Question

a. Calculate the equilibrium constant Kp for the dissociation of dinitrogen tetroxide at 25c: N2O4 = 2NO2 Given Delta H of rxn = 57.20 kJ/mol
b. What is the value of Kp at 55c?
c. Calculate the degree of dissociation (alpha) of N2O4 at 55c with a pressure of 1.000 bar.

The quantity of alpha is defined as the fraction of dimer molecules that have dissociated.

Delta G of formation for N2O4= 97.787 kJ/mol
Delta G of formation for NO2 = 51.258 kJ/mol

Answer 1

∆G of reaction is given by

∆G = ∆G(product) - ∆G(reactant)

∆G =2× ∆G(NO2) - ∆G(N2O4)

∆G = (2×51.258kJ/mol ) - (97.787kJ/mol)

∆G = (102.516kJ/mol) - (97.787kJ/mol)

∆G = 4.729kJ/mol = 4729J/mol

(a)

Using the formula

∆G = - RTln(K_p) = -2.303RTlog(K_p)

Where R = gas constant = 8.314 J/mol-K

T = 25°C = 297K

Putting the all value in formula

4729J/mol = - 2.303(8.314J/mol-K)(297K)log(K_p)

log(K_p) = - 0.8315

K_p = 10^-0.8315 = 0.1474

(b)

T = 55°C = 55+273 = 328K

4729J/mol = - 2.303(8.314J/mol-K)(328K)log(K_p)

logK_p = -0.762

Kp = 10^-0.762 = 0.1729

(c)

N2O4 ------------------------------>2NO2

Initially. 1 . 0

at equilibrium

1 - . 2   

Kp = [NO2]²/[[N2O4]

0.1729 = (2)²/(1-)

0.1729 - 0.1729 = 4²

4² + 0.1729 - 0.1729

on solving this equation we get

= 0.0350