Question

At a particular temperature, $\mathbf{K}=\mathbf{2}.\mathbf{0}\times {\mathbf{10}}^{-\mathbf{6}}$ for the reaction

$\mathbf{2}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\stackrel{}{\mathbf{\rightleftharpoons }}\mathbf{2}\mathbf{CO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

If 2.0 moles of CO₂ is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species.

Short answer

The equilibrium concentrations of ${\text{CO}}_{\text{2}}$, $\text{CO}$ and ${\text{O}}_{\text{2}}$ are 0.39 M, 0.0086 M and 0.0043 M, respectively.

Step-by-step solution

The expression for equilibrium constant

The equilibrium constant expression for the given reaction is as follows:

$\text{K=}\frac{{\left[\text{CO}\right]}^{\text{2}}\left[{\text{O}}_{\text{2}}\right]}{{\left[{\text{CO}}_{\text{2}}\right]}^{\text{2}}}$ …(1)

The initial number of moles of CO₂ is given as 2.0 mol and the given volume of the container is 5.0 L.

The initial concentration of can be determined as follows:

$\begin{array}{c}\left[{\text{CO}}_{\text{2}}\right]\text{=}\frac{{\text{Moles\hspace{0.17em}of\hspace{0.17em}CO}}_{\text{2}}}{\text{Volume}}\\ \text{=}\frac{\text{2.0\hspace{0.17em}mol}}{\text{5.0\hspace{0.17em}L}}\\ \text{=0.40\hspace{0.17em}M}\end{array}$

ICE table

Assume x as the change in concentration.

Setup the ICE table as:

$\begin{array}{cccccc}& {\text{2CO}}_{\text{2}}\left(\text{g}\right)& \stackrel{}{\rightleftharpoons }& \text{2CO}\left(\text{g}\right)& \text{+}& {\text{O}}_{\text{2}}\left(\text{g}\right)\\ \text{Iinitial}\left(\text{M}\right)& \text{0.40}& & \text{0}& & \text{0}\\ \text{Change}\left(\text{M}\right)& \text{\hspace{0.17em}–\hspace{0.17em}2x}& & \text{+2x}& & \text{+x}\\ \text{Equilibrium}\left(\text{M}\right)& \text{0.40\hspace{0.17em}–\hspace{0.17em}2x}& & \text{2x}& & \text{x}\end{array}$

Now, substitute these equilibrium concentration terms and the equilibrium constant value into equation (1) as shown below:

${\text{2.0×10}}^{\text{\hspace{0.17em}–\hspace{0.17em}6}}\text{=}\frac{{\left(\text{2x}\right)}^{\text{2}}\left(\text{x}\right)}{{\left(\text{0.40\hspace{0.17em}–\hspace{0.17em}2x}\right)}^{\text{2}}}$

Assume x << 0.40 because K is very small.

${\text{2.0×10}}^{\text{\hspace{0.17em}–\hspace{0.17em}6}}\text{=}\frac{{\left(\text{2x}\right)}^{\text{2}}\left(\text{x}\right)}{{\left(\text{0.40}\right)}^{\text{2}}}$

Solve for x.

$\text{x=0.0043}$

Therefore, the change in concentration, x is 0.0043 M.

The equilibrium concentrations

The equilibrium concentrations of all the chemical species can be calculated as follows:

$\begin{array}{c}\left[{\text{CO}}_{\text{2}}\right]\text{=0.40\hspace{0.17em}–\hspace{0.17em}2x}\\ \text{=0.40\hspace{0.17em}–\hspace{0.17em}2×0.0043}\\ \text{=0.39\hspace{0.17em}M}\end{array}$

$\begin{array}{c}\left[\text{CO}\right]\text{=2x}\\ \text{=2×0.0043}\\ \text{=0.0086\hspace{0.17em}M}\end{array}$

$\begin{array}{c}\left[{\text{O}}_{\text{2}}\right]\text{=x}\\ \text{=0.0043\hspace{0.17em}M}\end{array}$

Hence, the equilibrium concentrations of ${\text{CO}}_{\text{2}}$, $\text{CO}$ and ${\text{O}}_{\text{2}}$ are 0.39 M, 0.0086 M and 0.0043 M, respectively.