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    So, for (a) I did:

    n=\frac{vol(dm^{3})}{24}=\frac{1  000}{24}

    And then E_{mol}=\frac{49}{3}

    So: Energy release: \frac{1000}{24}\times\frac{49}{3  }=-681kJ

    I have no idea what to do for (c)!
    BTW: Part (b) was just asking me to draw an enthalpy profile diagram.
    Anybody give me some guidance?
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    (Original post by halpme)
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Size:  145.4 KBName:  2013-08-18 02_22_42-F322-01Jan13.pdf (SECURED) - Adobe Reader.jpg
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    So, for (a) I did:

    n=\frac{vol(dm^{3})}{24}=\frac{1  000}{24}

    And then E_{mol}=\frac{49}{3}

    So: Energy release: \frac{1000}{24}\times\frac{49}{3  }=-681kJ

    I have no idea what to do for (c)!
    BTW: Part (b) was just asking me to draw an enthalpy profile diagram.
    Anybody give me some guidance?
    The answer to part (a) would be positive 681kJ because its asking how much energy is released so it's positive. It would be negative if it asked for the enthalpy change.
    And for part (c) you're going to kick yourself when I tell you...
    Spoiler:
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    It's the same value as the original enthalpy change of reaction, but the opposite sign, so +49kJ
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    (Original post by sam1w2e)
    The answer to part (a) would be positive 681kJ because its asking how much energy is released so it's positive. It would be negative if it asked for the enthalpy change.
    And for part (c) you're going to kick yourself when I tell you...
    Spoiler:
    Show
    It's the same value as the original enthalpy change of reaction, but the opposite sign, so +49kJ
    . Without sound like a 4 year old... Why?
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    (Original post by halpme)
    . Without sound like a 4 year old... Why?
    Well it's the same for any equilibrium reaction, one reaction is essentially the reverse of the other.
    So for example: if you were to carry out a reaction which was endothermic, if you were to carry out the reverse reaction, in order to get back to the original reactants you would have to release the same amount of energy as was taken in from the forward reaction. So the magnitude of the enthalpy change of the forward and reverse reaction are the same, but opposite signs.
    I hope that helped :woo:
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    (Original post by sam1w2e)
    Well it's the same for any equilibrium reaction, one reaction is essentially the reverse of the other.
    So for example: if you were to carry out a reaction which was endothermic, if you were to carry out the reverse reaction, in order to get back to the original reactants you would have to release the same amount of energy as was taken in from the forward reaction. So the magnitude of the enthalpy change of the forward and reverse reaction are the same, but opposite signs.
    I hope that helped :woo:
    But it says energy released there: that means that bonds are being made and therefore energy is being released which makes \Delta H-ve, right? Or is \Delta H-ve but \Delta H_r +ve?

    EDIT: I think what I'm asking is how do I find out which reaction is endothermic/exothermic?
    Surely the forward reaction is exothermic because here is my answer to (b): Name:  2013-08-18 14_08_46-F322-01Jan13.pdf (SECURED) - Adobe Reader.jpg
Views: 55
Size:  46.7 KBIs that incorrect?
 
 
 
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