Question

For the reaction below at a certain temperature, it is found that the equilibrium concentrations in a 5.00-L rigid container are [H₂] = 0.0500 M, [F₂] = 0.0100 M, and [HF] = 0.400 M.

${\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{F}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\stackrel{}{\mathbf{\rightleftharpoons }}\mathbf{2}\mathbf{HF}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

If 0.200 mole of F₂ is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

Short answer

The equilibrium concentrations of ${\text{H}}_{\text{2}}$, ${\text{F}}_{\text{2}}$ and $\text{HF}$ are 0.0251 M, 0.0251 M and 0.450 M, respectively.

Step-by-step solution

The expression for equilibrium constant

The expression for the equilibrium constant, K of the given reaction in terms of concentrations is written below:

$\text{K=}\frac{{\left[\text{HF}\right]}^{\text{2}}}{\left[{\text{H}}_{\text{2}}\right]\left[{\text{F}}_{\text{2}}\right]}$ …(1)

The equilibrium concentrations are given as [H₂] = 0.0500 M, [F₂] = 0.0100 M, and [HF] = 0.400 M.

The volume of the container is given as 5.00 L.

The value of equilibrium constant

The equilibrium constant, K can be calculated from Equation (1) as follows:

$\begin{array}{c}\text{K=}\frac{{\left(\text{0.400}\right)}^{\text{2}}}{\left(\text{0.0500}\right)\left(\text{0.0100}\right)}\\ \text{=320}\end{array}$

Hence, the value of K is found to be 320.

New concentration of fluorine

The number of moles of fluorine gas added to the equilibrium mixture is 0.200 mol.

Hence, the new concentration of fluorine is,

$\begin{array}{c}\left[{\text{F}}_{\text{2}}\right]\text{=}\frac{\text{0.0100\hspace{0.17em}M×5.00\hspace{0.17em}L+0.200\hspace{0.17em}mol}}{\text{5.00\hspace{0.17em}L}}\\ \text{=}\frac{\text{0.0500\hspace{0.17em}mol+0.200\hspace{0.17em}mol}}{\text{5.00\hspace{0.17em}L}}\\ \text{=}\frac{\text{0.250\hspace{0.17em}mol}}{\text{5.00\hspace{0.17em}L}}\\ \text{=0.0500\hspace{0.17em}M}\end{array}$

The value of change in concentration “x”

Let x be the change in concentration.

Setup the ICE table as:

$\begin{array}{cccccc}& {\text{H}}_{\text{2}}\left(\text{g}\right)& \text{+}& {\text{F}}_{\text{2}}\left(\text{g}\right)& \stackrel{}{\rightleftharpoons }& \text{2HF}\left(\text{g}\right)\\ \text{Iinitial}\left(\text{M}\right)& \text{0.0500}& & \text{0.0500}& & \text{0.400}\\ \text{Change}\left(\text{M}\right)& \text{\hspace{0.17em}–\hspace{0.17em}x}& & \text{\hspace{0.17em}–\hspace{0.17em}x}& & \text{+2x}\\ \text{Equilibrium}\left(\text{M}\right)& \text{0.0500\hspace{0.17em}–\hspace{0.17em}x}& & \text{0.0500\hspace{0.17em}–\hspace{0.17em}x}& & \text{0.400+2x}\end{array}$

Insert the value of K and the equilibrium concentrations of all the gases from ICE table into equation (1) to find the value of x.

$\begin{array}{l}\text{320=}\frac{{\left(\text{0.400+2x}\right)}^{\text{2}}}{\left(\text{0.0500\hspace{0.17em}–\hspace{0.17em}x}\right)\left(\text{0.0500\hspace{0.17em}–\hspace{0.17em}x}\right)}\\ \text{320=}\frac{{\left(\text{0.400+2x}\right)}^{\text{2}}}{{\left(\text{0.0500\hspace{0.17em}–\hspace{0.17em}x}\right)}^2}\\ \sqrt{\text{320}}\text{=}\frac{\left(\text{0.400+2x}\right)}{\left(\text{0.0500\hspace{0.17em}–\hspace{0.17em}x}\right)}\end{array}$

Solve for x.

$\text{x=0.0249}$

Equilibrium concentrations

The equilibrium concentrations of all the gases can be calculated by using the x value as follows:

$\begin{array}{c}\left[{\text{H}}_{\text{2}}\right]\text{=0.0500\hspace{0.17em}–\hspace{0.17em}x}\\ \text{=0.0500\hspace{0.17em}–\hspace{0.17em}0.0249}\\ \text{=0.0251\hspace{0.17em}M}\end{array}$

$\begin{array}{c}\left[{\text{F}}_{\text{2}}\right]\text{=0.0500\hspace{0.17em}–\hspace{0.17em}x}\\ \text{=0.0500\hspace{0.17em}–\hspace{0.17em}0.0249}\\ \text{=0.0251\hspace{0.17em}M}\end{array}$

$\begin{array}{c}\left[\text{HF}\right]\text{=0.400+2x}\\ \text{=0.400+2×0.0249}\\ \text{=0.450\hspace{0.17em}M}\end{array}$

Hence, the equilibrium concentrations of ${\text{H}}_{\text{2}}$, ${\text{F}}_{\text{2}}$ and $\text{HF}$ are 0.0251 M, 0.0251 M and 0.450 M, respectively.