Chapter 15

General Chemistry: Principles and Modern Applications · 94 exercises

Problem 62

For the reaction $$ \mathrm{A}(\mathrm{s}) \rightleftharpoons \mathrm{B}(\mathrm{s})+2 \mathrm{C}(\mathrm{g})+\frac{1}{2} \mathrm{D}(\mathrm{g}) \quad \Delta H^{\circ}=0 $$ (a) Will \(K_{p}\) increase, decrease, or remain constant with temperature? Explain. (b) If a constant-volume mixture at equilibrium at 298 K is heated to 400 K and equilibrium re-established, will the number of moles of \(\mathrm{D}(\mathrm{g})\) increase, decrease, or remain constant? Explain.

3 step solution

Problem 63

What effect does increasing the volume of the system have on the equilibrium condition in each of the following reactions? (a) \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

3 step solution

Problem 64

For which of the following reactions would you expect the extent of the forward reaction to increase with increasing temperatures? Explain. (a) \(\quad \mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-90.2 \mathrm{kJ}\) (b) \(\quad \mathrm{SO}_{3}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=+98.9 \mathrm{kJ}\) (c) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-95.4 \mathrm{kJ}\) (d) \(\mathrm{COCl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=+108.3 \mathrm{kJ}\)

3 step solution

Problem 65

The following reaction represents the binding of oxygen by the protein hemoglobin (Hb): $$\mathrm{Hb}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{aq}) \rightleftharpoons \mathrm{Hb}: \mathrm{O}_{2}(\mathrm{aq}) \quad \Delta H<0$$ Explain how each of the following affects the amount of \(\mathrm{Hb}: \mathrm{O}_{2}:\) (a) increasing the temperature; (b) decreasing the pressure of \(\mathrm{O}_{2} ;\) (c) increasing the amount of 6 hemoglobin.

3 step solution

Problem 66

In the human body, the enzyme carbonic anahydrase catalyzes the interconversion of \(\mathrm{CO}_{2}\) and \(\mathrm{HCO}_{3}^{-}\) by either adding or removing the hydroxide anion. The overall reaction is endothermic. Explain how the following affect the amount of carbon dioxide: (a) increasing the amount of bicarbonate anion; (b) increasing the pressure of carbon dioxide; (c) increasing the amount of carbonic anhydrase; (d) decreasing the temperature.

4 step solution

Problem 67

A crystal of dinitrogen tetroxide (melting point, \(\left.-9.3^{\circ} \mathrm{C} ; \text { boiling point, } 21.3^{\circ} \mathrm{C}\right)\) is added to an equilibrium mixture of dintrogen tetroxide and nitrogen dioxide that is at \(20.0^{\circ} \mathrm{C} .\) Will the pressure of nitrogen dioxide increase, decrease, or remain the same? Explain.

3 step solution

Problem 68

When hydrogen iodide is heated, the degree of dissociation increases. Is the dissociation reaction exothermic or endothermic? Explain.

3 step solution

Problem 69

The standard enthalpy of reaction for the decomposition of calcium carbonate is \(\Delta H^{\circ}=813.5 \mathrm{kJmol}^{-1}\) As temperature increases, does the concentration of calcium carbonate increase, decrease, or remain the same? Explain.

3 step solution

Problem 70

Would you expect that the amount of \(\mathrm{N}_{2}\) to increase, decrease, or remain the same in a scuba diver's body as he or she descends below the water surface?

4 step solution

Problem 71

Explain why the percent of molecules that dissociate into atoms in reactions of the type \(\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) always increases with an increase in temperature.

3 step solution

Problem 72

A 1.100 L flask at \(25^{\circ} \mathrm{C}\) and 1.00 atm pressure contains \(\mathrm{CO}_{2}(\mathrm{g})\) in contact with \(100.0 \mathrm{mL}\) of a saturated aqueous solution in which \(\left[\mathrm{CO}_{2}(\mathrm{aq})\right]=3.29 \times 10^{-2} \mathrm{M}\) (a) What is the value of \(K_{c}\) at \(25^{\circ} \mathrm{C}\) for the equilibrium \(\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{aq}) ?\) (b) If 0.01000 mol of radioactive \(^{14} \mathrm{CO}_{2}\) is added to the flask, how many moles of the \(^{14} \mathrm{CO}_{2}\) will be found in the gas phase and in the aqueous solution when equilibrium is re-established? [Hint: The radioactive \(^{14} \mathrm{CO}_{2}\) distributes itself between the two phases in exactly the same manner as the nonradioactive \(\left.^{12} \mathrm{CO}_{2} .\right]\)

