Question

At a particular temperature, K = 3.75 for the reaction

${\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\stackrel{}{\mathbf{\rightleftharpoons }}{\mathbf{SO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

If all four gases had initial concentrations of 0.800 M, calculate the equilibrium concentrations of the gases.

Short answer

The equilibrium concentrations of ${\text{SO}}_{\text{2}}$, ${\text{NO}}_{\text{2}}$, ${\text{SO}}_{\text{3}}$and$\text{NO}$ are 0.54 M. 0.54 M, 1.06 M and 1.06 M respectively.

Step-by-step solution

Equillibrium constant

The equilibrium constant, K expression in terms of concentrations is shown below:

$\text{K=}\frac{\left[{\text{SO}}_{\text{3}}\right]\left[\text{NO}\right]}{\left[{\text{SO}}_{\text{2}}\right]\left[{\text{NO}}_{\text{2}}\right]}$ …(1)

The initial concentrations of each gas in the reaction is given as 0.800 M.

Change in concentration

Let x be the change in concentration.

Setup the ICE table:

$\begin{array}{cccccccc}& {\text{SO}}_{\text{2}}\left(\text{g}\right)& \text{+}& {\text{NO}}_{\text{2}}\left(\text{g}\right)& \stackrel{}{\rightleftharpoons }& {\text{SO}}_{\text{3}}\left(\text{g}\right)& \text{+}& \text{NO}\left(\text{g}\right)\\ \text{Iinitial}\left(\text{M}\right)& \text{0.800}& & \text{0.800}& & \text{0.800}& & \text{0.800}\\ \text{Change}\left(\text{M}\right)& \text{\hspace{0.17em}–\hspace{0.17em}x}& & \text{\hspace{0.17em}–\hspace{0.17em}x}& & \text{+x}& & \text{+x}\\ \text{Equilibrium}\left(\text{M}\right)& \text{0.800\hspace{0.17em}–\hspace{0.17em}x}& & \text{0.800\hspace{0.17em}–\hspace{0.17em}x}& & \text{0.800+x}& & \text{0.800+x}\end{array}$

Substitute these equilibrium concentration values and the equilibrium constant value into equation (1) as shown below.

$\text{3.75=}\frac{\left(\text{0.800\hspace{0.17em}+\hspace{0.17em}x}\right)\left(\text{0.800\hspace{0.17em}+\hspace{0.17em}x}\right)}{\left(\text{0.800\hspace{0.17em}–\hspace{0.17em}x}\right)\left(\text{0.800\hspace{0.17em}–\hspace{0.17em}x}\right)}$

$\text{3.75=}\frac{{\left(\text{0.800\hspace{0.17em}+\hspace{0.17em}x}\right)}^{\text{2}}}{{\left(\text{0.800\hspace{0.17em}–\hspace{0.17em}x}\right)}^{\text{2}}}$

$\sqrt{\text{3.75}}\text{=}\frac{\left(\text{0.800\hspace{0.17em}+\hspace{0.17em}x}\right)}{\left(\text{0.800\hspace{0.17em}–\hspace{0.17em}x}\right)}$

Solve for x.

$\text{x=0.26}$

Hence, the change in concentration is 0.26 M.

Equillibrium concentration of all species

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

$\begin{array}{c}\left[{\text{SO}}_{\text{2}}\right]\text{=0.800\hspace{0.17em}–\hspace{0.17em}x}\\ \text{=0.800\hspace{0.17em}–\hspace{0.17em}0.26}\\ \text{=0.54\hspace{0.17em}M}\end{array}$

$\begin{array}{c}\left[{\text{NO}}_{\text{2}}\right]\text{=0.800\hspace{0.17em}–\hspace{0.17em}x}\\ \text{=0.800\hspace{0.17em}–\hspace{0.17em}0.26}\\ \text{=0.54\hspace{0.17em}M}\end{array}$

$\begin{array}{c}\left[{\text{SO}}_{\text{3}}\right]\text{=0.800+x}\\ \text{=0.800+0.26}\\ \text{=1.06\hspace{0.17em}M}\end{array}$

$\begin{array}{c}\left[\text{NO}\right]\text{=0.800+x}\\ \text{=0.800+0.26}\\ \text{=1.06\hspace{0.17em}M}\end{array}$

Therefore, the equilibrium concentrations of ${\text{SO}}_{\text{2}}$, ${\text{NO}}_{\text{2}}$,${\text{SO}}_{\text{3}}$ and$\text{NO}$ are 0.54 M. 0.54 M, 1.06 M and 1.06 M respectively.