Chapter 14

Consider a point during a chemical reaction at which the rate of the forward reaction is less than the rate of the reverse reaction. (a) Is the reaction at equilibrium at that point? (b) Which way does the overall reaction appear to be running?

From a practical point of view, why would you want a reaction equilibrium to lie very far to the right?

Why can we ignore equilibrium for reactions that go to completion?

Write the equilibrium constant expression for the reaction \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftarrows \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\)

Write the equilibrium constant expression for the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{HI}(g)\)

Write the equilibrium constant expression for the reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftarrows \mathrm{Fe}(\mathrm{SCN})^{2+}(a q)\)

At the start of the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{HI}(g)\) the concentrations are \(\left[\mathrm{H}_{2}\right]=0.100 \mathrm{M},\left[\mathrm{I}_{2}\right]=0.100 \mathrm{M},[\mathrm{HI}]=0.000 \mathrm{M}\). At \(427^{\circ} \mathrm{C}\), the equilibrium concentrations are \([\mathrm{HI}]=0.158 \mathrm{M},\left[\mathrm{H}_{2}\right]=0.021 \mathrm{M},\left[\mathrm{I}_{2}\right]=0.021 \mathrm{M}\). Calculate \(K_{\mathrm{eq}}\) for this reaction.

The equilibrium concentrations for the reaction \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftarrows \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) are \(\left[\mathrm{CS}_{2}\right]=6.10 \times 10^{-3} \mathrm{M},\left[\mathrm{H}_{2}\right]=1.17 \times 10^{-3} \mathrm{M},\left[\mathrm{CH}_{4}\right]=2.35 \times 10^{-3} \mathrm{M},\left[\mathrm{H}_{2} \mathrm{~S}\right]=2.93 \times\) \(10^{-3}\) M. Calculate \(K_{\mathrm{eq}}\) for this reaction.

If you added \(\mathrm{SO}_{3}\) to a vessel in which the reaction \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftarrows 2 \mathrm{SO}_{3}\) is at equilibrium, which way would the reaction shift?

Ethyl acetate, a solvent used as nail polish remover, is produced by the reaction $$ \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightleftarrows \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \Delta E=+180.5 \mathrm{~kJ} $$ Ethyl acetate (a) Rewrite this reaction with the word heat in it. (b) Does the amount of ethyl acetate in the equilibrium mixture increase or decrease when the temperature is raised?

(a) Rewrite this reaction with the word heat in it: $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{CO}_{2}(g) \quad \Delta E=-563.5 \mathrm{~kJ} $$ (b) Which way does the reaction shift when the temperature is raised? Explain your answer.

Write the equilibrium constant expression for the reaction \(\mathrm{CaCO}_{3}(s) \leftrightarrows \mathrm{CaO}(s)+\) \(\mathrm{CO}_{2}(g)\)

The process of photosynthesis in plants converts carbon dioxide and water to glucose and oxygen: \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \quad \Delta E_{\mathrm{rxn}}=2801 \mathrm{~kJ}\) (a) Write the equilibrium constant expression for this conversion. (b) How would the equilibrium be affected if \(\mathrm{CO}_{2}(g)\) were added? (c) How would the equilibrium be affected if \(\mathrm{H}_{2} \mathrm{O}(l)\) were added? (d) How would the equilibrium be affected if the reaction vessel were warmed? (e) How would the equilibrium be affected if a catalyst were added?

(a) Write the equilibrium constant expression for the reaction $$ \mathrm{PbI}_{2}(s) \leftrightarrows \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q) $$ (b) How would the equilibrium be affected if \(\mathrm{PbI}_{2}(s)\) were added? (c) How would the equilibrium be affected if \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s)\) were added? (Hint: Don't forget that \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) is a water-soluble salt.)

Sparingly soluble \(\mathrm{PbCl}_{2}\) dissolves in water to yield an equilibrium \(\mathrm{Pb}^{2+}(a q)\) concentration of \(0.039 \mathrm{M}\). (a) Write the balanced equilibrium equation for \(\mathrm{PbCl}_{2}(\) s) dissolving in water. (b) Write the \(K_{\text {sp }}\) expression for \(\mathrm{PbCl}_{2}\). (c) What is the equilibrium concentration of chloride ion? (d) Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{PbCl}_{2}\) (show your calculation).

Sparingly soluble calcium phosphate dissolves in water to yield an equilibrium calcium ion concentration of \(7.8 \times 10^{-6} \mathrm{M}\). (a) Write the balanced equilibrium equation for calcium phosphate dissolving in water. (b) Write the \(K_{\text {sp }}\) expression for calcium phosphate. (c) What is the equilibrium concentration of phosphate ion? (d) Calculate the value of \(K_{s p}\) for calcium phosphate (show your calculation).

