Asystem was prepared with nh3 = o2 = 3.60 m as the only components initially. at equilibrium n2o4 is 0.60m. calculate the value of the equilibrium constant kc for the reaction

Kc = 0.0216

Explanation:

To do this, we need to write the equilibrium reaction which is the following:

NH3 + O2 <> N2O4 + H2O

And now, let's balance the equation

4NH3 + 7O2 <> 2N2O4 + 6H2O

The general expression for Kc is:

Kc = [N2O4]²[H2O]⁶ / [O2]⁷[NH3]⁴

Now, in order to know the concentrations, we need to do an ICE chart here, so we can know the final concentrations of all species here:

         4NH3 + 7O2 <> 2N2O4 + 6H2O

I:          3.6        3.6                      0            0

C:         -4x       -7x                      +2x        +6x

E:      3.6 - 4x   3.6 - 7x              0.6          +6x

According to this, we know that N2O4 is 0.6M, but in the change we have that it gained 2x so the value of x will be:

2x = 0.6

x = 0.6/2 = 0.3

Therefore, the value of x is 0.3 ans we can know now the concentrations of all species in equilibrium, and then, the kc:

[NH3] = 3.6 - 4(0.3) = 2.4 M

[O2] = 3.6 - 7(0.3) = 1.5 M

[H2O] = 6(0.3) = 1.8 M

Now the value of Kc would be:

Kc = (0.6)² (1.8)⁶ / (2.4)⁴ (1.5)⁷

Kc = 0.0216

5.6⋅10−^19  j

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