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How do I figure out the change in enthalpy?

  1. Mar 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Given the data:
    N2(g) + O2(g) → 2NO(g) ΔH = 180.7kJ
    2NO(g) + O2(g) → 2NO2(g) ΔH = -113.1kJ
    2N2O(g) → 2N2(g) + O2(g) ΔH = -163.2kJ

    use Hess's law to calculate ΔH for the reaction
    N2O(g) + NO2(g) → 3NO(g)

    2. Relevant equations
    "Hess's law states that if a reaction is carried out in a series of steps, ΔH for the overall reaction equals the sum of the enthalpy changes for the individual steps."

    3. The attempt at a solution
    ½(N2(g) + O2(g) → 2NO(g)) ΔH = 90.35kJ
    ½(2N2O(g) → 2N2(g) + O2(g)) ΔH = -81.6kJ

    3(N(g) + O(g) → NO(g)) ΔH = 271.05 kJ
    ½(2NO(g) + O2(g) → 2NO2(g)) ΔH = -56.55kJ
    N2(g) + O(g) → N2O(g) ΔH = 81.6kJ

    271.05 - (-56.55 + 81.6) = 246kJ
     
    Last edited: Mar 11, 2015
  2. jcsd
  3. Mar 11, 2015 #2

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    Product, good.
    One reactant, good.
    You omitted a step with this reactant.
     
  4. Mar 11, 2015 #3
    Did I forget to distribute the half?

    NO(g) + O(g) → NO2(g) ΔH = -56.55kJ
     
    Last edited: Mar 11, 2015
  5. Mar 11, 2015 #4

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    You forgot to "synthesize" the NO as far as getting a heat of formation for NO2.
     
  6. Mar 11, 2015 #5
    So -56.55 kJ isn't the heat of formation for nitrogen dioxide?
     
  7. Mar 11, 2015 #6

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    It's the heat of formation of NO2 from NO and oxygen, but not from nitrogen and oxygen, which would be the heat of formation as you've chosen to solve the problem.
     
  8. Mar 12, 2015 #7
    You need to find the coefficients of a linear combination of the three reactions that gives you the desired reaction. The third reaction is the only one that contains N2O, so it's coefficient must be 1/2. The second reaction is the only one that contains NO2, but it on the right hand side, so it's coefficient must be -1/2. To make good on these, the coefficient of the first reaction must be 1.

    Chet
     
  9. Mar 13, 2015 #8
    So is it (-113.1 - 180.7) = -293.8kJ?

    I don't understand. I just added the coefficients so as to match the number of each gas in the desired equation. What are you talking about?
     
  10. Mar 13, 2015 #9
    ##\frac{1}{2}(-163.2)-\frac{1}{2}(-113.1)+1(180.7)=155.65##

    Here's a check on my answer:

    Heat of formation of NO = 90.25

    Heat of formation of NO2 = 33.18

    Heat of formation of N2O = 82.05

    3(90.25) - 33.18 - 82.05 = 155.52

    How does that grab ya?

    Chet
     
    Last edited: Mar 13, 2015
  11. Mar 16, 2015 #10
    I don't understand why you divided both the 2NO2 and 2N2O by two, whereas you left the 2NO untouched. I checked with a tutor and he came up with the same thing. So why aren't we multiplying the heat of formation of 2NO by 3/2?
     
  12. Mar 16, 2015 #11
    Because N2 and O2 are not involved in the reaction for which you are trying to find the heat of reaction.

    Suppose you had 3 algebraic equations involving the 5 unknowns a, b, c, d, and e. The three equations are:

    a + b = 2c

    2c + b = 2d

    2e = 2a +b

    You are asked to form a linear combination of these three equations by multiplying each equation by a constant and then adding the resulting (new) three relationships together to obtain:

    e + d = 3c

    Are you able to figure out what the three constants would have to be in order for you to accomplish this?

    Chet
     
  13. Mar 17, 2015 #12
    Let's see:

    ½3(a + b) = ½3(2c)
    ½3a + ½3b = 3c
    The constant is ½3.

    ½(2c + b) = ½(2d)
    c + ½b = d

    ½(2e) = ½(2a + b)
    e = a + ½b
    The constant for the latter two is ½.

    (a + ½b) + (c + ½b) = (½3a + ½3b)
    a + b + c = (½3a + ½3b)
    -½a - ½b + c = 0
     
  14. Mar 17, 2015 #13
    Well, that correctly gives a/2 + b/2 = c, which would give the heat of formation of NO (parameter c). Of course, that could have been obtained just you multiplying the first equation by 1/2.

    But, the problem statement asks for the heat of the reaction e + d = 3c.

    What do you get if you multiply the first equation by 1, the second equation by -1/2, and the third equation by +1/2 and add them together?

    Chet
     
  15. Mar 17, 2015 #14
    a + b = 2c

    -½(2c + b) = -½(2d)
    -c - ½b = -d
    c + ½b = d

    ½2e = ½(2a + b)
    e = a + ½b

    (a + ½b) + (c + ½b) + a + b = 2a + 2b + c

    I'm still having trouble with where you're going with this.
     
  16. Mar 17, 2015 #15
    a + b = 2c
    -c - ½b = -d
    e = a + ½b

    Adding your 3 equations together:

    (a + b) - (c + ½b) + e = 2c - d + (a + ½b)

    Combining terms:

    e + d = 3c

    Chet
     
  17. Mar 17, 2015 #16
    Oh, so I had to insert both sides of each equation into the final one?
     
  18. Mar 18, 2015 #17
    Yes. That's what we have done. And note that the operations we carried out on the three equations are exactly the same as the operations we should perform on the three heats of reaction to arrive at the heat of the desired reaction.

    Chet
     
  19. Mar 25, 2015 #18
    Okay, thanks.

    I appreciate the help.
     
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