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Reaction Feasibility question

  1. Apr 17, 2017 #1
    PzoYX.png

    How would one approach part I and II? In terms of thermodynamics I'm not sure how I can show that this hypothesis is true. Could I work out the equilibrium constant and make a decision based on its magnitude? For example if it is >>> 1 then the hypothesis is true, would that be correct?

    For the second part I am completely lost.

    Any help will be appreciated, thanks.
     
  2. jcsd
  3. Apr 22, 2017 #2
    You are correct regarding part (i). The first step is to determine the equilibrium constant at 800 K. Then you determine the conversion of the proposed mixture at equilibrium. See if any significant conversion is possible, based on the equilibrium constant. This gives you the thermodynamic answer.
     
  4. Apr 23, 2017 #3

    Thanks Chet, any tips on how to approach part ii?
     
  5. Apr 23, 2017 #4
    Part ii could depend on what we see about the answer to part i. Let's worry about part ii later.
     
  6. Apr 24, 2017 #5
    So for the first part, I'm getting a strange answer. First getting the standard entropy change:

    $$
    \Delta S^o = \frac{\Delta G^o - \Delta H^o}{-298} = -421.6 \frac{kJ}{kmol K}
    $$

    Getting change in heat capacities:
    $$
    \Delta C_p = C_p(CH_4) - 2C_p(H_2) - C_p(C) = -44.237 + 0.03027T \frac{kJ}{kmol K}
    $$

    Getting the parameters at 800K:
    $$
    \Delta H^o_{800} = \Delta H^o_{298} + \int^{800}_{298} \Delta C_p dt = -88704.6225 \frac{kJ}{kmol}
    $$
    Entropy:

    $$
    \Delta S^o_{800} = \Delta S^o_{298} + \int^{800}_{298} \frac{\Delta C_p}{T}dt = -450.1 \frac{kJ}{kmol}
    $$

    Calculating change in Gibbs at 800K
    $$
    \Delta G^o_{800} = \Delta H^o_{800} - T\Delta S^o_{800} = 271378.3775\frac{kJ}{kmol}
    $$

    This is a very high change in Gibbs free energy, K is pretty much 0, this doesn't seem right at all, have I made a mistake?
     
    Last edited: Apr 24, 2017
  7. Apr 24, 2017 #6
    I roughly confirm your calculations. I got a ##\Delta G## at 800 K of 271346 kJ/kmol. I also did it by using $$\frac{d(\Delta G/T)}{d(1/T)}=\Delta H$$with an average value of ##\Delta H## over the temperature interval of -81767 kJ/kmol. This gave about 274000 kJ/kmol. For an equilibrium constant, I got 1.25E-9 bars^(-1) at 298 K and 1.93E-18 bars^(-1) at 800 K.
     
  8. Apr 24, 2017 #7
    I checked, and the gibbs free energy of formation of methane is -50794 kJ/kmol, not +50794 kJ/kmol.
     
  9. Apr 25, 2017 #8
    Damn, sorry about that.

    Recalculating with the new Gibbs free energy of formation I arrive at a K value of 1.23 bars^{-1} which is more appropriate, the conversion will also be large. So the reaction is definitely feasible/spontaneous at 800K, I can work out the conversion but I think it will be close to one.

    Which leaves me with the second part, I don't understand how changing the composition will change the feasibility of the reaction, do you have any hints/tips?
     
  10. Apr 25, 2017 #9
    Why don't you first work out the equilibrium mole fractions? I don't think the conversion is going to be close to one.
     
  11. Apr 25, 2017 #10
    I'll try..
    $$
    K = 1.23 bars = \bigg[ \frac{P_{CH_{4}}}{P_{H^2_{2}}}\bigg] = \frac{1}{15} \bigg[ \frac{y_{CH_{4}}}{y^2_{H_{2}}} \bigg]
    $$

    This is where I might start being wrong, we start with C + 2H2 --> CH4, but the question states that the 'feed' is 30% hydrogen and 70% methane. So taking the feed as one mol..:

    $$
    \text{Starting Moles: } H_2 = 0.3 \qquad CH_4 = 0.7 \\
    \text{End Moles: } H_2 = 0.3 - \zeta \qquad CH_4 = 0.7 + 0.5\zeta \\
    \text{Total End Moles: } 1 -0.5\zeta
    $$

    Mole fractions:

    $$
    y_{CH_{4}} = \frac{0.7+0.5\zeta}{1-0.5\zeta} \qquad y^2_{H_{2}} = \frac{(0.3-\zeta)^2}{(1-0.5\zeta )^2} \implies 1.23\times 15 = 18.45 = \frac{(0.7+0.5\zeta)(1-0.5\zeta )}{(0.3-\zeta)^2}
    $$

    Can solve the above and get extent of reaction values of 0.1 and 0.495, I discard 0.495 because that's greater than the number of starting moles, so that leaves me with a extent of reaction value of 0.1 or a conversion of 33.3%, does my working look rightish?
     
  12. Apr 25, 2017 #11
    I got 0.104 for ##\zeta##. What do you get for the final mole fractions?
     
  13. Apr 25, 2017 #12

    0.8 for methane and 0.2 for hydrogen, to 1 DP.
     
  14. Apr 25, 2017 #13
    I have an idea for the second part, we don't want the reaction to go forward, so we're looking for a negative extent of reaction or an extent of reaction of 0. So if we go back to my original statement for extent:
    $$
    18.45 = \frac{(0.7+0.5\zeta)(1-0.5\zeta )}{(0.3-\zeta)^2}
    $$
    I can let the mole amount of hydrogen = some constant, say alpha and solve? For example:
    $$
    18.45 = \frac{(1-\alpha)}{\alpha} \implies \alpha = 0.05 \text{moles}
    $$
    So if they were to distil that mixture, they would need to make it so the hydrogen composition was at max 5%. What do you think?
     
    Last edited: Apr 25, 2017
  15. Apr 25, 2017 #14
    I think you've shown that the equilibrium mole fraction of H2 is 0.2 and that, if the mole fraction is more than that, the graphite will erode. So this is the maximum mole fraction that can be tolerated in the feed. Also, distilling the snow to a lower mole fraction than that will require more energy utilization.

    You didn't answer the second part of question (i), related to what additional information would be required.
     
  16. Apr 25, 2017 #15
    -
     
    Last edited: Apr 25, 2017
  17. Apr 25, 2017 #16
    My bad, forgetting the basics.

    For the second part I'm not sure, perhaps the enthalpy of vaporisation/fusion, assuming the enthalpy of reaction is quoted in terms of the substances in their gas forms, but the question says the feed is frozen, we would need the enthalpy of vaporisation and fusion to adjust?
     
  18. Apr 25, 2017 #17
    My guess is that distillation is not the focus. I assume that they are saying that part of the methane would just be discarded, or possibly used as a heat source for the reboiler.
     
  19. Apr 26, 2017 #18
    I see, that sounds more appropriate.

    Thanks again for your help.
     
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