Question 32cda

Jul 15, 2017

${K}_{c} = 1.77 \times {10}^{-} 3$

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:

${K}_{c} = \left({\left[{\text{NO"]^2["O"_2])/(["NO}}_{2}\right]}^{2}\right)$

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

Initial Concentrations:

• ${\text{NO}}_{2}$: $\left(0.240 \textcolor{w h i t e}{l} \text{mol")/(2.00color(white)(l)"L}\right) = 0.120 M$

• $\text{NO}$: $0$

• ${\text{O}}_{2}$: $0$

because there are no $\text{NO}$ or ${\text{O}}_{2}$ initially.

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

Change in concentration:

• ${\text{NO}}_{2}$: $- 2 x$

• $\text{NO}$: $+ 2 x$

• ${\text{O}}_{2}$: $+ x$

Knowing that the equilibrium concentration of ${\text{NO}}_{2}$ is

$\left(0.179 \textcolor{w h i t e}{l} \text{mol")/(2.00color(white)(l)"L}\right) = 0.0895 M$

We can calculate the actual change in concentration of ${\text{NO}}_{2}$:

$\text{change} = 0.120 M - 0.0895 M = 0.0305 M$

Which means $x$ is equal to

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

So the equilibrium concentrations are

Equilibrium Concentrations:

• ${\text{NO}}_{2}$: $0.0895 M$

• $\text{NO}$: $2 \left(\textcolor{red}{0.01525 M}\right) = 0.0305 M$

• ${\text{O}}_{2}$: 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).