Unit 3 Equilibrium
3.4 Exercises
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Section 3.4 Exercises
- A reaction is represented by this equation: [latex]\text{A}(aq)\;+\;2\text{B}(aq)\;{\rightleftharpoons}\;2\text{C}(aq)\;\;\;\;\;\;\;K_\text{c} = 1\;\times\;10^3[/latex]
(a) Write the mathematical expression for the equilibrium constant.
(b) Using concentrations ≤ 1 M, make up two sets of concentrations that describe a mixture of A, B, and C at equilibrium. - What is the value of the equilibrium constant at 500 °C for the formation of NH3 according to the following equation?
[latex]\text{N}_2(g)\;+\;3\text{H}_2(g)\;{\rightleftharpoons}\;2\text{NH}_3(g)[/latex]
An equilibrium mixture of NH3(g), H2(g), and N2(g) at 500 °C was found to contain 1.35 M H2, 1.15 M N2, and 4.12 × 10−1 M NH3.
- A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated to a certain temperature. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.40 mol of Cl2(g). Write the balanced equilibrium equation and calculate the value of the equilibrium constant for the decomposition of PCl5 to PCl3 and Cl2 at this temperature.
- Calculate the value of the equilibrium constant Kp for the reaction [latex]2\text{NO}(g)\;+\;\text{Cl}_2(g)\;{\rightleftharpoons}\;2\text{NOCl}(g)[/latex] from these equilibrium pressures: NO, 0.050 atm; Cl2, 0.30 atm; NOCl, 1.2 atm.
- A sample of ammonium chloride was heated in a closed container.
[latex]\text{NH}_4\text{Cl}(s)\;{\rightleftharpoons}\;\text{NH}_3(g)\;+\;\text{HCl}(g)[/latex]
At equilibrium, the pressure of NH3(g) was found to be 1.75 atm. What is the value of the equilibrium constant Kp for the decomposition at this temperature?
- Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.
(a) [latex]\begin{array}{lcccc} 2\text{SO}_3(g) & {\rightleftharpoons} & 2\text{SO}_2(g) & + & \text{O}_2(g) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & +x \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & 0.125\;\text{M} \end{array}[/latex]
(b) [latex]\begin{array}{lcccccc} 4\text{NH}_3(g) & + & 3\text{O}_2(g) & {\rightleftharpoons} & 2\text{N}_2(g) & + & 6\text{H}_2\text{O}(g) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & +3x & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & 0.24\;\text{M} & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]
(c) Change in pressure:[latex]\begin{array}{lcccc} 2\text{CH}_4(g) & {\rightleftharpoons} & \text{C}_2\text{H}_2(g) & + & 3\text{H}_2(g) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & +x & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & 25\;\text{torr} & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]
(d) Change in pressure:[latex]\begin{array}{lcccccc} \text{CH}_4(g) & + & \text{H}_2\text{O}(g) & {\rightleftharpoons} & \text{CO}(g) & + & 3\text{H}_2(g) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & +x & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & 5\;\text{atm} & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]
(e) [latex]\begin{array}{lcccc} \text{NH}_4\text{Cl}(s) & {\rightleftharpoons} & \text{NH}_3(g) & + & \text{HCl}(g) \\[0.5em] & & +x & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] & & 1.03\;\times\;10^{-4}\;\text{M} & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]
(f) Change in pressure:[latex]\begin{array}{lcccc} \text{Ni}(s) & + & 4\text{CO}(g) & {\rightleftharpoons} & \text{Ni(CO)}_4(g) \\[0.5em] & & +4x & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] & & 0.40\;\text{atm} & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]
- (a) Why are there no changes specified for Ni in Exercise 6, part (f)? What property of Ni does change?
