Consider the equilibrium reaction: 2NO_2(g) N_2O_4(g) You will obtain all your n

Question

Consider the equilibrium reaction: 2NO_2(g) N_2O_4(g) You will obtain all your needed data from standard thermodynamic tables. Compute Delta H_r Times n at 298.15 K. Compute Delta G for the reaction and use it to compute K_P and K_c at 298.15 K. Using the heat capacities of both species, obtain an analytical expression for Delta H_r Times n as a function of temperature. Produce analytical expressions for K_p and K_c as a function of temperature and plot both simultaneously as ln(K) vs. 1/T in the range of 298 K to 1000 K. Why are the plots nearly linear and justify our not using Cp's as functions of temperature. One mole of NO_2(g) is placed in a 2 L constant volume container. Use K_c to obtain analytical expressions for the equilibrium concentrations of both species. Using a log scale, plot these concentrations over the same temperature range as above. Obtain and complete the MCAD worksheet: Gibbs_Rxn_Min_NO2_N2O4_Stud_Excrcise_G_given.xmcd

Explanation / Answer

2 NO2 (g) <-----------> N2O4 (g)

a)

Horxn = Hoproducts - Horeactants

            = HoN2O4 - 2 x HoNO2

           =   9.16 - 2 x 33.18

           = - 57.2 kJ/mol

Horxn   = - 57.2 kJ/mol

b)

Gorxn = Goproducts - Goreactants

            =   GoN2O4 - 2 x GoNO2

                 =   97.82 - 2 x 51.30

            = - 4.78 kJ/mol

Gorxn = - 4.78 kJ/mol

G= RT lnKc

- 4.78 = - 8.314 x 10^-3 x 298.15 x ln Kc

Kc = 6.876

Kp = Kc (RT)Dn               (Dn = -1)

Kp = 6.876 x (0.0821 x 298.15)-1

     = 0.281

Kp = 0.281

e)

2 NO2 (g) <-----------> N2O4 (g)

0.5                                 0

0.5 - 2x                            x

Kc = [N2O4] / [NO2]^2

6.876 = x / (0.5-2x)^2

x = 0.232

equilibrium concentration of

[N2O4] = x = 0.232 M

[NO2] = 0.5 - 2x = 0.5 - 2 x 0.232

           = 0.036 M

[NO2] = 0.036 M

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