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Answer 1 · 2017-07-15T03:59:06

$K_c = 1.77 xx 10^-3$

We're asked to calculate the $K_c$ for a reaction at a certain temperature.

Let's first write the equilibrium constant expression for this reaction:

$K_c = (["NO"]^2["O"_2])/(["NO"_2]^2)$

We can create a makeshift I.C.E. chart via bullet points as follows:

Initial Concentrations:

because there are no $"NO"$ or $"O"_2$ initially.

From the coefficients of the reaction equation, we expect the changes to be

Change in concentration:

Knowing that the equilibrium concentration of $"NO"_2$ is

$(0.179color(white)(l)"mol")/(2.00color(white)(l)"L") = 0.0895M$

We can calculate the actual change in concentration of $"NO"_2$ :

$"change" = 0.120M - 0.0895M = 0.0305M$

Which means $x$ is equal to

$(0.0305M)/(2) = color(red)(0.01525M$

So the equilibrium concentrations are

Equilibrium Concentrations:

We now plug these into the $K_c$ expression:

$K_c = ((0.0305M)^2(0.01525M))/((0.0895M)^2) = color(blue)(1.77 xx 10^-3$

Which agrees with our calculations that equilibrium lies to the left (more reactants present).