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Homework Help: Chemistry Uncertainty

  1. Jan 9, 2005 #1
    Hello all:

    Given the following data:

    H2(g) + 1/2 O2(g) --> H20 (l) ∆H = -285.8 kJ

    N2O5(g) + H20(l) --> 2HNO3(l) ∆H = -76.6 kJ

    1/2 N2(g) + 3/2O2(g) + 1/2 H2 (g) --> HNO3(l) ∆ = -174.1 kJ

    Calculate the ∆H for the reaction

    2N2(g) + 5O2(g) --> 2N2O5(g)

    My Solution

    from N2: 2 = 1/2 *k3
    from O2: 5 = 1/2 * k1 + 3/2 * k3
    from N2O5: -2 = -k2

    Hence ∆comb = ∆H1 * k1 + ∆H2 * k2 + ∆H3 * k3. I am not sure if this is right. Can someone please see if I made a mistake in my steps?

    Thanks a lot
     
  2. jcsd
  3. Jan 9, 2005 #2
    everything seems right...
     
  4. Jan 9, 2005 #3
    Why is the answer 28.4 kJ when I get -201.8?
     
  5. Jan 9, 2005 #4
    do you have the correct signs? Can you show all your work...
     
  6. Jan 9, 2005 #5
    from N2: 2 = 1/2 * k3 k3 = 4
    from O2: 5 = 1/2*k1 + 6. 1/2*k1 = -1, k1 = -2
    from N2O5: -2 = -k2, k2 = 1

    ∆Comb = -2(-285.8) + 1( -76.6) + 4(-174.1)
     
  7. Jan 9, 2005 #6
    k2 does not equal one

    in fact

    should be - k2 = 2 ( my bad.... )

    so k2 = -2..
     
  8. Jan 9, 2005 #7

    Gokul43201

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    Staff Emeritus
    Science Advisor
    Gold Member

    yes, that's the error : k2 = -2

    Besides relying on the math, make sure the values you get make logical sense. Equation #2 has the N2O5 on the wrong side, so k2 must be -ve.
     
    Last edited: Jan 9, 2005
  9. Jan 9, 2005 #8
    thanks a lot guys!
     
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