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

chemistry- chemical equilibrium

  • 16-11-2010 8:02pm
    #1
    hatethisfeeling
    Closed Accounts Posts: 185 ✭✭


    Really cant figure out how to work out how many moles is left at chemical equilibrium. The quesstion im looking at has 0.05 moles of hydrogen and 0.05 moles of iodine reacting to form 2HI.

    At equilibrium there is 0.08moles of 2HI left. I thought you would take away 0.02 from both 0.05 's to work out how much of each was used in the reaction but apparently you have to take 0.04 from each one?Im so confused!

    Has anyonegot any chemistry websites that explain this?thanks!


Comments

  • #2
    grinds
    Banned (with Prison Access) Posts: 70 ✭✭grinds


    hatethisfeeling wrote: »
    Really cant figure out how to work out how many moles is left at chemical equilibrium. The quesstion im looking at has 0.05 moles of hydrogen and 0.05 moles of iodine reacting to form 2HI.

    At equilibrium there is 0.08moles of 2HI left. I thought you would take away 0.02 from both 0.05 's to work out how much of each was used in the reaction but apparently you have to take 0.04 from each one?Im so confused!

    Has anyonegot any chemistry websites that explain this?thanks!

    This is what you started with.. Draw out the table below. These are the three numbers of moles you mentioned.


    H2 + I2
    2HI
    Moles initially 0.05 0.05
    Change
    Moles at equilibrium 0.08

    We can place 0 underneathe 2HI because there are 0 moles here initially... i.e. the H2 and the I2 haven't reacted yet.

    Now calculate the one change which is +0.08. Use the numbers in the equation to calculate the two other changes.

    0.08 goes with the 2 before the 2HI
    There is a 1 before H2 and I2 so the change here is 0.04

    H2 + I2
    2HI
    Moles initially 0.05 0.05 0
    Change -0.04 -0.04 +0.08
    Moles at equilibrium 0.08

    Final step is a little addition/subtraction and throw these "moles at equilibrium" numbers into your Kc equation:

    H2 + I2
    2HI
    Moles initially 0.05 0.05 0
    Change -0.04 -0.04 +0.08
    Moles at equilibrium 0.01 0.01 0.08


    Hope this helps!

    Sorry... just noticed the numbers didn't come out underneathe the appropriate part of the equation... Just line them up and it will all become clear!


  • #3
    hatethisfeeling
    Closed Accounts Posts: 185 ✭✭hatethisfeeling


    you are a legend:D THANK YOU!


  • #4
    grinds
    Banned (with Prison Access) Posts: 70 ✭✭grinds


    Very welcome... Good luck!


  • #5
    OisinOGorman
    Registered Users Posts: 10 ✭OisinOGorman


    Thank you so much!!


Advertisement