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    Given the data:
    4NH3(g) + 3O2(g)  2N2(g) + 6H2O(l), H = -1530kJmol-1
    H2(g) + 1/2O2(g)  H2O(l), H = -288 kJmol-1

    Calculate the enthalpy of formation of ammonia (NH3)

    Answer: -49.5kJmol-1

    Did you get this answer, if so how? Thank you
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    (Original post by Frappé)
    Given the data:
    4NH3(g) + 3O2(g)  2N2(g) + 6H2O(l), H = -1530kJmol-1
    H2(g) + 1/2O2(g)  H2O(l), H = -288 kJmol-1

    Calculate the enthalpy of formation of ammonia (NH3)

    Answer: -49.5kJmol-1

    Did you get this answer, if so how? Thank you
    So you know H2 + 0.5Os = H2O has a delta H value of -288, in the first equation there are 6 H2O. so 6 x -288 = -1728

    therefore, 4NH3 = 2N2 where delta H now equals -1728 + 1530
    this gives 4NH3 = 2N2 a delta H value of -198
    since there are 4 NH3, -198 divided by 4 gives -49.5

    hope this helps!
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    The only thing you get closer to when trying to solve a problem concerning Hess's law is death.


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    1) Write the equation for formation of NH3.
    So, N2+3H2>2NH3

    2) Look at the equations given; there is w 4NH3 in the first eqn so scale up your equation from 1:
    2N2+6H2>4NH3

    3) write both equations given then try work out how to make the above equation in 2)

    4) Reverse first eqn so 4NH3 is in the right. This will reverse H = 1530KJ

    5) Multiply 2nd eqn by 6 to get 6H2
    H= 6*-288= -1728Kj

    6) Now it should cancel to give your desired eqn of formation of Ammonia.
    H=1530-1728=-198Kj

    7) However, this is for 4 moles of Ammonia but you are looking for just 1 mol of ammonia because its formation.

    8) Scale it right down so you are only producing NH3 not 4NH3

    9) You should now get -49.5 Kjmol-1

    If you don't get that^

    Spoiler:
    Show
    Divide by the eqn by 4 to get 1/2N2+3/2H2>NH3. -198/4=-49.5kj
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    (Original post by Amellia123)
    So you know H2 + 0.5Os = H2O has a delta H value of -288, in the first equation there are 6 H2O. so 6 x -288 = -1728

    therefore, 4NH3 = 2N2 where delta H now equals -1728 + 1530
    this gives 4NH3 = 2N2 a delta H value of -198
    since there are 4 NH3, -198 divided by 4 gives -49.5

    hope this helps!
    Thank you!!
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    (Original post by Mystery.)
    1) Write the equation for formation of NH3.
    So, N2+3H2>2NH3

    2) Look at the equations given; there is w 4NH3 in the first eqn so scale up your equation from 1:
    2N2+6H2>4NH3

    3) write both equations given then try work out how to make the above equation in 2)

    4) Reverse first eqn so 4NH3 is in the right. This will reverse H = 1530KJ

    5) Multiply 2nd eqn by 6 to get 6H2
    H= 6*-288= -1728Kj

    6) Now it should cancel to give your desired eqn of formation of Ammonia.
    H=1530-1728=-198Kj

    7) However, this is for 4 moles of Ammonia but you are looking for just 1 mol of ammonia because its formation.

    8) Scale it right down so you are only producing NH3 not 4NH3

    9) You should now get -49.5 Kjmol-1

    If you don't get that^
    Spoiler:
    Show
    Divide by the eqn by 4 to get 1/2N2+3/2H2>NH3. -198/4=-49.5kj
    Thank you!!
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    (Original post by Frappé)
    Thank you!!
    You're welcome!
    Hope it made sense
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    (Original post by Frappé)
    Thank you!!
    You're welcome! I hope it made sense!
 
 
 
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