(b) Using these equations and the associated Kp,[latex]\begin{align*} 2\text{ CH}_4(g)\;&{\rightleftharpoons}\;\text{C}_2\text{H}_2(g)\;+\;3\text{ H}_2(g) && K_{\text{p}_1}\\ \text{CH}_4(g)\;+\;\text{H}_2\text{O}(g)\;&{\rightleftharpoons}\;\text{CO}(g)\;+\;3\text{ H}_2(g) && K_{\text{p}2} \end{align*}[/latex]
write an expression for Kp in terms of Kp1 and Kp2 for the equation
[latex]\begin{align*} \text{C}_2\text{H}_2(g)\;+\;2\text{ H}_2\text{O}(g)\;&{\rightleftharpoons}\;2\text{ CO}(g)\;+\;3\text{ H}_2(g) && K_{\text{p}}=? \end{align*}[/latex]
- Analysis of the gases in a sealed reaction vessel containing NH3, N2, and H2 at equilibrium at 400 °C established the concentration of N2 to be 1.2 M and the concentration of H2 to be 0.24 M.
[latex]\text{N}_2(g)\;+\;3\text{H}_2(g)\;{\rightleftharpoons}\;2\text{NH}_3(g)\;\;\;\;\;\;\;K_\text{c} = 0.50\;\text{at}\;400\;^{\circ}\text{C}[/latex]
Calculate the equilibrium molar concentration of NH3.
- What is the pressure of BrCl in an equilibrium mixture of Cl2, Br2, and BrCl if the pressure of Cl2 in the mixture is 0.115 atm and the pressure of Br2 in the mixture is 0.450 atm?
[latex]\text{Cl}_2(g)\;+\;\text{Br}_2(g)\;{\rightleftharpoons}\;2\text{BrCl}(g)\;\;\;\;\;\;\;K_\text{p} = 4.7\;\times\;10^{-2}[/latex]
- Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.
[latex]\text{CoO}(s)\;+\;\text{CO}(g)\;{\rightleftharpoons}\;\text{Co}(s)\;+\;\text{CO}_2(g)\;\;\;\;\;\;\;K_\text{c} = 4.90\;\times\;10^2\;\text{at}\;550\;^{\circ}\text{C}[/latex]
What concentration of CO remains in an equilibrium mixture with [CO2] = 0.100 M?
- Sodium sulfate 10−hydrate, Na2SO4·10H2O, dehydrates according to the equation
[latex]\text{Na}_2\text{SO}_4{\cdot}10\text{H}_2\text{O}(s)\;{\rightleftharpoons}\;\text{Na}_2\text{SO}_4(s)\;+\;10\text{H}_2\text{O}(g)\;\;\;\;\;\;\;K_\text{p} = 4.08\;\times\;10^{-25}\;\text{at}\;25\;^{\circ}\text{C}[/latex]
What is the pressure of water vapor at equilibrium with a mixture of Na2SO4·10H2O and Na2SO4?
- A student solved the following problem and found the equilibrium concentrations to be [SO2] = 0.590 M, [O2] = 0.0450 M, and [SO3] = 0.260 M. How could this student check the work without reworking the problem? The problem was: For the following reaction at 600 °C:
[latex]2\text{SO}_2(g)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;2\text{SO}_3(g)\;\;\;\;\;\;\;K_\text{c} = 4.32[/latex]
At 580 °C, when initial [SO3] = 0.500 M, initial [O2] = 0.350 M, initial [SO2] = 0 M and equilibrium [SO3] = 0.300 M, calculate the new Kc. Is this reaction exothermic or endothermic?
- Assume that the change in concentration of N2O4 is small enough to be neglected in the following problem.
(a) Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N2O4 with chloroform as the solvent.[latex]\text{N}_2\text{O}_4(g)\;{\rightleftharpoons}\;2\text{NO}_2(g)\;\;\;\;\;\;\;K_\text{c} = 1.07\;\times\;10^{-5}[/latex] in chloroform
(b) Show that the change is small enough to be neglected. - Assume that the change in pressure of H2S is small enough to be neglected in the following problem.
(a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H2S with an initial pressure of 0.824 atm.
