Question

What is the numerical value of Kc for the following reaction if the equilibrium mixture contains 0.034 M N2O4 and 0.20 M NO2 ?N2O4(g)⇌2NO2(g) Express your answer to two significant figures.

Answer 1

The equilibrium constant Kc is the value that relates the molar concentrations of reactants and products. For a reaction with the following general balanced equation: aA+bB-->cC+dD, the value of the constant in equilibrium will be:

$Kc=\frac{{}[C\rbrack^c[D\rbrack^d}{[A]^a[B]^b}$

The square brackets represent the molar concentrations of the compounds.

Now, for this reaction we have 1 reactant and 1 product (aA--->cC), the equilibrium equation will then be:

$Kc={}\frac{\lbrack NO_2\rbrack^2}{\lbrack N_2O_4\rbrack^1}$

If we replace the molar concentrations we will then have Kc:

$Kc=\frac{\lbrack0.20M\rbrack^2}{\lbrack0.034M\rbrack}$$Kc=\frac{\lbrack0.20M\rbrack^2}{\lbrack0.034M\rbrack}=1.2M$

Answer: The numerical value of Kc of the reaction is 1.2M