Q171CP - Chemistry: Molecular Nature Of Matter And Change

Question

The Group 5A(15) hydrides react with boron trihalides in a reversible Lewis acid-base reaction. When 0.15 mol of ${\mathbf{\text{PH}}}_{\mathbf{\text{3}}}{\mathbf{\text{BCl}}}_{\mathbf{\text{3}}}$  (s) is introduced into a  3.0 -Container at a certain temperature, $\mathbf{\text{8.4}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{ - 3}}}$ mol of  ${\mathbf{\text{PH}}}_{\mathbf{\text{3}}}$ is present at equilibrium: 

${\mathbf{\text{PH}}}_{\mathbf{\text{3}}}{\mathbf{\text{BCl}}}_{\mathbf{\text{3}}}\mathbf{\text{(s)}}\mathbf{\rightleftharpoons }{\mathbf{\text{PH}}}_{\mathbf{\text{3}}}{\mathbf{\text{(g) + BCl}}}_{\mathbf{\text{3}}}\mathbf{\text{(g)}}$

(a) Find ${\text{K}}_{\text{c}}$for the reaction at this temperature. 

(b) Draw a Lewis structure for the reactant

Step-by-Step Solution

Answer

a) ${\text{K}}_{\text{c}}\text{ = 7.8}\times {\text{10}}^{\text{ - 6}}$

b) The Lewis structure for the reactant is as shown below.

Step 1(a): Find the K c for the reaction

Given information:

$0.15 \text{mol} {\text{PH}}_3{\text{BCl}}_3$

$$\begin{gathered} 8.4\times {10}^{-3} \mathrm{m} \\ \mathrm{V} =3.00 \mathrm{L} \end{gathered}$$

First, write the ${\text{K}}_{\text{c}}$  expression for the reaction. We will not include the solid chemical species since this will be constant.

$$\begin{gathered} {\text{K}}_{\text{c}}\text{ = }\frac{\text{[ product ]}}{\text{[ reactant ]}} \\ {\text{K}}_{\text{c}}\text{ = }\left[{\text{PH}}_{\text{3}}\right]\left[{\text{BCl}}_{\text{3}}\right] \end{gathered}$$

Next, solve for the concentration of the ${\text{PH}}_3,$ which is also equal to the concentration of ${\text{BCl}}_3$ since they are produced in the reaction with 1 mol each. We will not solve for the concentration of ${\text{PH}}_3{\text{BCl}}_3$  since we do not need it to solve for ${\text{K}}_{\text{c}}$

$\begin{array}{rcl}& & {\text{K}}_{\text{c}}\text{ = } \left[{\text{PH}}_{\text{3}}\right]\left[{\text{BCl}}_{\text{3}}\right]\\ & & \text{ = } {\left[{\text{PH}}_{\text{3}}\right]}^{\text{2}}\\ & & \text{ = } {\left(\text{2.8}\times {\text{10}}^{\text{ - 3}}\right)}^{\text{2}}\\ & & \text{ = } \text{7.8}\times {\text{10}}^{\text{ - 6}}.\end{array}$

Note that ${\text{PH}}_3{\text{BCl}}_3$ Solve for ${\text{K}}_{\text{c}}$

 

Hence, ${\text{K}}_{\text{c}}\text{ = 7.8}\times {\text{10}}^{\text{ - 6}}$.

Step 2(b): Draw a Lewis structure for the reactant

To draw the Lewis structure, calculate the total electrons that need to be distributed first.

$$\begin{gathered} \text{Total electrons = } \text{B + 3Cl + P + 3H = 3 + 3(7) + 5 + 3(1)} \\ \text{ = } \text{32 electrons.} \end{gathered}$$

The boron and phosphorus should be set as the central atoms because they can form multiple bonds with other atoms unlike chlorine and oxygen. Then write their valence electrons.

Connect the lone electrons. Connect the chlorine lone electrons to the lone electrons of boron. Then connect the lone electrons of hydrogen to the lone electrons phosphorus. Lastly, the lone pair of electrons in phosphorus will be donated to the boron giving the boron a negative charge, and the phosphorus a positive charge. Despite the charges, the molecule will not be ionic since their net charge will still be equal to zero. Hence, the Lewis structure for the reactant is as shown below.