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Chapter 14: Problem 16

Write equilibrium constant expressions for the following reactions: $$\begin{array}{l}{\text { a. } 2 \mathrm{NO}_{2}(g) \rightleftarrows \mathrm{N}_{2} \mathrm{O}_{4}(g)} \\ {\text { b. } \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \operatorname{COCl}_{2}(g)} \\\ {\text { c. } \mathrm{AgCl}(s) \rightleftarrows \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)}\end{array}$$ $$\begin{array}{c}{\mathrm{d} \cdot \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows} \\ {\mathrm \quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad {H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)}\end{array}$$

Short Answer

Expert verified
The equilibrium constant expressions for the given reactions are: \[ K_{c} = \frac{[N_{2}O_{4}]}{[NO_{2}]^2}, K_{c} = \frac{[COCl_{2}]}{[CO][Cl_{2}]}, K_{sp} = [Ag^{+}][Cl^{-}], K_{a} = \frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} \] respectively.

Step by step solution

01

Write down the equilibrium expression for a

For the reaction: \(2NO_{2(g)} \rightleftarrows N_{2}O_{4(g)}\), the equilibrium constant expression, denoted as \(K_{c}\) (since it's in terms of molar concentration), is given by: \[ K_{c} = \frac{[N_{2}O_{4}]}{[NO_{2}]^2} \] where [N_{2}O_{4}] and [NO_{2}] represent the equilibrium concentrations of the given components in moles per liter.
02

Write down the equilibrium expression for b

For the reaction: \(CO_{(g)} + Cl_{2(g)} \rightleftarrows COCl_{2(g)}\), the equilibrium constant expression is given by: \[ K_{c} = \frac{[COCl_{2}]}{[CO][Cl_{2}]} \]
03

Write down the equilibrium expression for c

For the reaction: \(AgCl_{(s)} \rightleftarrows Ag^{+}_{(aq)} + Cl^{-}_{(aq)}\), the equilibrium constant expression is given by: \[ K_{sp} = [Ag^{+}][Cl^{-}] \] Here, it is important to note that solids are not included in the equilibrium constant expression, hence the name \(K_{sp}\), meaning solubility product constant.
04

Write down the equilibrium expression for d

For the reaction: \(CH_{3}COOH_{(aq)} + H_{2}O_{(l)} \rightleftarrows H_{3}O^{+}_{(aq)} + CH_{3}COO^{-}_{(aq)}\), the equilibrium constant expression is given by: \[ K_{a} = \frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} \] Here, it is important to note that water, being pure liquid, is not included in the equilibrium constant expression. Therefore, the equilibrium expression is often termed as the \(K_{a}\) or acid dissociation constant.

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Most popular questions from this chapter

What information may be conveyed by the knowledge that the equilibrium constant of a reaction is very small?

Methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) can be prepared in the presence of a catalyst by the reaction of \(\mathrm{H}_{2}\) and \(\mathrm{CO}\) at high temperatures according to the following equation: $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}(g)$$ What is the concentration of \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) in moles per liter if the concentration of \(\mathrm{H}_{2}=0.080 \mathrm{mol} / \mathrm{L}\) , the concentration of \(\mathrm{CO}=0.025 \mathrm{mol} / \mathrm{L},\) and \(K_{e q}=290\) at 700 \(\mathrm{K} ?\)

At \(450^{\circ} \mathrm{C}\) the value of the equilibrium constant for the following system is \(6.59 \times 10^{-3}\) . If \(\left[\mathrm{NH}_{3}\right]=1.23 \times 10^{-4}\) and \(\left[\mathrm{H}_{2}\right]=2.75 \times 10^{-3}\) at equilibrium, determine the concentration of \(\mathrm{N}_{2}\) at that point. $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NH}_{3}(\mathrm{g})$$

What is the concentration of \(\mathrm{F}^{-}\) ions in a saturated solution that is 0.10 \(\mathrm{M}\) in \(\mathrm{Ca}^{2+}\) ? The \(K_{s p}\) of \(\mathrm{CaF}_{2}\) is \(1.6 \times 10^{-10} .\)

What is the distinction between a static equilibrium and a dynamic equilibrium? In which class do chemical equilibria fit?
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