Question

The Group \(5 \mathrm{~A}(15)\) hydrides react with boron trihalides in a reversible Lewis acid-base reaction. When \(0.15 \mathrm{~mol}\) of \(\mathrm{PH}_{3} \mathrm{BCl}_{3}(s)\) is introduced into a \(3.0-\mathrm{L}\) container at a certain temperature, \(8.4 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{PH}_{3}\) is present at equilibrium: $$ \mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g) $$ (a) Find \(K_{c}\) for the reaction at this temperature. (b) Draw a Lewis structure for the reactant.

Short answer

(a) \( K_c = 7.84 \times 10^{-6} \). (b) Lewis structure involves a bond between \( \mathrm{P} \) with three \( \mathrm{H} \) and \( \mathrm{BCl}_3 \).

Step-by-step solution

- Set up the equation and known values

The equilibrium reaction is given: \[ \mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g) + \mathrm{BCl}_{3}(g) \] We know the initial amount of \( \mathrm{PH}_{3} \mathrm{BCl}_{3} \) is 0.15 mol in a 3.0-L container, and at equilibrium, \( 8.4 \times 10^{-3} \) mol of \( \mathrm{PH}_{3} \) is present.

- Determine concentration at equilibrium

To find concentrations, we use the formula: \[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \] The concentration of \( \mathrm{PH}_{3} \) at equilibrium is: \[ \frac{8.4 \times 10^{-3} \text{ mol}}{3.0 \text{ L}} = 2.8 \times 10^{-3} \text{ M} \]

- Relationship of species at equilibrium

For every mole of \( \mathrm{PH}_{3} \), there is one mole of \( \mathrm{BCl}_{3} \). Therefore, the concentration of \( \mathrm{BCl}_{3} \) at equilibrium is also \( 2.8 \times 10^{-3} \text{ M} \).

- Calculate the equilibrium constant \( K_c \)

The equilibrium constant expression for the reaction is: \[ K_c = [\mathrm{PH}_{3}][\mathrm{BCl}_{3}] \] Plugging in the equilibrium concentrations, we get: \[ K_c = (2.8 \times 10^{-3}\text{ M})(2.8 \times 10^{-3}\text{ M}) = 7.84 \times 10^{-6} \]

- Draw the Lewis structure for \( \mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \)

The Lewis structure for \( \mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \) involves: - \( \mathrm{P} \) with three single bonds to \( \mathrm{H} \) and a lone pair on \( \mathrm{P} \). - \( \mathrm{B} \) attached to three \( \mathrm{Cl} \) atoms via single bonds, and bonded to the lone pair of \( \mathrm{P} \). Here is the structure: \[ H\ H\ H \ n-\ n\ |\ -P - -B - Cl - - H\ H\ H \ n- \ n \]

Key concepts

Equilibrium Constant

In chemistry, the equilibrium constant \( K_c \) is a measure of the ratio of the concentration of products to reactants at equilibrium in a reversible chemical reaction. The equilibrium expression depends on the balanced chemical equation.
To solve for \( K_c \) in the given reaction \( \text{PH}_3 \text{BCl}_3(s) \rightleftharpoons \text{PH}_3(g) + \text{BCl}_3(g) \), you start by determining the concentrations of \( \text{PH}_3 \) and \( \text{BCl}_3 \) at equilibrium. These values are used in the equilibrium expression:
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