Problem: Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equationN2O4(g) ⇌ 2NO2(g).If at equilibrium N2O4 is 28.0% dissociated, what is the value of the equilibrium constant, Kc, for the reaction under these conditions?A) 0.44B) 2.3C) 0.31D) 0.78E) 0.11

FREE Expert Solution

We’re being asked to calculate the equilibrium constant of HA if a 0.0538 M solution is 3.57% ionized.


Recall that the percent ionization is given by:


% dissociation =[A]initial - [A]final[A]initial×100


The dissociation is as follows:


N2O4(g) ⇌ 2NO2(g)


From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table and Ka expression.



The Kc expression for is:


Kc=productsreactants=[NO2]2[N2O4]


Note that each concentration is raised by the stoichiometric coefficient: [NO2] is raised to 2 and [N2O4] is raised to 1.


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Problem Details

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation

N2O4(g) ⇌ 2NO2(g).

If at equilibrium N2O4 is 28.0% dissociated, what is the value of the equilibrium constant, Kc, for the reaction under these conditions?

A) 0.44

B) 2.3

C) 0.31

D) 0.78

E) 0.11

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