3 step solution

Problem 74

In the equilibrium described in Example \(15-12,\) the percent dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) can be expressed as $$\frac{3.00 \times 10^{-3} \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4}}{0.0240 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4} \text { initially }} \times 100 \%=12.5 \%$$ What must be the total pressure of the gaseous mixture if \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\) is to be \(10.0 \%\) dissociated at \(298 \mathrm{K} ?\) $$ \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.113 \text { at } 298 \mathrm{K} $$

4 step solution

Problem 75

Starting with \(\mathrm{SO}_{3}(\mathrm{g})\) at \(1.00 \mathrm{atm},\) what will be the total pressure when equilibrium is reached in the following reaction at \(700 \mathrm{K} ?\) \(2 \mathrm{SO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=1.6 \times 10^{-5}\)

5 step solution

Problem 76

A sample of air with a mole ratio of \(\mathrm{N}_{2}\) to \(\mathrm{O}_{2}\) of 79: 21 is heated to 2500 K. When equilibrium is established in a closed container with air initially at 1.00 atm, the mole percent of \(\mathrm{NO}\) is found to be \(1.8 \% .\) Calculate \(K_{\mathrm{p}}\) for the reaction. $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$$

4 step solution

Problem 79

One of the key reactions in the gasification of coal is the methanation reaction, in which methane is produced from synthesis gas-a mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\). $$\begin{aligned} \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons & \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ \Delta H &=-230 \mathrm{kJ} ; K_{\mathrm{c}}=190 \mathrm{at} 1000 \mathrm{K} \end{aligned}$$ (a) Is the equilibrium conversion of synthesis gas to methane favored at higher or lower temperatures? Higher or lower pressures? (b) Assume you have 4.00 mol of synthesis gas with a 3:1 mol ratio of \(\mathrm{H}_{2}(\mathrm{g})\) to \(\mathrm{CO}(\mathrm{g})\) in a 15.0 L flask. What will be the mole fraction of \(\mathrm{CH}_{4}(\mathrm{g})\) at equilibrium at \(1000 \mathrm{K} ?\)

5 step solution

Problem 80

A sample of pure \(\mathrm{PCl}_{5}(\mathrm{g})\) is introduced into an evacuated flask and allowed to dissociate. $$ \mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ If the fraction of \(\mathrm{PCl}_{5}\) molecules that dissociate is denoted by \(\alpha,\) and if the total gas pressure is \(P\) show that $$ K_{\mathrm{p}}=\frac{\alpha^{2} P}{1-\alpha^{2}} $$

4 step solution

Problem 81

Nitrogen dioxide obtained as a cylinder gas is always a mixture of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) .\) A \(5.00 \mathrm{g}\) sample obtained from such a cylinder is sealed in a \(0.500 \mathrm{L}\) flask at \(298 \mathrm{K}\). What is the mole fraction of \(\mathrm{NO}_{2}\) in this mixture? $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.61 \times 10^{-3}$$

4 step solution

Problem 82

What is the apparent molar mass of the gaseous mixture that results when \(\mathrm{COCl}_{2}(\mathrm{g})\) is allowed to dissociate at \(395^{\circ} \mathrm{C}\) and a total pressure of 3.00 atm? $$\begin{aligned} \operatorname{COCl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+& \mathrm{Cl}_{2}(\mathrm{g}) \\ & K_{\mathrm{p}}=4.44 \times 10^{-2} \mathrm{at} 395^{\circ} \mathrm{C} \end{aligned}$$ Think of the apparent molar mass as the molar mass of a hypothetical single gas that is equivalent to the gaseous mixture.