[latex]2\text{H}_2\text{S}(g)\;{\rightleftharpoons}\;2\text{H}_2(g)\;+\;\text{S}_2(g)\;\;\;\;\;\;\;K_\text{p} = 2.2\;\times\;10^{-6}[/latex]
(b) Show that the change is small enough to be neglected. - What are the concentrations of PCl5, PCl3, and Cl2 in an equilibrium mixture produced by the decomposition of a sample of pure PCl5 with [PCl5] = 2.00 M?
[latex]\text{PCl}_5(g)\;{\rightleftharpoons}\;\text{PCl}_3(g)\;+\;\text{Cl}_2(g)\;\;\;\;\;\;\;K_\text{c} = 0.0211[/latex]
- Calculate the equilibrium concentrations of NO, O2, and NO2 in a mixture at 250 °C that results from the reaction of 0.20 M NO and 0.10 M O2. (Hint: K is large; assume the reaction goes to completion then comes back to equilibrium. After the reaction has been assumed to have gone to completion, calculate the % change in [NO2] to reach equilibrium.)
[latex]2\text{NO}(g)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;2\text{NO}_2(g)\;\;\;\;\;\;\;K_\text{c} = 2.3\;\times\;10^5\;\text{at}\;250\;^{\circ}\text{C}[/latex]
- One of the important reactions in the formation of smog is represented by the equation
[latex]\text{O}_3(g)\;+\;\text{NO}(g)\;{\rightleftharpoons}\;\text{NO}_2(g)\;+\;\text{O}_2(g)\;\;\;\;\;\;\;K_\text{p} = 6.0\;\times\;10^{34}[/latex]
What is the pressure of O3 remaining after a mixture of O3 with a pressure of 1.2 × 10−8 atm and NO with a pressure of 1.2 × 10−8 atm comes to equilibrium? (Hint: Kp is large, similar to the large Kc in Exercise 16; assume the reaction goes to completion then comes back to equilibrium.) - Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H2 and 63.5 g of iodine at 448 °C.
[latex]\text{H}_2(g)\;+\;\text{I}_2(g)\;{\rightleftharpoons}\;2\text{HI}(g)\;\;\;\;\;\;\;K_\text{c} = 50.2\;\text{at}\;448\;^{\circ}\text{C}[/latex]
The volume does not matter when the moles of gas are the same on both sides of the balanced equation. To simplify the calculation, it is acceptable to assume that the volume is 1.00 L. - What is the minimum mass of CaCO3 required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (Kc) is 0.050 for the decomposition reaction of CaCO3 at that temperature?
[latex]\text{CaCO}_3(s)\;{\rightleftharpoons}\;\text{CaO}(s)\;+\;\text{CO}_2(g)[/latex]
- At 25 °C and at 1 atm, the partial pressures in an equilibrium mixture of N2O4 and NO2 are [latex]P_{\text{N}_2\text{O}_4} = 0.70\;\text{atm}[/latex] and [latex]P_{\text{NO}_2} = 0.30\;\text{atm}[/latex].
(a) Predict how the pressures of NO2 and N2O4 will change if the total pressure increases to 9.0 atm. Will they increase, decrease, or remain the same?
(b) Calculate the partial pressures of NO2 and N2O4 when they are at equilibrium at 9.0 atm and 25 °C. Hint: The total pressure at equilibrium is 9.0 atm. If [latex]P_{\text{NO}_2}=x[/latex], what would be [latex]P_{\text{N}_2\text{O}_4}[/latex]? - The equilibrium constant (Kc) for this reaction is 5.0 at a given temperature.
[latex]\text{CO}(g)\;+\;\text{H}_2\text{O}(g)\;{\rightleftharpoons}\;\text{CO}_2(g)\;+\;\text{H}_2(g)[/latex]
(a) On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H2 in a liter. How many moles of CO2 were there in the equilibrium mixture?
(b) Maintaining the same temperature, additional H2 was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H2 in a liter. How many moles of CO2 were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established. - Consider the reaction between H2 and O2 at 1000 K
[latex]2\text{H}_2(g)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;2\text{H}_2\text{O}(g)\;\;\;\;\;\;\;K_\text{p} = \frac{(P_{\text{H}_2\text{O}})^2}{(P_{\text{O}_2})(P_{\text{H}_2})^2} = 1.33\;\times\;10^{20}[/latex]If 0.500 atm of H2 and 0.500 atm of O2 are allowed to come to equilibrium at this temperature, what are the partial pressures of the components?
