Dissociation of Dinitrogen Tetra oxide

Dinitrogen Tetra oxide

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Dissociation of N₂O₄ (Dinitrogen Tetra oxide)

N₂O₄    2NO₂

Let   ‘a’  moles of N₂O₄ in a closed vessel of Volume ‘v’  litre

N₂O₄        2NO₂

Initial Concⁿ (mole/l)                 a/V                         0

Concⁿ at equi. (mole/l)             ( a-x)/V                   2x/V

Applying law of mass action,

Kc =[NO2]2 /[N2O4]

=  [2x/V ]2 /[ ( a-x)/V]

=4×2 / [( a-x)/V]

Factors affecting Equilibrium –

1. Effect of Pressure-

If pressure is increased the value of  ‘V’ decreases  to maintain the constt. value of Kc,  x  must also decrease hence the dissociation of N₂O₄ decreases with increase of pressure. Low pressure favours the dissociation of N₂O₄

2. Effect of Temperature-
3. N₂O4  2NO₂ – QK.Cal

            The dissociation of N2O4 is an endothermic reaction, thus the values of equilibrium constant increases with increase in temperature (because  Kc =Kf/Kb )

It is concluded that dissociation of N₂O₄ increases with increase in temperature.

3. Effect of Concⁿ–

If N₂O₄ is added in equilibrium mixture the rate of forward reaction increases. Therefore dissociation of N₂O₄ increases. In case NO₂ is added the rate of backward reacⁿ increases & dissociation of N₂O₄ decreases.

Q – The vapour density of N2O4 at acertain temp. is 30. calculate % of dissociation  of N2O4 at this temp.?

Ans.                N2O4   2NO2

initial mole      1                    0

moles at equi.   1-x              2x

total moles at equi. =1-x +2x =1+x

( 1 +x) /1 =D/d

d =30

D = mol.wt. /2 =92 /2 =46

(1 +x) /1 =D/d

( 1+x) /1 =46/30

x = 0.533

% of x =53.3 % Ans.