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Calculate the enthalpy of formation of water (Bond Enthalpies) watch

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    Hi guys, I need help with this question
    'Calculate the enthalpy of formation for water'
    Bond enthalpies:
    H-H = 436KJmol-1
    O-O=146KJmol-1
    O-H=463KJmol-1

    What I Did:
    H2+ 1/2 O2 -> H2O
    H-H + O -> H-O-H
    436 – (2x463) = -490KJmol-1
    ΔHf = -490KJmol-1
    Is this correct as I have a feeling its not
    Apologies for formatting, I don't regularly post
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    (Original post by Alvaro.Morata9)
    Hi guys, I need help with this question
    'Calculate the enthalpy of formation for water'
    Bond enthalpies:
    H-H = 436KJmol-1
    O-O=146KJmol-1
    O-H=463KJmol-1

    What I Did:
    H2+ 1/2 O2 -> H2O
    H-H + O -> H-O-H
    436 – (2x463) = -490KJmol-1
    ΔHf = -490KJmol-1
    Is this correct as I have a feeling its not
    Apologies for formatting, I don't regularly post
    Step one: Word equation
    Hydrogen + Oxygen -> Water

    Step two: Balanced symbol equation
    2H2+ O2 -> 2H2O

    Step three: Stick in the bonds
    (H-H + H-H) + O=O -> H-O-H + H-O-H

    Step four: Substitute in the bond enthalpy figures
    (436 + 436) + 146 -> 4 x 463

    Step five: Find your figures
    1018 -> 1744

    Step six: Work out the bond enthalpy change
    Enthalpy change = energy in - energy out
    = 1018 - 1744
    = -726 KJmol-1
    So exothermic reaction as the bond enthalpy change is negative.

    The problem with your working out was when you balanced the equation. You should not make fewer atoms or molecules and keep the final products the same - you can change the number of products, as I did. You would never do half an atom or molecule - you go up, not down.

    Also, whilst Oxygen is O2, it is a double bond (O=O instead of the single bond H-H for example), so we don't need to multiply the bond enthalpies by two like with Hydrogen.

    If you have any questions, feel free to ask.
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    (Original post by Koalifications)
    Step one: Word equation
    Hydrogen + Oxygen -> Water

    Step two: Balanced symbol equation
    2H2+ O2 -> 2H2O

    Step three: Stick in the bonds
    (H-H + H-H) + O=O -> H-O-H + H-O-H

    Step four: Substitute in the bond enthalpy figures
    (436 + 436) + 146 -> 4 x 463

    Step five: Find your figures
    1018 -> 1744

    Step six: Work out the bond enthalpy change
    Enthalpy change = energy in - energy out
    = 1018 - 1744
    = -726 KJmol-1
    So exothermic reaction as the bond enthalpy change is negative.

    The problem with your working out was when you balanced the equation. You should not make fewer atoms or molecules and keep the final products the same - you can change the number of products, as I did. You would never do half an atom or molecule - you go up, not down.

    Also, whilst Oxygen is O2, it is a double bond (O=O instead of the single bond H-H for example), so we don't need to multiply the bond enthalpies by two like with Hydrogen.

    If you have any questions, feel free to ask.
    Sorry, I'm finding it hard to understand, the question says calculate enthalpy of formation of water so I thought enthalpy of formation was ' Enthalpy change when 1 mole of a substance is formed from elements' so isn't 1 mole of water H2O and wouldn't 2 moles of water be H2O2?
    Thank you for taking the time to reply btw
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    (Original post by Alvaro.Morata9)
    Sorry, I'm finding it hard to understand, the question says calculate enthalpy of formation of water so I thought enthalpy of formation was ' Enthalpy change when 1 mole of a substance is formed from elements' so isn't 1 mole of water H2O and wouldn't 2 moles of water be H2O2?
    Thank you for taking the time to reply btw
    Hey, so the enthalpy change basically means the change in energy - so for example, in an exothermic reaction, how much energy is given out. That is why we do energy in - energy out, to see the enthalpy (which means change). So, the formation of water enthalpy is the energy brought in when bonds are broken minus the energy given out when new bonds are made.

    On your second point about the moles, I can see why you would think that, but if you balance the equation then it would be equal to one mole, as both the products and reactants are being multiplied by the same amount(s). For example, 1/2 is equal to 2/4, because you multiply the fraction by two on both sides. Well, we do the same; Oxygen has two atoms, so we times the products by two (to get 2H2O) so it is equal.

    If you are still confused, feel free to ask again.
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    (Original post by Koalifications)
    Hey, so the enthalpy change basically means the change in energy - so for example, in an exothermic reaction, how much energy is given out. That is why we do energy in - energy out, to see the enthalpy (which means change). So, the formation of water enthalpy is the energy brought in when bonds are broken minus the energy given out when new bonds are made.

    On your second point about the moles, I can see why you would think that, but if you balance the equation then it would be equal to one mole, as both the products and reactants are being multiplied by the same amount(s). For example, 1/2 is equal to 2/4, because you multiply the fraction by two on both sides. Well, we do the same; Oxygen has two atoms, so we times the products by two (to get 2H2O) so it is equal.

    If you are still confused, feel free to ask again.
    Ah i understand now, so for questions like this should I make sure i get whole numbers in equations instead of halves? sorry if it sound like a dumb questions
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    It's not a dumb question at all - I also found these questions hard when I first learnt about bond enthalpy changes. Yes, you should try to get whole numbers when balancing equations instead of halving the formulae to get the exact original product formulae. If you had H2 for example and wanted to halve it to fit the products (whatever that product is), you actually would not have a bond because H-H would become a single H atom, which has no bonds. So just remember, when you balance equations, make more products/reactants instead of less to balance.

    For example:
    Cu + O2 → CuO

    If you wanted to balance that equation, you would need to see that in the reactants, there are two Oxygen atoms (as shown by O2), but only one Oxygen atom in the products (as shown by CuO). So, you would need to increase the number of Oxygens in the products. You have to, therefore, put a two at the front of CuO (making it 2CuO) because you cannot split the products up to CuO2 or something like that. So we get:

    Cu + O2 → 2CuO

    But now, we have to get two Cu atoms in the reactants, because whilst Oxygen is now balanced (two Oxygen atoms on each side of the equation), there is only one Cu atom in the reactants, but two in the products. So, stick a 2 in front of Cu.

    2Cu + O2 → 2CuO

    Now, this is balanced. We have two Cu atoms on either side of the equation, and two Oxygen atoms on either side of the equation.

    If you have any more questions, feel free to ask. BBC Bitesize has good resources for these topics. If you want one on balancing equations, here it is: http://www.bbc.co.uk/schools/gcsebit...ntalrev5.shtml
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    (Original post by Alvaro.Morata9)
    Thank you, i understand now; one last question
    can you check if question 2c is right, I think it should be -207 but i keep getting +207
    Haven't had college for 3 days due to the weather & I'm our chemistry class is so far behind!! So I'm trying to self teach, which doesn't seem to be going too well
    Enthalpy change = -110 –(-75) + (-242) = +207 kJ mol-1

    Yep, 2c is correct. I haven't been in either since Monday because of the snow, although we aren't very behind - self-teaching can be hard, have you tried using revision guides/videos? They work for me quite well.
 
 
 
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