2NO2(g) N2O4(g) a) At a total pressure of 1 bar and T=298.15 K, the partial pres

Question

2NO2(g) N2O4(g)

a) At a total pressure of 1 bar and T=298.15 K, the partial pressure of NO2(g) and N2O4(g) in an equilibrium mixture of the two gases are found to be 0.3181 and 0.6819 bar respectively. Calculate the equilibrium constant and o for the reaction.

b) Suppose you insert one mole of NO2(g) at 298.15 K into a different 1.0 L container. Determine the number of moles of NO2(g) and N2O4(g) present in the container after reaching equilibrium. In addition, compute G and A for the process. Is the dimerization spontaneous in this system?

Explanation / Answer

2NO2(g) N2O4(g)

1. Kp= PN2O4/(PNO2)2 = 0.6819/0.31812 = 6.74

dG0 = n0 = -RTlnKp

or, 2 * 0 = - 8.314*298.15*ln6.74

0 = -2.36kJ/mol

2. 2NO2(g) N2O4(g)

    1mol                 0

   1-2x                   x

Kp = KC(RT) ....( is change of mols in reaction, here the value is 1)

So, Kc = 0.0027 = x/(1-2x)2

Solving this we get, x= 0.049 M

So, [N2O4] = 0.049M

[NO2] = 1-2*0.049 = 0.9 M

dG = -RTlnKC

= -8.314*29815*ln0.049/0.92

=6.95 kJ

As the value is positive, the reaction is not spontaneous.

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