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Chapter 18: Problem 168

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

Expert verified
(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

01

- 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.
02

- 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: \[ \frac{8.4 \times 10^{-3} \text{ mol}}{3.0 \text{ L}} = 2.8 \times 10^{-3} \text{ M} \]
03

- 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 \( 2.8 \times 10^{-3} \text{ M} \).
04

- 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: \[ K_c = (2.8 \times 10^{-3}\text{ M})(2.8 \times 10^{-3}\text{ M}) = 7.84 \times 10^{-6} \]
05

- 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: \[ H\ H\ H \ n-\ n\ |\ -P - -B - Cl - - H\ H\ H \ n- \ n \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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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Most popular questions from this chapter

Drinking water is often disinfected with \(\mathrm{Cl}_{2},\) which hydrolyzes to form \(\mathrm{HClO},\) a weak acid but powerful disinfectant: $$ \mathrm{Cl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HClO}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{Cl}^{-}(a q) $$ The fraction of \(\mathrm{HClO}\) in solution is defined as $$ \frac{[\mathrm{HClO}]}{[\mathrm{HClO}]+\left[\mathrm{ClO}^{-}\right]} $$ (a) What is the fraction of \(\mathrm{HClO}\) at \(\mathrm{pH} 7.00\left(K_{\mathrm{a}}\right.\) of \(\mathrm{HClO}=\) \(\left.2.9 \times 10^{-8}\right) ?\) (b) What is the fraction at pH \(10.00 ?\)

Human urine has a normal \(\mathrm{pH}\) of \(6.2 .\) If a person eliminates an average of \(1250 . \mathrm{mL}\) of urine per day, how many \(\mathrm{H}^{+}\) ions are climinated per week?

Chloroacetic acid, \(\mathrm{ClCH}_{2} \mathrm{COOH}\), has a \(\mathrm{p} K_{\mathrm{a}}\) of 2.87 . What are \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right], \mathrm{pH},\left[\mathrm{ClCH}_{2} \mathrm{COO}^{-}\right],\) and \(\left[\mathrm{ClCH}_{2} \mathrm{COOH}\right]\) in \(1.25 \mathrm{M}\) \(\mathrm{ClCH}_{2} \mathrm{COOH} ?\)

The antimalarial properties of quinine \(\left(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{~N}_{2} \mathrm{O}_{2}\right)\) saved thousands of lives during the construction of the Panama Canal. This substance is a classic example of the medicinal wealth that tropical forests hold. Both \(\mathrm{N}\) atoms are basic, but the \(\mathrm{N}\) (colored) of the \(3^{\circ}\) amine group is far more basic \(\left(p K_{b}=5.1\right)\) than the \(N\) within the aromatic ring system \(\left(p K_{b}=9.7\right)\) (a) A saturated solution of quinine in water is only \(1.6 \times 10^{-3} M\). What is the pH of this solution? (b) Show that the aromatic N contributes negligibly to the pH of the solution. (c) Because of its low solubility, quinine is given as the salt quinine hydrochloride \(\left(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{~N}_{2} \mathrm{O}_{2} \cdot \mathrm{HCl}\right),\) which is 120 times more soluble than quinine. What is the pH of \(0.33 M\) quinine hydrochloride? (d) An antimalarial concentration in water is \(1.5 \%\) quinine hydrochloride by mass \((d=1.0 \mathrm{~g} / \mathrm{mL}) .\) What is the \(\mathrm{pH} ?\)

Chloral \(\left(\mathrm{Cl}_{3} \mathrm{C}-\mathrm{CH}=\mathrm{O}\right)\) forms a monohydrate, chloral hydrate, the sleep-inducing depressant called "knockout drops" in old movies. (a) Write two possible structures for chloral hydrate, one involving hydrogen bonding and one that is a Lewis adduct. (b) What spectroscopic method could be used to identify the real structure? Explain.
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