hr (standard) otherwise the examiner may deduct marks.

    Make sure in the diagram you write "C(s) + O2(g) + H2(g)". I wrote by C(g) by mistake in the diagram.
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    (Original post by dip0)
    Draw a hesses law cycle.
    The first red arrow is going against the direction of the first black arrow so you must change the sign from - to +.
    Also remember for the final answer write Delta Hr (standard) otherwise the examiner may deduct marks.

    Make sure in the diagram you write "C(s) + O2(g) + H2(g)". I wrote by C(g) by mistake in the diagram.
    thanks but how did you know to use them similar to enthalpies of formation?
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    (Original post by Bertybassett)
    thanks but how did you know to use them similar to enthalpies of formation?

    I don't understand what you mean, could you please rephrase the question?
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    (Original post by dip0)
    I don't understand what you mean, could you please rephrase the question?
    how did you know to format the hess's cycle in the way that you did. you gave no explanation as to how you knew what to do.
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    (Original post by Bertybassett)
    how did you know to format the hess's cycle in the way that you did. you gave no explanation as to how you knew what to do.
    When one looks at the enthalpy data for H2 they will see that it is 0. Therefore the enthalpy data must be referring to the enthalpy of formation data.
    If you look at the definition for standard enthapy of formation it is "enthalpy change when one mole of a compound is formed from its constituent elements under standard conditions. All reactants and products in their standard states."

    You can't really have a enthalpy of formation for H2, because the hypothetical reaction would be:
    H2(g) => H2(g) (as you can see there is no reaction occuring so you can easily deduce the enthalpy data for H2 is 0).

    If however, we had enthalpy of combustion data, then for H2 this wouldn't be 0 as there would be a chemical reaction occuring - 2H2(g) + 02(g) =>2H2O(l)


    So now we know that the enthalpy data is refering to enthalpy of formation. We construct the hesses law cycle as in the diagram and work out Delta H r. as per Hesses law. This is how I knew to construct the hesses law cycle.

    Now we we work out the enthalpy of formation for both the reactants and products (which is the stuff next to the black arrows in the diagram) and then we work out a "route" to get from reactants to products (red arrows).

    Please do let me know if there is anything else needing explaining.
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    (Original post by dip0)
    When one looks at the enthalpy data for H2 they will see that it is 0. Therefore the enthalpy data must be referring to the enthalpy of formation data.
    If you look at the definition for standard enthapy of formation it is "enthalpy change when one mole of a compound is formed from its constituent elements under standard conditions. All reactants and products in their standard states."

    You can't really have a enthalpy of formation for H2, because the hypothetical reaction would be:
    H2(g) => H2(g) (as you can see there is no reaction occuring so you can easily deduce the enthalpy data for H2 is 0).

    If however, we had enthalpy of combustion data, then for H2 this wouldn't be 0 as there would be a chemical reaction occuring - 2H2(g) + 02(g) =>2H2O(l)


    So now we know that the enthalpy data is refering to enthalpy of formation. We construct the hesses law cycle as in the diagram and work out Delta H r. as per Hesses law. This is how I knew to construct the hesses law cycle.

    Now we we work out the enthalpy of formation for both the reactants and products (which is the stuff next to the black arrows in the diagram) and then we work out a "route" to get from reactants to products (red arrows).

    Please do let me know if there is anything else needing explaining.
    I understand now - thank you so much for the help. But why didn't they just say it was enthalpy of formation?
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    (Original post by Bertybassett)
    I understand now - thank you so much for the help. But why didn't they just say it was enthalpy of formation?
    I don't know - it confused me at first sight as well. It took me a while to realise that the data was enthalpy of formation by careful inspection of the numbers.
    Maybe they could do this in the exam to confuse the candidates or something?
 
 
 

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