Sparingly soluble magnesium hydroxide dissolves in water to yield an equilibrium magnesium ion concentration of \(1.44 \times 10^{-4} \mathrm{M}\). (a) Write the balanced equilibrium equation for magnesium hydroxide dissolving in water. (b) Write the \(K_{\mathrm{sp}}\) expression for magnesium hydroxide. (c) What is the equilibrium concentration of hydroxide ion? (d) Calculate the value of \(K_{\mathrm{sp}}\) for magnesium hydroxide (show your calculation).

Sparingly soluble aluminum hydroxide dissolves in water to yield an equilibrium hydroxide ion concentration of \(8.58 \times 10^{-9} \mathrm{M}\). (a) Write the balanced equilibrium equation for aluminum hydroxide dissolving in water. (b) Write the \(K_{\mathrm{sp}}\) expression for aluminum hydroxide. (c) What is the equilibrium concentration of aluminum ion? (d) Calculate the value of \(K_{\mathrm{sp}}\) for aluminum hydroxide (show your calculation).

How can you quickly determine the saturation solubility of a sparingly soluble \(1: 1\) salt at \(25^{\circ} \mathrm{C}\) ?

The saturation solubility of \(\mathrm{Ag}_{2} \mathrm{~S}\) at \(25^{\circ} \mathrm{C}\) is \(1.14 \times 10^{-17} \mathrm{M}\). What are the equilibrium concentrations of the cation and anion?

What does the "equi" portion of the word equilibrium refer to?

What do we call a reaction when we say that it can proceed in both the forward and reverse directions? In principle, which chemical reactions can proceed in both directions?

What do we mean by the position of a reaction's equilibrium, and what practical consequence can it have?

Where on the reaction coordinate is the equilibrium point for a reaction that appears not to occur?

Sometimes reactions are written with two arrows pointing in opposite directions instead of a single arrow going from reactants to products. What do the two arrows mean?

A chemist runs a reaction that is known to proceed very rapidly and keeps isolating product that is contaminated with starting material, even though he is following the stoichiometry of the reaction and giving the reaction enough time to run. How might this be explained?

When a reaction vessel is loaded with just reactants, the reverse reaction initially has a rate of zero. Explain why this is so.

Once a reaction begins, the rate of the reverse reaction gradually speeds up. Explain why this is so.

Suppose a reaction vessel is loaded only with the products of a reaction. Which would be faster at the moment after loading, the forward reaction or the reverse reaction? Explain your answer.

The water in a beaker of water left in a room will slowly evaporate until the beaker is dry. However, place that same beaker in a sealed box and the water level in the beaker will drop a bit but then remain constant. Is the latter case an example of equilibrium? Explain your answer.

Write the definition of equilibrium in terms of the general rate laws Rate \(_{\text {forward rxn }}=k_{\text {forward rxn }}[\text { Reactants }]^{\text {order }}\) and Rate \(_{\text {reverse rxn }}=k_{\text {reverse rxn }}[\text { Products }]^{\text {order }}\)

Using the definition of equilibrium, show how \(k_{\mathrm{f}} / k_{\mathrm{r}}\) for the one-step reaction \(\mathrm{R} \rightleftarrows \mathrm{P}\) is equal to the ratio \([\mathrm{P}] /[\mathrm{R}]\).

At a given temperature, why is the ratio \(k_{\mathrm{f}} / k_{\mathrm{r}}\) constant for a given reaction?

What symbol and name are used to replace the ratio \(k_{\mathrm{f}} / k_{\mathrm{r}}\) for a reaction?

If \(k_{\mathrm{f}}>k_{\mathrm{r}}\), will \(K_{\mathrm{eq}}\) be less than 1 or greater than 1? Explain your answer.

Suppose you have a reaction with many reactants. When you write the equilibrium expression for the reaction, do the reactant concentrations all go in the numerator or in the denominator? What mathematical operation(s) should be used for these concentrations?

Consider the gas-state reaction \(2 \mathrm{~A}(g)+3 \mathrm{~B}(g) \rightleftarrows \mathrm{C}(g)+\mathrm{D}(g)\) (a) Write the equilibrium constant expression for the reaction. (b) Write the reaction in reverse. (c) Write the equilibrium constant expression for the reverse reaction that you wrote for part (b). (d) Compare your answers to \((\mathrm{a})\) and \((\mathrm{c})\). What conclusion can you draw from the comparison? (e) Suppose \(K_{\mathrm{eq}}\) for a reaction \(=10.0 .\) What will the value of \(K_{\text {eq }}\) be for the reverse reaction?

