Question

Question #32cda

Answer 1

`K_c = 1.77 xx 10^-3`

Explanation:

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:

• `"NO"_2`: `(0.240color(white)(l)"mol")/(2.00color(white)(l)"L") = 0.120M`

• `"NO"`: `0`

• `"O"_2`: `0`

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:

• `"NO"_2`: `-2x`

• `"NO"`: `+2x`

• `"O"_2`: `+x`

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:

• `"NO"_2`: `0.0895M`

• `"NO"`: `2(color(red)(0.01525M)) = 0.0305M`

• `"O"_2`: `color(red)(0.01525M`

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).