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Chemical equilibrium: Enthalpies and equilibrium contsants

  1. Feb 21, 2008 #1
    1. The problem statement, all variables and given/known data
    For the system 2 CO2 = 2 CO + O2, ∆H= 510 kJ
    the percentage decomposition of CO2 changes w/ temperature as follows.
    Temp, K.....% Decomposition
    Calculate the equilibrium constants, plot lnK vs. 1/T. In the graph, find the slope and confirm ∆H. Also, using average bond enthalpies, calculate ∆H(it should come out the same for both approaches).

    2. Relevant equations
    1.) Slope = ∆H / R
    2.) R(constant) = 8.314 J/k-mol
    3.) ∆Hrxn = Sum of bond enthalpies of broken bonds - Sum of bond enthalpies of formed bonds

    3. The attempt at a solution
    I started by calculating K at each temperature. We also did this part in class so I know all the K's are correct.
    K at 1500=5.5 x 10-9
    K at 2500=4.01 x 10-1
    K at 3000=4.03 x 10

    I plugged all these #s into the calculator to figure out lnK and 1/T. I plotted them on a graph and got a straight line with a negative slope. This threw me because lnK vs. 1/T should not be straight because this isn't a first order reaction. The slope was something like -40,000, which would then be multiplied by the constant R and ∆H came out to be -330,000(wrong).

    Then I tried the 2nd approach. Using equation #3 above, I got ∆H= (4x799)-((2x1072)+495). I'm well aware of how to draw Lewis structures, and am positive I got all the bond enthalpies right and ∆H came out to be 557kJ, which is much closer to the 510kJ given in the problem description.

    Where I am stuck is getting ∆H through the graph. Supposedly ∆H=Slope x R(constant)... but with such a small different in 1/T's or the ∆x value or the denominator in the rise/run slope formula, the slope always comes out enormous. How am I supposed to get ∆H for the first method? I looked over the numbers many times and they all seem correct.

    Any help please?
  2. jcsd
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