Write the expression for \(K_{\mathrm{eq}}\) for the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g)\).

Write the expression for \(K_{\text {eq }}\) for the reaction \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

Would the value you obtain for \(K_{\text {eq }}\) for a reaction depend on the initial concentrations of reactants and products you use? Explain your answer.

The equilibrium concentrations for the reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) at \(2000^{\circ} \mathrm{C}\) are \(\left[\mathrm{N}_{2}\right]=\) \(0.25 \mathrm{M} ;\left[\mathrm{O}_{2}\right]=1.2 \mathrm{M} ;[\mathrm{NO}]=0.011 \mathrm{M} .\) What is the value of \(K_{\mathrm{eq}}\) for this reaction?

What does a value of \(K_{\text {eq }}\) greater than \(10^{3}\) imply? Prove that your answer is correct by using the general expression \(K_{\mathrm{eq}}=[\) Products \(] /[\) Reactants \(]\).

Suppose a reaction has a \(K_{\text {eq }}\) value of \(2.05\). When we write the reaction, can we use a single arrow to the right instead of a double set of equilibrium arrows? Explain your answer.

What does a value of \(K_{\text {eq }}\) less than \(10^{-3}\) imply? Prove that your answer is correct by using the general expression \(K_{\text {eq }}=\) [Products \(] /[\) Reactants \(]\).

A certain reaction has a \(K_{\text {eq }}\) value of \(1.5 \times 10^{-6}\). (a) Would this be a practical reaction from which to isolate pure product? Explain your answer.

The reaction \(\mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) is run in a 10.0-L vessel. The vessel is loaded with 1 mole of \(\mathrm{CO}\) and 3 moles of \(\mathrm{H}_{2}\). At equilibrium, the amounts are \(0.613\) mole of \(\mathrm{CO}, 1.839\) moles of \(\mathrm{H}_{2}, 0.387\) mole of \(\mathrm{CH}_{4}\), and \(0.387\) mole of \(\mathrm{H}_{2} \mathrm{O}\). What is the value of the equilibrium constant for this reaction? Describe the position of the equilibrium.

One way to calculate the value of a reaction's equilibrium constant is to perform the reaction, let it come to equilibrium, measure the concentration of all the reactants and products, and then plug those concentrations into the equilibrium constant expression and calculate its value. (a) A student performs the reaction \(\mathrm{A}(a q)+2 \mathrm{~B}(a q) \rightleftarrows \mathrm{C}(a q)\) starting with \(2.0 \mathrm{M} \mathrm{A}\) and \(4.0 \mathrm{M} \mathrm{B}\). He finds that at equilibrium, the concentrations of \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) are \(0.020 \mathrm{M}, 0.040 \mathrm{M}\), and \(1.98\) \(\mathrm{M}\), respectively. What is the value of this reaction's equilibrium constant (write your answer using scientific notation)? (b) Next, the student repeats the experiment, but this time he starts with \(3.0 \mathrm{M} \mathrm{A}\) and 5.0 M B. What value will he get for \(K_{\text {eq }}\) when he measures the equilibrium concentrations and plugs them into the equilibrium constant expression? (Hint: Think about why equilibrium constant is called a constant.)

For the reaction \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftarrows \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) \(K_{\text {eq }}=3.59\) at \(900^{\circ} \mathrm{C}\). After the reaction has run for 10 min at \(900^{\circ} \mathrm{C}\), the concentrations are \(\left[\mathrm{CH}_{4}\right]=\) \(1.15 \mathrm{M} ;\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.20 \mathrm{M} ;\left[\mathrm{CS}_{2}\right]=1.51 \mathrm{M} ;\left[\mathrm{H}_{2}\right]=\) \(1.08 \mathrm{M}\). Is this reaction at equilibrium?

On the basis of \(K_{\text {eq }}\) values, which reaction goes essentially to completion? How would you describe the other reaction? (a) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \leftrightarrows 2 \mathrm{H}_{2} \mathrm{O}(g)\); \(K_{\mathrm{eq}}=3 \times 10^{81}\) (b) \(2 \mathrm{HF}(g) \rightleftarrows \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) \(K_{\mathrm{eq}}=1 \times 10^{-95}\)

Write the balanced chemical equation for the reaction that goes with the equilibrium constant \(K_{\mathrm{eq}}=\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]^{2} \times\left[\mathrm{Cl}_{2}\right]^{2}}{[\mathrm{HCl}]^{4} \times\left[\mathrm{O}_{2}\right]}\)