4 step solution

Problem 83

Show that in terms of mole fractions of gases and total gas pressure the equilibrium constant expression for $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})$$ is $$K_{\mathrm{p}}=\frac{\left(x_{\mathrm{NH}_{3}}\right)^{2}}{\left(x_{\mathrm{N}_{2}}\right)\left(x_{\mathrm{H}_{2}}\right)^{2}} \times \frac{1}{\left(P_{\mathrm{tot}}\right)^{2}}$$

3 step solution

Problem 85

A mixture of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) and \(\mathrm{CH}_{4}(\mathrm{g})\) in the mole ratio 2: 1 was brought to equilibrium at \(700^{\circ} \mathrm{C}\) and a total pressure of 1 atm. On analysis, the equilibrium mixture was found to contain \(9.54 \times 10^{-3} \mathrm{mol} \mathrm{H}_{2} \mathrm{S} .\) The \(\mathrm{CS}_{2}\) pre- sent at equilibrium was converted successively to \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and then to \(\mathrm{BaSO}_{4} ; 1.42 \times 10^{-3} \mathrm{mol} \mathrm{BaSO}_{4}\) was obtained. Use these data to determine \(K_{\mathrm{p}}\) at \(700^{\circ} \mathrm{C}\) for the reaction $$\begin{aligned} 2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{g})+& 4 \mathrm{H}_{2}(\mathrm{g}) \\\ & K_{\mathrm{p}} \text { at } 700^{\circ} \mathrm{C}=? \end{aligned}$$

4 step solution

Problem 89

The formation of nitrosyl chloride is given by the following equation: \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g})\) \(K_{\mathrm{c}}=4.6 \times 10^{4}\) at \(298 \mathrm{K} .\) In a \(1.50 \mathrm{L}\) flask, there are \(4.125 \mathrm{mol}\) of \(\mathrm{NOCl}\) and \(0.1125 \mathrm{mol}\) of \(\mathrm{Cl}_{2}\) present at equilibrium (298 K). (a) Determine the partial pressure of \(\mathrm{NO}\) at equilibrium. (b) What is the total pressure of the system at equilibrium?

6 step solution

Problem 90

At \(500 \mathrm{K}\), a 10.0 L equilibrium mixture contains 0.424 \(\mathrm{mol} \mathrm{N}_{2}, 1.272 \mathrm{mol} \mathrm{H}_{2},\) and \(1.152 \mathrm{mol} \mathrm{NH}_{3} .\) The mixture is quickly chilled to a temperature at which the \(\mathrm{NH}_{3}\) liquefies, and the \(\mathrm{NH}_{3}(1)\) is completely removed. The 10.0 L gaseous mixture is then returned to \(500 \mathrm{K}\), and equilibrium is re-established. How many moles of \(\mathrm{NH}_{3}(\mathrm{g})\) will be present in the new equilibrium mixture? $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3} \quad K_{\mathrm{c}}=152 \text { at } 500 \mathrm{K}$$

3 step solution

Problem 91

Recall the formation of methanol from synthesis gas, the reversible reaction at the heart of a process with great potential for the future production of automotive fuels (page 663 ). $$\begin{aligned} \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 14.5 \mathrm{at} 483 \mathrm{K} \end{aligned}$$ A particular synthesis gas consisting of 35.0 mole percent \(\mathrm{CO}(g)\) and 65.0 mole percent \(\mathrm{H}_{2}(\mathrm{g})\) at a total pressure of 100.0 atm at \(483 \mathrm{K}\) is allowed to come to equilibrium. Determine the partial pressure of \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) in the equilibrium mixture.

4 step solution

Problem 92

A classic experiment in equilibrium studies dating from 1862 involved the reaction in solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) to produce ethyl acetate and water. $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}$$ The reaction can be followed by analyzing the equilibrium mixture for its acetic acid content. $$\begin{array}{r} 2 \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightleftharpoons \\ \mathrm{Ba}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In one experiment, a mixture of 1.000 mol acetic acid and 0.5000 mol ethanol is brought to equilibrium. A sample containing exactly one-hundredth of the equilibrium mixture requires \(28.85 \mathrm{mL} 0.1000 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. Calculate the equilibrium constant, \(K_{c}\), for the ethanol-acetic acid reaction based on this experiment.

