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If only 0.500 mol of NO2(g) is placed in a 1.0 L container and the reaction is allowed to come to equilibrium, 0.186 mol of N2O4(g) is formed. Find the value of Keq.
Solve using ICE table.

Sagot :

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The equilibrium constant of the system is 11.3.

What is the value of Keq?

We know that the equilibrium constant shows the extent to which reactants are converted into products.

Now;

2NO2 ⇄ N2O4

[NO2] = 0.500 mol /1 L =  0.500 M

[N2O4] =  0.186 mol  /1 L =  0.186 M

The ICE table is;

          2NO2 ⇄ N2O4

I           0.5           0

C         -2x              +x

E          0.5 - 2x     0.186

The concentration of NO2 at equilibrium = 0.5 - 2(0.186) = 0.128

Keq =  0.186/( 0.128)^2

Keq = 11.3

Learn more about equilibrium constant:https://brainly.com/question/10038290

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