- Consider the equilibrium
[latex]4\text{NO}_2(g)\;+\;6\text{H}_2\text{O}(g)\;{\rightleftharpoons}\;4\text{NH}_3(g)\;+\;7\text{O}_2(g)[/latex]
(a) What is the expression for the equilibrium constant (Kc) of the reaction?
(b) How must the concentration of NH3 change to reach equilibrium if the reaction quotient is less than the equilibrium constant?
(c) If the reaction were at equilibrium, how would a decrease in pressure (from an increase in the volume of the reaction vessel) affect the pressure of NO2?
(d) If the change in the pressure of NO2 is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O2 change? - The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose.
[latex]\text{C}_{12}\text{H}_{22}\text{O}_{11}(aq)\;+\;\text{H}_2\text{O}(l)\;{\longrightarrow}\;\text{C}_6\text{H}_{12}\text{O}_6(aq)\;+\;\text{C}_6\text{H}_{12}\text{O}_6(aq)[/latex]
Rate = k[C12H22O11]
In neutral solution, k = 2.1 × 10−11/s at 27 °C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation—the products of the reaction, glucose and fructose, have the same molecular formulas, C6H12O6, but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 × 105 at 27 °C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 M aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1. - Liquid N2O3 is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO2.
[latex]\text{N}_2\text{O}_3(g) \rightleftharpoons \text{NO}(g) + \text{NO}_2(g)[/latex]
At 25 °C, a value of Kp = 1.91 has been established for this decomposition. If 0.236 moles of N2O3 are placed in a 1.52-L vessel at 25 °C, calculate the equilibrium partial pressures of N2O3(g), NO2(g), and NO(g). To start, we can calculate the initial pressure of N2O3 in atm.
Solutions
- [latex]K_\text{c} = \frac{[\text{C}]^2}{[\text{A}][\text{B}]^2}[/latex]. [A] = 0.1 M, [B] = 0.1 M, [C] = 1 M; and [A] = 0.01 M, [B] = 0.250 M, [C] = 0.791 M
- Kc = 6.00 × 10−2
- PCl5(g) ⇌ PCl3(g) + Cl2(g), Kc = 0.50
- The equilibrium equation is Kp = 1.9 × 103
- Kp = 3.06
- (a) −2x, 2x, −0.250 M, 0.250 M; (b) 4x, −2x, −6x, 0.32 M, −0.16 M, −0.48 M; (c) −2x, 3x, −50 torr, 75 torr; (d) x, − x, −3x, 5 atm, −5 atm, −15 atm; (e) x, 1.03 × 10−4 M; (f) x, 0.1 atm.
- (a) Activities of pure crystalline solids equal 1 and are constant; however, the mass (and equivalently, moles) of Ni does change.
(b) Notice that C2H2(g) is in the reactant side. Reverse the first equation and reciprocate Kp1.[latex]\begin{align*} \text{C}_2\text{H}_2(g)\;+\;3\text{ H}_2(g)\;&{\rightleftharpoons}\;2\text{ CH}_4(g) && \frac{1}{K_{\text{p}_1}} \end{align*}[/latex]
Notice that CH4(g) does not appear in the net equation, and 2 CO(g) is needed. Multiply the second equation by 2 and put exponent 2 on Kp2.
