1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculation of enthalpy and internal energy or air in chemical equilibrium

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider air in chemical equilibrium at 0.1 atm and T=4500 K. The chemical species are O2, O, N2, N (Ignore NO). Calculate the enthalpy and internal energy per unit mass of the mixture. Neglect electronic excitation in your calculations.

    [tex]K_{p,O_2}=12.19 atm[/tex]
    [tex]K_{p,N_2}=0.7899*10^{-4}atm[/tex]

    [tex](\Delta H_f^o)_O=2.47*10^8 J/(kg*mol)[/tex]
    [tex](\Delta H_f^o)_N=4.714*10^8 J/(kg*mol)[/tex]

    [tex](\Theta_v)_{N_2}=3390 K[/tex]
    [tex](\Theta_v)_{O_2}=2270 K[/tex]

    2. Relevant equations
    [tex](1)h=\Sigma ^n_{i=1}c_ih_i + \Sigma ^n_{i=1}c_i \Delta h_{f_i}[/tex]

    [tex]c_i=X_i\frac{\mu_i}{\mu}[/tex]
    [tex]X_i=\frac{P_i}{P}[/tex]

    [tex]R_{air}=\frac{R_u}{\mu_{air}}[/tex]
    [tex]\mu_{air}=X_{O_2}*\mu_{O_2} + X_O*\mu_O + X_N*\mu_N + X_{N_2}*\mu_{N_2}[/tex]

    Diatomic Gas:
    [tex]h=e_{sens}+RT[/tex]
    [tex]e_{sens}=3/2RT + RT +\frac{\frac{\Theta_v}{T}}{e^{\frac{\Theta_v}{T}}-1}RT[/tex]

    Monatomic Gas:
    [tex]h=5/2RT[/tex]
    [tex]=e_{sens}=3/2RT[/tex]


    3. The attempt at a solution

    The weight of each species is as follows:
    [tex]O=16kg/(kg*mol)[/tex]
    [tex]N=14kg(kg*mol)[/tex]
    [tex]O_2=32kg/(kg*mol)[/tex]
    [tex]N_2=28kg/(kg*mol)[/tex]

    I then found what the gas constant was for each species with Ru=8314 J/(kg*K).
    [tex]R_O=519.625[/tex]
    [tex]R_N=593.8571[/tex]
    [tex]R_{O_2}=259.8125[/tex]
    [tex]R_{N_2}=296.9286[/tex]

    I then calculated the sensible internal enthalpy for each species.
    [tex]e_{sense,O}=3/2(519.625)(4500)=3.5075*10^6[/tex]
    [tex]e_{sense,N}=3/2(593.8571)(4500)=4.0085*10^6[/tex]
    [tex]e_{sense,O_2}=3/2(259.8125)(4500) + (259.8125)(4500) +\frac{\frac{2270}{4500}}{e^{\frac{2270}{4500}}-1}(259.8125)(4500)=3.8218*10^6[/tex]
    [tex]e_{sense,N_2}=3/2(296.9286)(4500) + (296.9286)(4500) +\frac{\frac{3390}{4500}}{e^{\frac{3390}{4500}}-1}(296.9286)(4500)=4.2359*10^6[/tex]

    Next the enthalpy per unit mass of each species was calculated.
    [tex]h_O=5/2(519.625)(4500)=5.8458*10^6[/tex]
    [tex]h_N=5/2(593.8571)(4500)=6.6809*10^6[/tex]
    [tex]h_{O_2}=3.8218*10^6+259.8125(4500)=4.9910*10^6[/tex]
    [tex]h_{N_2}=4.2359*10^6+296.9286(4500)=5.5721*10^6[/tex]

    This is where I get stuck. I believe I use equation (1) in some way but I do not see how since I am not given the partial pressures to find [tex]c_i[/tex]. I'm not sure how the equilibrium constants work into this problem.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Calculation of enthalpy and internal energy or air in chemical equilibrium
Loading...