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-   -   Work against the external pressure (https://www.askmehelpdesk.com/showthread.php?t=575607)

  • May 10, 2011, 10:50 AM
    pop000
    work against the external pressure
    in the picture you can see the reaction.
    http://p1cture.me/images/23518895930282025474.jpg
    it's known that 2 mol nitrogen reaction with 6 mol hydrogen gas to create ammonia.
    what is the work against the external pressure in 1 Atm in 25 degrees Celsius?

    I think I need to use this formula:W=-PexdeltaV.
    but I can't know what is deltaV.

    I just know that -Pex=-1atm. 1atm=101,325 Pa
    and is a exothermic reaction.

    how can I keep on from here?
  • May 11, 2011, 09:54 AM
    Unknown008

    Um... I think that you have to consider the gases as being ideal for there are two things going on simultaneously concerning the volume...

    First, the volume is going from 4 units (4moles of gas) to 2 units (2 moles of gas)

    Secondly, the gases are being heated, hence expand. Assuming completeness of reaction, your final 'cool' volume is V/2.

    Let's say there is 1 mole of gas at this point (it will mean we started with 0.5 mol of N2 and 1.5 mol of H2).

    The pressure of that 'cool' gas is given by PV = nRT

    When you get P in terms of V, you'll have to use

    The final pressure is 1 atm in pascals, the initial volume is V/2, the initial pressure is the one you got previously. Find the final volume.

    The difference between that volume and V is your delta V.
  • May 12, 2011, 05:33 AM
    pop000
    Comment on Unknown008's post
    well at the start I got 6 mol hydrogen gas and 2 mol nitrogen gas sp I can do 6+2=8 - 2 mol ammoniagas= 5 mol now I can multiply it in 22.5 and than I get delta v? Is correct way?
    thank you.
  • May 12, 2011, 06:00 AM
    pop000
    Comment on Unknown008's post
    OH I mean 8-2=6 mol ammonia gas now I can multiply it in 22.4
  • May 12, 2011, 09:59 AM
    Unknown008

    Oh, so you had 6 mol of H2 at the start with 2 mol of N2! Okay that changes things.

    Okay, using your equation, 3 mol of H2 reacts with 1 mol of N2 to give 2 mol of NH3, you'll have, by proportion:

    6 mol of H2 reacting with 2 mol of N2 giving 4 mol of NH3.

    This means all your 8 moles of initial gas react and you're left with 4 mol of ammonia gas.

    From this, you know that you initially have (8*22.4) L of gas at the start.

    Now, you assume that when the reaction takes place at room temperature, there is no change in volume, that is the gases are enclosed and reacted in a closed vessel with fixed volume. You try finding the pressure now.

    I forgot one step in my previous post. This is not the pressure when the gas is heated up, so you'll have to do that. Use

    P1 is the pressure you just got,
    T1 is room temperature,
    P2 is the pressure when heated,
    T2 is the final temperature of the gas.

    You have 4 mol of produced ammonia, which is twice the amount in the equation. Hence you have (92.6*2) kJ which is used to heat the gas. Use Q = mcT to get the final temperature.

    Then you use that in P1V1 = P2V2

    =S

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