[latex]\begin{align*} 2\text{ CH}_4(g)\;+\;2\text{ H}_2\text{O}(g)\;&{\rightleftharpoons}\; 2\text{ CO}(g)\;+\;6\text{ H}_2(g) && (K_{\text{p}2})^2 \end{align*}[/latex]
Add the modified equations together
[latex]\text{C}_2\text{H}_2(g)\;+\;2\text{ H}_2\text{O}(g)\;{\rightleftharpoons}\;2\text{ CO}(g)\;+\;3\text{ H}_2(g)[/latex]
[latex]K_\text{p} = \frac{1}{K_{\text{p}1}}(K_{\text{p}2})^2=\frac{(K_{\text{p}2})^2}{K_{\text{p}1}}[/latex]
- [NH3] = 9.1 × 10−2 M
- PBrCl = 4.9 × 10−2 atm
- [CO] = 2.0 × 10−4 M
- To obtain the tenth root, raise both sides of the equation to [latex]\frac{1}{10}[/latex]. A shortcut to entering 4.08 × 10−25 on a calculator is to use the EXP button for the power of 10. In this case, 4.08 × 10−25 can be entered as 4.08EXP−25.
[latex]\begin{eqnarray} K_{\text{p}}&=&(P_{\text{H}_2\text{O}})^{10} \\ 4.08\times10^{-25}&=&(10x)^{10} \\ \left(4.08\times10^{-25}\right)^{1/10}&=&\left[(10x)^{\cancel{10}}\right]^{\cancel{1/10}} \\ 10x&=&3.64\times10^{-3} \text{ atm} = P_{\text{H}_2\text{O}} \end{eqnarray}[/latex] - Calculate Q based on the calculated concentrations and see if it is equal to Kc. Because Q does equal 4.32, the system must be at equilibrium.
The new Kc is 5.0. The reaction is exothermic. - (a) [NO2] = 1.17 × 10−3 M
[N2O4] = 0.128 M
(b) Percent error [latex]= \frac{5.87\;\times\;10^{-4}}{0.129}\;\times\;100\% = 0.455\%[/latex]. The change in concentration of N2O4 is far less than the 5% maximum allowed. - (a) The partial pressures are 0.810 atm H2S, 0.014 atm H2, and 0.0072 atm S2
(b) The 2x is dropped from the equilibrium calculation because 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring 2x is [latex]\frac{0.014}{0.824}\;\times\;100\% = 1.7\%[/latex], which is less than allowed by the “5% test.” The error is, indeed, negligible. - [PCl5] = 1.80 M; [PCl3] = 0.195 M; [Cl2] = 0.195 M.
- The % change in [NO2] to reach equilibrium is 3.5%, less than 5%, confirming that 0.20 – 2x could be approximated to 0.20.
[NO2] = 0.19 M
[NO] = 0.0070 M
[O2] = 0.0035 M - [latex]P_{\text{O}_3} = 4.9\;\times\;10^{-26}\;\text{atm}[/latex]
- 507 g
- 32 g
- (a) Both gases must increase in pressure.
(b) [latex]P_{\text{N}_2\text{O}_4} = 8.0\;\text{atm }[/latex] and [latex];P_{\text{NO}_2} = 1.0\;\text{atm}[/latex] - (a) 0.33 mol
(b) 0.50 mol of CO2. Added H2 forms some water to compensate for the removal of water vapor and as a result of a shift to the left after H2 is added. - [latex]P_{\text{H}_2} = 8.67\;\times\;10^{-11}\;\text{atm}[/latex]
[latex]P_{\text{O}_2} = 0.250\;\text{atm}[/latex]
[latex]P_{\text{H}_2\text{O}} = 0.500\;\text{atm}[/latex] - (a) [latex]K_\text{c} = \frac{[\text{NH}_3]^4[\text{O}_2]^7}{[\text{NO}_2]^4[\text{H}_2\text{O}]^6}[/latex]. (b) [NH3] must increase for Qc to reach Kc. (c) That decrease in pressure would decrease [NO2]. (d) [latex]P_{\text{O}_2} = 49\;\text{torr}[/latex]
- [glucose] = 0.150 M, [fructose] = 0.15 M, [sucrose] = 1.65 × 10-7 M
- The initial pressure of N2O3 is 3.80 atm. At equilibrium, [latex]P_{\text{N}_2\text{O}_3} = 1.90 \text{ atm}[/latex] and [latex]P_{\text{NO}} = P_{\text{NO}_2} = 1.90\;\text{atm}[/latex]