Question

At$\mathbf{298}\mathbf{}\mathbf{\text{K}}$, unequal amounts of${\mathbf{\text{BCl}}}_{\mathbf{3}}\mathbf{(}\mathbf{}\mathbf{\text{g}}\mathbf{)}$and${\mathbf{\text{BF}}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$were mixed in a container. The gases reacted to form${\mathbf{\text{BFCl}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$and${\mathbf{\text{BClF}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$. When equilibrium was finally reached, the four gases were present in these relative chemical amounts:${\mathbf{\text{BCl}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{90}\mathbf{\right)}$,${\mathbf{\text{BF}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{470}\mathbf{\right)}$ ,${\mathbf{\text{BClF}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{200}\mathbf{\right)}$ ,${\mathbf{\text{BFCl}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{45}\mathbf{\right)}$ .

1. Determine the equilibrium constants at$\mathbf{298}\mathit{K}$of the two reactions

role="math" localid="1663593368437" $$\begin{gathered} \mathbf{2}{\mathbf{\text{BCl}}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{BF}}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftarrows }\mathbf{3}{\mathbf{\text{BFCl}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)} \\ {\mathbf{\text{BCl}}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{\text{BF}}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{3}{\mathbf{\text{BClF}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)} \end{gathered}$$

1. Determine the equilibrium constant of the reaction

${\mathbf{\text{BCl}}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{BF}}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftarrows }{\mathbf{\text{BFCl}}}_{\mathbf{2}}\mathbf{+}{\mathbf{\text{BClF}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

and explain why knowing this equilibrium constant really adds nothing to what you knew in part (a)

Short answer

(a) The equilibrium constants at $298\text{K}$of the two reactions are$0.024$ and$0.402$

(b) The equilibrium constant of the reaction is$0.213$

Step-by-step solution

Given information

The reaction form${\text{BClF}}_2$and${\text{BFCl}}_2$from unknown amounts of ${\text{BCl}}_3$and ${\text{BF}}_3$When equilibrium was reached these were the chemical amounts :${\text{BCl}}_3\left(90\right)$,${\text{BF}}_3\left(470\right)$,${\text{BClF}}_2\left(200\right)$${\text{BClF}}_2\left(200\right)$,${\text{BFCl}}_2\left(45\right)$

Concept of equilibrium constant

The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium, a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.

Calculate the equilibrium constant

(a)

The expression for equilibrium constant for reaction$n$reactants$\leftrightharpoons m$products is

$K=\frac{{\left[\text{products}\right]}^m}{{\left[\text{reactanls}\right]}^n}$

The equilibrium constant expression is$2{\text{BCl}}_3\left(g\right)+{\text{BF}}_3\left(g\right)\leftrightharpoons 3{\text{BFCl}}_2\left(g\right)$:is calculated below.

$$\begin{gathered} K_1=\frac{{\left[{\text{BFCl}}_2\right]}^3}{{\left[{\text{BCl}}_3\right]}^2\left[{\text{BF}}_3\right]} \\ {K}_1=\frac{{45}^3}{{90}^2\times 470} \\ {K}_1=0.024 \end{gathered}$$

The equilibrium constant for${\text{BCl}}_3\left(g\right)+2{\text{BF}}_3\left(g\right)\leftrightharpoons 3{\text{BClF}}_2\left(g\right)$is calculated as,

$$\begin{gathered} K_2=\frac{{\left[BCl{F}_2\right]}^3}{{\left[BC{l}_3\left]\right[B{F}_3\right]}^2} \\ {K}_2=\frac{{200}^3}{90\times {470}^2} \\ {K}_2=0.402 \end{gathered}$$

Evaluate the value of equilibrium constant

(b)

The reaction is given below

${\text{BCl}}_3\left(g\right)+{\text{BF}}_3\left(g\right)\leftrightharpoons {\text{BCIF}}_2\left(g\right)+{\text{BFCl}}_2\left(g\right)$

This equilibrium constant expression is a combination of equations 1 and 2

$$\begin{gathered} 1.2{\text{BCl}}_3\left(g\right)+{\text{BF}}_3\left(g\right)\rightleftarrows 3{\text{BFCl}}_2\left(g\right) \\ {\text{2.BCl}}_3\left(g\right)+2{\text{BF}}_3\left(g\right)\rightleftarrows 3{\text{BClF}}_2\left(g\right) \end{gathered}$$

The final equation becomes.

$$\begin{gathered} 3{\text{BCl}}_3\left(g\right)+3{\text{BF}}_3\left(g\right)\leftrightharpoons 3{\text{BFCl}}_2\left(g\right)+3{\text{BClF}}_2\left(g\right) \\ {\text{BCl}}_3\left(g\right)+{\text{BF}}_3\left(g\right)\leftrightharpoons {\text{BClF}}_2\left(g\right)+{\text{BFCl}}_2\left(g\right) \end{gathered}$$

The equilibrium constant for final reaction would be

$$\begin{gathered} K=\sqrt[3]{K_1{K}_2} \\ K=\sqrt[3]{0.024\times 0.402} \\ K=0.213 \end{gathered}$$