Highly toxic disulfurdecafluoride decomposes by a free radical process: ${\mathbf{\text{S}}}_{\mathbf{2}}\mathbf{}{\mathbf{\text{F}}}_{\mathbf{10}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{\text{SF}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{SF}}}_{\mathbf{6}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$ . In a study of the decomposition,${\mathbf{\text{S}}}_{\mathbf{2}}\mathbf{}{\mathbf{\text{F}}}_{\mathbf{10}}$ was placed in a$\mathbf{2}\mathbf{.}\mathbf{0}$ $\mathbf{L}$- flask and heated to${\mathbf{100}}^{\mathbf{°}}\mathbf{\text{C}}\mathbf{;}\mathbf{[}{\mathbf{\text{S}}}_{\mathbf{2}}\mathbf{}{\mathbf{\text{F}}}_{\mathbf{10}}\mathbf{]}$ was$\mathbf{0}\mathbf{.50}\mathbf{\text{M}}$ at equilibrium. More${\mathbf{\text{S}}}_{\mathbf{2}}\mathbf{}{\mathbf{\text{F}}}_{\mathbf{10}}$ was added, and when equilibrium was retained,$\mathbf{[}{\mathbf{\text{S}}}_{\mathbf{2}}\mathbf{}{\mathbf{\text{F}}}_{\mathbf{10}}\mathbf{]}$ was$\mathbf{2}\mathbf{.5}\mathbf{\text{M}}$ . How diddata-custom-editor="chemistry" $\mathbf{\left[}{\mathbf{\text{SF}}}_{\mathbf{4}}\mathbf{\right]}$and$\mathbf{\left[}{\mathbf{\text{SF}}}_{\mathbf{6}}\mathbf{\right]}$ change from the original to the new equilibrium position after the addition of more data-custom-editor="chemistry" ${\mathbf{\text{S}}}_{\mathbf{2}}\mathbf{}{\mathbf{\text{F}}}_{\mathbf{10}}$?

Chemical equilibrium is a state in which no net change in the amounts of reactants and products occurs during a reversible chemical reaction.

In this issue, we must account for the change in product concentrations when an extra reactant is added while maintaining equilibrium. The reaction is depicted in the diagram below.

${\text{S}}_2{\text{F}}_{10}\left(g\right)\rightleftharpoons {\text{SF}}_4\left(g\right)+{\text{SF}}_6\left(g\right)$

First, write the expression for the equilibrium constant of the reaction in terms of concentration.

${\mathrm{K}}_{\mathrm{c}}=\frac{\left[\text{products}\right]}{\left[\text{reactants}\right]}$

Therefore, ${\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\text{SF}}_4\right]\left[{\text{SF}}_6\right]}{[{\text{S}}_2{\text{F}}_{10}]}$ .

During equilibrium, the amount of product are in factors of x multiplied by its coefficient since our products are 1 mol each, then $\left[{\text{SF}}_4\right]=\left[{\text{SF}}_6\right]$ . Solve for their concentration by denoting them as [product ] when $[{\text{S}}_2{\text{F}}_{10}]=0.50\text{M}$.

$\begin{array}{c}{\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\text{SF}}_4\right]\left[{\text{SF}}_6\right]}{[{\text{S}}_2{\text{F}}_{10}]}\\ {\mathrm{K}}_{\mathrm{c}}=\frac{{\left[\text{product}\right]}^2}{0.50}\\ {\left[\text{product}\right]}^2=0.50\cdot {\mathrm{K}}_{\mathrm{c}}\\ \left[\text{product}\right]=\sqrt{0.50{\mathrm{K}}_{\mathrm{c}}}\end{array}$

Next, do the same method but using the new equilibrium concentration of ${\text{S}}_2{\text{F}}_{10}$ which is$2.5\text{M}$ $\begin{array}{c}{\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\text{SF}}_4\right]\left[{\text{SF}}_6\right]}{[{\text{S}}_2{\text{F}}_{10}]}\\ {\mathrm{K}}_{\mathrm{c}}=\frac{{\left[\text{product}\right]}^2}{2.5}\\ \left[\text{product}\right]=2.5\cdot {\mathrm{K}}_{\mathrm{c}}\end{array}$.

Therefore, $\left[\text{product}\right]=\sqrt{2.5{\mathrm{K}}_{\mathrm{c}}}$

To solve for the change in concentration, we need to get the ratio of the two equations.

$1=\sqrt{0.50{\mathrm{K}}_{\mathrm{c}}}$

data-custom-editor="chemistry" ${\left[\text{product}\right]}_2=\sqrt{2.5{\mathrm{K}}_{\mathrm{c}}}$

data-custom-editor="chemistry" $\frac{[{\text{product}}_{\text{2}}}{[{\text{product}}_{\text{1}}}=\frac{\sqrt{2.5K_c}}{\sqrt{0.50K_c}}$

Therefore, the concentration of our products changed by a factor of data-custom-editor="chemistry" $2.2$.