4 step solution

Problem 93

The decomposition of \(\mathrm{HI}(\mathrm{g})\) is represented by the equation $$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g})$$ \(\mathrm{HI}(\mathrm{g})\) is introduced into five identical \(400 \mathrm{cm}^{3}\) glass bulbs, and the five bulbs are maintained at \(623 \mathrm{K}\) Each bulb is opened after a period of time and analyzed for \(I_{2}\) by titration with \(0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq})\) $$\begin{array}{l} \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \longrightarrow \\ \quad \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}(\mathrm{aq})+2 \mathrm{NaI}(\mathrm{aq}) \end{array}$$ Data for this experiment are provided in the table below. What is the value of \(K_{\mathrm{c}}\) at \(623 \mathrm{K} ?\) $$\begin{array}{llll} \hline & & & \text { Volume } \\ & \text { Initial } & \text { Time } & 0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \\ \text { Bulb } & \text { Mass of } & \text { Bulb } & \text { Required for } \\\ \text { Number } & \mathrm{Hl}(\mathrm{g}), \mathrm{g} & \text { Opened, } \mathrm{h} & \text { Titration, in } \mathrm{mL} \\ \hline 1 & 0.300 & 2 & 20.96 \\ 2 & 0.320 & 4 & 27.90 \\ 3 & 0.315 & 12 & 32.31 \\ 4 & 0.406 & 20 & 41.50 \\ 5 & 0.280 & 40 & 28.68 \\ \hline \end{array}$$

3 step solution

Problem 94

In one of Fritz Haber's experiments to establish the conditions required for the ammonia synthesis reaction, pure \(\mathrm{NH}_{3}(\mathrm{g})\) was passed over an iron catalyst at \(901^{\circ} \mathrm{C}\) and 30.0 atm. The gas leaving the reactor was bubbled through 20.00 mL of a HCl(aq) solution. In this way, the \(\mathrm{NH}_{3}(\mathrm{g})\) present was removed by reaction with HCl. The remaining gas occupied a volume of 1.82 L at STP. The \(20.00 \mathrm{mL}\) of \(\mathrm{HCl}(\mathrm{aq})\) through which the gas had been bubbled required \(15.42 \mathrm{mL}\) of \(0.0523 \mathrm{M} \mathrm{KOH}\) for its titration. Another \(20.00 \mathrm{mL}\) sample of the same HCl(aq) through which no gas had been bubbled required \(18.72 \mathrm{mL}\) of \(0.0523 \mathrm{M} \mathrm{KOH}\) for its titration. Use these data to obtain a value of \(K_{\mathrm{p}}\) at \(901^{\circ} \mathrm{C}\) for the reaction \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})\)

3 step solution

Problem 95

The following is an approach to establishing a relationship between the equilibrium constant and rate constants mentioned in the section on page 660 \(\bullet\)Work with the detailed mechanism for the reaction. \(\bullet\) Use the principle of microscopic reversibility, the idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter \(14,\) we treated some reaction steps as reversible and others as going to completion. However, as noted in Table \(15.3,\) every reaction has an equilibrium constant even though a reaction is generally considered to go to completion if its equilibrium constant is very large.) \(\bullet\) Use the idea that when equilibrium is attained in an overall reaction, it is also attained in each step of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the mechanism, similar to what we did with the steady-state assumption in describing reaction mechanisms. \(\bullet\)Combine the \(K_{\mathrm{c}}\) expressions for the elementary steps into a \(K_{\mathrm{c}}\) expression for the overall reaction. The numerical value of the overall \(K_{c}\) can thereby be expressed as a ratio of rate constants, \(k\) Use this approach to establish the equilibrium constant expression for the overall reaction, $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ The mechanism of the reaction appears to be the following: Fast: \(\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) Slow: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

3 step solution

Problem 96

The following two equilibrium reactions can be written for aqueous carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) $$ \begin{array}{ll} \mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{HCO}_{3}^{-}(\mathrm{aq}) & K_{1} \\ \mathrm{HCO}_{3}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq}) & K_{2} \end{array} $$ For each reaction write the equilibrium constant expression. By using Le Châtelier's principle we may naively predict that by adding \(\mathrm{H}_{2} \mathrm{CO}_{3}\) to the system, the concentration of \(\mathrm{CO}_{3}^{2-}\) would increase. What we observe is that after adding \(\mathrm{H}_{2} \mathrm{CO}_{3}\) to the equilibrium mixture, an increase in the concentration of \(\mathrm{CO}_{3}^{2-}\) occurs when \(\left[\mathrm{CO}_{3}^{2-}\right] \ll \mathrm{K}_{2}\) however, the concentration of \(\mathrm{CO}_{3}^{2-}\) will decrease when \(\left[\mathrm{CO}_{3}^{2-}\right] \gg K_{2} .\) Show that this is true by considering the ratio of \(\left[\mathrm{H}^{+}\right] /\left[\mathrm{HCO}_{3}^{-}\right]\) before and after adding a small amount of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) to the solution, and by using that ratio to calculate the \(\left[\mathrm{CO}_{3}^{2-}\right]\)

