Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    I missed a couple of lessons on Born-Haber cycles so i don't understand it fully but I've been doing some reading up and i can answer most questions. I've just had a bit of trouble with this working out the enthalpy of formation in this question:

    Calculate the enthalpy of formation of sodium iodide given that its lattice enthalpy is -648 kJmol-1. This is the cycle i made:



    /\Hf = -(+109)-(+107)-(+494)+(-314)+(-684)
    =-1708 kJ mol-1

    But the answer is supposed to be -288 kj mol-1. Can anyone help please?
    Offline

    17
    ReputationRep:
    (Original post by Ridium)
    I missed a couple of lessons on Born-Haber cycles so i don't understand it fully but I've been doing some reading up and i can answer most questions. I've just had a bit of trouble with this working out the enthalpy of formation in this question:

    Calculate the enthalpy of formation of sodium iodide given that its lattice enthalpy is -648 kJmol-1. This is the cycle i made:



    /\Hf = -(+109)-(+107)-(+494)+(-314)+(-684)
    =-1708 kJ mol-1

    But the answer is supposed to be -288 kj mol-1. Can anyone help please?
    Eformation = everything else added together
    = -288 KJmol^-1

    It's not that difficult O_O
    • Community Assistant
    • Study Helper
    Offline

    15
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by Ridium)
    I missed a couple of lessons on Born-Haber cycles so i don't understand it fully but I've been doing some reading up and i can answer most questions. I've just had a bit of trouble with this working out the enthalpy of formation in this question:

    Calculate the enthalpy of formation of sodium iodide given that its lattice enthalpy is -648 kJmol-1. This is the cycle i made:



    /\Hf = -(+109)-(+107)-(+494)+(-314)+(-684)
    =-1708 kJ mol-1

    But the answer is supposed to be -288 kj mol-1. Can anyone help please?
    just at a glance you have started off with I2(g) when it's a solid. You have to add in the enthalpy of vaporisation of iodine (x1/2).
    Offline

    17
    ReputationRep:
    Iodine is meant to start off solid at the start, just notice you already had enthalpy of atomisation for Iodine lol. Anyway just add all the values and you get -288.
    Offline

    12
    ReputationRep:
    All clockwise arrows=anti clockwise arrows
    So, enthalpy formation=109+107+494+(-314}+(-684)
    = -288 kj mol-1
    • Community Assistant
    • Study Helper
    Offline

    15
    ReputationRep:
    Community Assistant
    Study Helper
    The following are endothermic

    Step 1: Na(s) + ½I2(s) --> Na(s) + ½I2(g) this is ½ the enthalpy of vaporisation of iodine

    Step 2: Na(s) + ½I2(g) --> Na(s) + I(g) this is ½ the bond enthalpy of iodine
    Step 3: Na(s) + I(g) --> Na(g) + I(g) this is the enthalpy of atomisation of sodium
    Step 4: Na(g) + I(g) --> Na+(g) + I(g) this is the ionisation of sodium
    ---------------------------------------------

    Then the exothermic steps:

    Step 5: Na+(g) + I(g) --> Na+(g) + I-(g) this is the electron affinity of iodine

    Step 6: Na+(g) + I-(g) --> NaI(s) this is the lattice enthalpy

    ----------------------------------------------

    add up the endothermic steps and the exothermic steps keeping the signs correct and you will get the enthapy of formation.
    Offline

    18
    ReputationRep:
    (Original post by charco)
    just at a glance you have started off with I2(g) when it's a solid. You have to add in the enthalpy of vaporisation of iodine (x1/2).
    At A-Level, I think we presume atomisation assumes vaporisation as well.
    • Community Assistant
    • Study Helper
    Offline

    15
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by Hippysnake)
    At A-Level, I think we presume atomisation assumes vaporisation as well.
    That's not the issue here. Iodine is a solid under standard conditions so the ΔHf equation MUST read:

    Na(s) + ½I2(s) --> NaI(s)

    Atomisation of iodine would include ½(vaporisation + bond dissociation) but you have to start from solid iodine.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Yeah i just realised that i put I2 as gas, i was meant to put it as solid and why didn't i get -288 in the start it was so obvious :rolleyes:. I mistakenly went anti-clockwise :p: Thanks for the help guys.
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.