Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Leaving Cert Chemistry HL 2004 Q9

  • 03-02-2015 9:08pm
    #1
    SLyn9745
    Registered Users, Registered Users 2 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, Registered Users 2 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! :)


Advertisement