Calculation of enthelpy and internal energy or air in chemical equilibrium

In summary: R_{air} = R_u/\mu_{air} = 8314/0.3908 = 21245 J/(kg*K)Using equation (1), we can now calculate the enthalpy and internal energy per unit mass of the mixture:h = c_{O_2} * h_{O_2} + c_{N_2} * h_{N_2} = X_{O_2} * \mu_{O_2} * h_{O_2} + X_{N_2} * \mu_{N_2} * h_{N_2} = (0.01219)(32)(3.
  • #1
roldy
237
2

Homework Statement


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]

Homework 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]


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.
 
Physics news on Phys.org
  • #2
I also do not see where the given values for the enthalpies of formation and vibrational temperatures come into play. Any help would be appreciated.

Thank you for your question. To calculate the enthalpy and internal energy per unit mass of the mixture, we need to first determine the composition of the mixture. This can be done using the equilibrium constants provided for O2 and N2.

Using the ideal gas law, we can calculate the partial pressures of O2 and N2:
P_{O_2} = K_{p,O_2} * P = 12.19 * 0.1 = 1.219 atm
P_{N_2} = K_{p,N_2} * P = 0.7899 * 10^{-4} * 0.1 = 7.899 * 10^{-6} atm

Next, we can use the given values for the enthalpies of formation to calculate the enthalpy of each species at 4500 K:
h_{O_2} = (3/2)RT + (\Delta H_f^o)_O = (3/2)(259.8125)(4500) + 2.47 * 10^8 = 3.8219 * 10^6 J/kg
h_{N_2} = (3/2)RT + (\Delta H_f^o)_N = (3/2)(296.9286)(4500) + 4.714 * 10^8 = 4.2365 * 10^6 J/kg

To calculate the enthalpy and internal energy of the mixture, we first need to determine the mole fractions of each species:
X_{O_2} = P_{O_2}/P = 1.219/0.1 = 0.01219
X_{N_2} = P_{N_2}/P = 7.899 * 10^{-6}/0.1 = 7.899 * 10^{-5}

Next, we can use the mole fractions to calculate the molecular weight and specific gas constant of the mixture:
\mu_{air} = X_{O_2} * \mu_{O_2} + X_{N_2} * \mu_{N_2} = (0.01219)(32) + (7.899 * 10^{-
 

1. What is enthalpy and internal energy?

Enthalpy and internal energy are thermodynamic properties that describe the total energy of a system. Enthalpy is the sum of the internal energy and the product of the pressure and volume of the system, while internal energy is the total energy of the system at constant volume.

2. How is enthalpy and internal energy calculated?

Enthalpy and internal energy can be calculated using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. The equation for enthalpy is H = U + PV, where H is enthalpy, U is internal energy, P is pressure, and V is volume.

3. What is chemical equilibrium?

Chemical equilibrium is a state in which the forward and reverse reactions of a chemical reaction occur at equal rates. At equilibrium, the concentrations of reactants and products remain constant over time.

4. How is the enthalpy and internal energy of air in chemical equilibrium calculated?

The enthalpy and internal energy of air in chemical equilibrium can be calculated using the Gibbs free energy equation, which takes into account the temperature, pressure, and concentration of each component in the system. This calculation is often done using specialized software or thermodynamic tables.

5. What factors affect the enthalpy and internal energy of air in chemical equilibrium?

The enthalpy and internal energy of air in chemical equilibrium are affected by the temperature, pressure, and composition of the system. Changes in these factors can alter the equilibrium state of the system and therefore impact the enthalpy and internal energy values. Additionally, the type of chemical reaction and the presence of catalysts can also influence these values.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
723
  • Other Physics Topics
Replies
4
Views
1K
  • Biology and Chemistry Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Biology and Chemistry Homework Help
Replies
4
Views
6K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Materials and Chemical Engineering
Replies
4
Views
4K
Replies
7
Views
4K
  • Advanced Physics Homework Help
Replies
2
Views
3K
Replies
1
Views
2K
Back
Top