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    6. (a) (i) Draw a fully-labelled Born–Haber cycle forthe formation of solid barium chloride, BaCl2, from its elements.Include state symbols for all species involved.

    (ii) Use your Born–Haber cycle and the standardenthalpy data given below to calculate a value for the electron affinity ofchlorine. Enthalpy of atomisation of barium +180 kJ mol–1
    Enthalpy of atomisation of chlorine +122kJ mol–1
    Enthalpy of formation of barium chloride –859kJ mol–1
    First ionisation enthalpy of barium +503kJ mol–1
    Second ionisation enthalpy of barium +965kJ mol–1
    Lattice formation enthalpy of barium chloride –2056kJ mol–1

    (ii) Use your Born–Haber cycle and the standardenthalpy data given below to calculate a value for the electron affinity ofchlorine. Enthalpy of atomisation of barium +180 kJ mol–1
    Enthalpy of atomisation of chlorine +122kJ mol–1
    Enthalpy of formation of barium chloride –859kJ mol–1
    First ionisation enthalpy of barium +503kJ mol–1
    Second ionisation enthalpy of barium +965kJ mol–1
    Lattice formation enthalpy of barium chloride –2056kJ mol–1



    Ive had a go at this Q and know the answer to it - it came up as -347kj/mol which was half the value initially got of -695 - so i missed the step to half that value but i dont understand why i need to half it?
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    (Original post by ahsan_ijaz)
    6. (a) (i) Draw a fully-labelled Born–Haber cycle forthe formation of solid barium chloride, BaCl2, from its elements.Include state symbols for all species involved.

    (ii) Use your Born–Haber cycle and the standardenthalpy data given below to calculate a value for the electron affinity ofchlorine. Enthalpy of atomisation of barium +180 kJ mol–1
    Enthalpy of atomisation of chlorine +122kJ mol–1
    Enthalpy of formation of barium chloride –859kJ mol–1
    First ionisation enthalpy of barium +503kJ mol–1
    Second ionisation enthalpy of barium +965kJ mol–1
    Lattice formation enthalpy of barium chloride –2056kJ mol–1

    (ii) Use your Born–Haber cycle and the standardenthalpy data given below to calculate a value for the electron affinity ofchlorine. Enthalpy of atomisation of barium +180 kJ mol–1
    Enthalpy of atomisation of chlorine +122kJ mol–1
    Enthalpy of formation of barium chloride –859kJ mol–1
    First ionisation enthalpy of barium +503kJ mol–1
    Second ionisation enthalpy of barium +965kJ mol–1
    Lattice formation enthalpy of barium chloride –2056kJ mol–1



    Ive had a go at this Q and know the answer to it - it came up as -347kj/mol which was half the value initially got of -695 - so i missed the step to half that value but i dont understand why i need to half it?
    Since you have 2 Cl's, you found 2 times the electron affinity of Chlorine.
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    (Original post by SeanFM)
    Since you have 2 Cl's, you found 2 times the electron affinity of Chlorine.
    hi

    Thats why am confused as we have 2CL thus why do we divide by 2? as the total electron affinity should be the number i got but the mark scheme had final step / by 2.
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    (Original post by ahsan_ijaz)
    hi

    Thats why am confused as we have 2CL thus why do we divide by 2? as the total electron affinity should be the number i got but the mark scheme had final step / by 2.
    By using the cycle you have found the energy for 2Cl + 2e- ->2Cl-, and since the electron affinity of chlorine is the energy from Cl + 2e -> Cl-, you divide by two since you are only interest in one, not two lots of the equation.
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    (Original post by SeanFM)
    By using the cycle you have found the energy for 2Cl + 2e- ->2Cl-, and since the electron affinity of chlorine is the energy from Cl + 2e -> Cl-, you divide by two since you are only interest in one, not two lots of the equation.
    Hm i get it to the part were intrested in one as the compound is bacl2 - chlorine must have 2 electron affinities?
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    (Original post by ahsan_ijaz)
    Hm i get it to the part were intrested in one as the compound is bacl2 - chlorine must have 2 electron affinities?
    Up to where your answer is is correct, you are just missing the step where you have to half it.

    In the cycle there are 2 lots of electron affinities going on because there are two Chlorines gaining electrons (and this is the value before you halve it) and since electron affinity is the energy when one Chlorine gains an electron, you need to halve the energy you calculated because you're interested in (1/2)* (energy from 2Cl + 2e- -> 2Cl-) = 1/2 * -695 = -397.
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    (Original post by SeanFM)
    Up to where your answer is is correct, you are just missing the step where you have to half it.

    In the cycle there are 2 lots of electron affinities going on because there are two Chlorines gaining electrons (and this is the value before you halve it) and since electron affinity is the energy when one Chlorine gains an electron, you need to halve the energy you calculated because you're interested in (1/2)* (energy from 2Cl + 2e- -> 2Cl-) = 1/2 * -695 = -397.
    Thanks a lot - i know i may be being stupid but i want to understand this instead of getting shy to ask

    from what the text book say electron affinity is when one mole of gaseous 1- ion is formed from gaseous atom - and their can be a first elctron affinity and second- thats why i didnt half as i thought it would be the 2 electron affinities added so why half?

    Shouldnt we have 2 chlorine gaining electrons? As we need 2cl- to join with ba2+ to form bacl2 ?
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    (Original post by ahsan_ijaz)
    Thanks a lot - i know i may be being stupid but i want to understand this instead of getting shy to ask

    from what the text book say electron affinity is when one mole of gaseous 1- ion is formed from gaseous atom - and their can be a first elctron affinity and second- thats why i didnt half as i thought it would be the 2 electron affinities added so why half?

    Shouldnt we have 2 chlorine gaining electrons? As we need 2cl- to join with ba2+ to form bacl2 ?
    No worries, ask as many questions as you want

    There are no second electron affinities going on here (Cl isn't going to Cl^{2-}), there are two chlorines (Cl) and each gains one electron. Then each Cl- bonds with the Ba2+ to give BaCl2.

    So in the electron affinity for Chlorine stage, you get the first Cl gaining an electron to make Cl- then the second Cl gaining an electron to get Cl- as well. So overall -695 energy is being released, hence we can deduce that half of that is the energy being released from one of those equations (and they are both equal, as they are the same).

    And the definition of the electron affinity of Cl is the energy when Cl gains an electron, which is exactly what we have found. A confusing thing here might be that chlorine refers to Cl rather than Cl2, which is chlorine gas.
 
 
 
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