3 step solution

Problem 97

In organic synthesis many reactions produce very little yield, that is \(K \ll 1 .\) Consider the following hypothetical reaction: \(\mathrm{A}(\mathrm{aq})+\mathrm{B}(\mathrm{aq}) \longrightarrow \mathrm{C}(\mathrm{aq}), K=1 \times 10^{-2}\) We can extract product, \(\mathrm{C}\), from the aqueous layer by adding an organic layer in which \(\mathrm{C}(\mathrm{aq}) \longrightarrow \mathrm{C}(\mathrm{or})\), \(K=15 .\) Given initial concentrations of \([\mathrm{A}]=0.1 \mathrm{M}\) \([\mathrm{B}]=0.1,\) and \([\mathrm{C}]=0.1,\) calculate how much \(\mathrm{C}\) will be found in the organic layer. If the organic layer was not present, how much C would be produced?

3 step solution

Problem 98

In your own words, define or explain the following terms or symbols: (a) \(K_{\mathrm{p}} ;\) (b) \(Q_{\mathrm{c}} ;\) (c) \(\Delta n_{\text {gas }}\)

3 step solution

Problem 99

Briefly describe each of the following ideas or phenomena: (a) dynamic equilibrium; (b) direction of a net chemical change; (c) Le Châtelier's principle; (d) effect of a catalyst on equilibrium.

4 step solution

Problem 100

Explain the important distinctions between each pair of terms: (a) reaction that goes to completion and reversible reaction; (b) \(K_{\mathrm{c}}\) and \(K_{\mathrm{p}} ;\) (c) reaction quotient (Q) and equilibrium constant expression ( \(K\) ); (d) homogeneous and heterogeneous reaction.

4 step solution

Problem 101

In the reversible reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{HI}(\mathrm{g}),\) an initial mixture contains \(2 \mathrm{mol} \mathrm{H}_{2}\) and 1 mol I \(_{2} .\) The amount of HI expected at equilibrium is (a) \(1 \mathrm{mol} ;\) (b) \(2 \mathrm{mol} ;\) (c) less than \(2 \mathrm{mol}\); (d) more than 2 mol but less than 4 mol.

4 step solution

Problem 102

Equilibrium is established in the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\) \(\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) \quad\) at \(\quad \mathrm{a} \quad\) temperature \(\quad\) where \(K_{\mathrm{c}}=100 .\) If the number of moles of \(\mathrm{SO}_{3}(\mathrm{g})\) in the equilibrium mixture is the same as the number of moles of \(\mathrm{SO}_{2}(\mathrm{g}),\) (a) the number of moles of \(\mathrm{O}_{2}(\mathrm{g})\) is also equal to the number of moles of \(\mathrm{SO}_{2}(\mathrm{g}) ;\) (b) the number of moles of \(\mathrm{O}_{2}(\mathrm{g})\) is half the number of moles of \(\mathrm{SO}_{2} ;\) (c) \(\left[\mathrm{O}_{2}\right]\) may have any of several values; (d) \(\left[\mathrm{O}_{2}\right]=0.010 \mathrm{M}\)

3 step solution

Problem 103

The volume of the reaction vessel containing an equilibrium mixture in the reaction \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) is increased. When equilibrium is re-established, (a) the amount of \(\mathrm{Cl}_{2}\) will have increased; (b) the amount of \(\mathrm{SO}_{2}\) will have decreased; (c) the amounts of \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) will have remained the same; (d) the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) will have increased.

