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Leaving Cert Chemistry HL 2004 Q9

  • 03-02-2015 10:08pm
    #1
    SLyn9745
    Registered Users Posts: 29 ✭


    Hi, can any help explain Q9 in the 2004 Chemistry HL papers on Chemical Equilibrium. It's N2+3H2<->2NH3 . Checked marking scheme but still confused! ..


Comments

  • #2
    Corkgirl18
    Closed Accounts Posts: 894 ✭✭✭Corkgirl18


    For the last part? The calculation?

    N2 + 3H2 -> 2NH3
    Initially: 6 moles 18 moles 0 moles
    6 – x 18 – 3x 2x = 6 ⇒ x = 3

    At equilibrium: 3 moles 9 moles 6 moles
    Conc. at equil. [0.6] [1.8] [1.2]

    Kc = [NH3]2/[N2][H2]3
    =
    (1.2)^2/(0.6)(1.8)3
    = 0.41

    Does that help?


  • #3
    SLyn9745
    Registered Users Posts: 29 ✭SLyn9745


    Corkgirl18 wrote: »
    For the last part? The calculation?

    N2 + 3H2 -> 2NH3
    Initially: 6 moles 18 moles 0 moles
    6 – x 18 – 3x 2x = 6 ⇒ x = 3

    At equilibrium: 3 moles 9 moles 6 moles
    Conc. at equil. [0.6] [1.8] [1.2]

    Kc = [NH3]2/[N2][H2]3
    =
    (1.2)^2/(0.6)(1.8)3
    = 0.41

    Does that help?

    Yes, thanks for explaining it better! :)


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