3 step solution

Problem 104

For the reaction \(2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\) \(K_{\mathrm{c}}=1.8 \times 10^{-6}\) at \(184^{\circ} \mathrm{C} . \mathrm{At} 184^{\circ} \mathrm{C},\) the value of \(K_{\mathrm{c}}\) for the reaction \(\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g})\) is (a) \(0.9 \times 10^{6}\) (b) \(7.5 \times 10^{2}\) (c) \(5.6 \times 10^{5}\) (d) \(2.8 \times 10^{5}\)

3 step solution

Problem 105

For the dissociation reaction \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{g})+\) \(\mathrm{S}_{2}(\mathrm{g}), K_{\mathrm{p}}=1.2 \times 10^{-2}\) at \(1065^{\circ} \mathrm{C} .\) For this same reaction at \(1000 \mathrm{K},\) (a) \(K_{\mathrm{c}}\) is less than \(K_{\mathrm{p}} ;\) (b) \(K_{\mathrm{c}}\) is greater than \(K_{\mathrm{p}} ;(\mathrm{c}) K_{\mathrm{c}}=K_{\mathrm{p}} ;\) (d) whether \(K_{\mathrm{c}}\) is less than, equal to, or greater than \(K_{\mathrm{p}}\) depends on the total gas pressure.

3 step solution

Problem 106

The following data are given at \(1000 \mathrm{K}: \mathrm{CO}(\mathrm{g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) ; \quad \Delta H^{\circ}=-42 \mathrm{k} \mathrm{J}\) \(K_{\mathrm{c}}=0.66 .\) After an initial equilibrium is established in a \(1.00 \mathrm{L}\) container, the equilibrium amount of \(\mathrm{H}_{2}\) can be increased by (a) adding a catalyst; (b) increasing the temperature; (c) transferring the mixture to a 10.0 L container; (d) in some way other than (a), (b), Or (c).

5 step solution

Problem 107

Equilibrium is established in the reversible reaction \(2 \mathrm{A}+\mathrm{B} \rightleftharpoons 2 \mathrm{C} .\) The equilibrium concentrations are \([\mathrm{A}]=0.55 \mathrm{M},[\mathrm{B}]=0.33 \mathrm{M},[\mathrm{C}]=0.43 \mathrm{M}\) What is the value of \(K_{c}\) for this reaction?

4 step solution

Problem 108

The Deacon process for producing chlorine gas from hydrogen chloride is used in situations where \(\mathrm{HCl}\) is available as a by-product from other chemical processes. $$\begin{aligned} 4 \mathrm{HCl}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{Cl}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&-114 \mathrm{kJ} \end{aligned}$$ A mixture of \(\mathrm{HCl}, \mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{Cl}_{2}\) is brought to equilibrium at \(400^{\circ} \mathrm{C}\). What is the effect on the equilibrium amount of \(\mathrm{Cl}_{2}(\mathrm{g})\) if (a) additional \(\mathrm{O}_{2}(\mathrm{g})\) is added to the mixture at constant volume? (b) \(\mathrm{HCl}(\mathrm{g})\) is removed from the mixture at constant volume? (c) the mixture is transferred to a vessel of twice the volume? (d) a catalyst is added to the reaction mixture? (e) the temperature is raised to \(500^{\circ} \mathrm{C} ?\)

5 step solution

Problem 109

For the reaction \(\mathrm{SO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{aq}), K=1.25 \mathrm{at}\) \(25^{\circ} \mathrm{C} .\) Will the amount of \(\mathrm{SO}_{2}(\mathrm{g})\) be greater than or less than the amount of \(\mathrm{SO}_{2}(\mathrm{aq}) ?\)

2 step solution

Problem 110

In the reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}), K=1.0 \times\) \(10^{5}\) at \(25^{\circ} \mathrm{C} .\) Would you expect a greater amount of product or reactant?

3 step solution

Problem 111

An equilibrium mixture of \(\mathrm{SO}_{2}, \mathrm{SO}_{3},\) and \(\mathrm{O}_{2}\) gases is maintained in a \(2.05 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=35.5\) for the reaction $$2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})$$ (a) If the numbers of moles of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) in the flask are equal, how many moles of \(\mathrm{O}_{2}\) are present? (b) If the number of moles of \(\mathrm{SO}_{3}\) in the flask is twice the number of moles of \(\mathrm{SO}_{2}\), how many moles of \(\mathrm{O}_{2}\) are present?

4 step solution

Problem 112

Using the method in Appendix \(\mathrm{E}\), construct a concept map of Section \(15-6,\) illustrating the shift in equilibrium caused by the various types of disturbances discussed in that section.

5 step solution

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