Unit 7 Topic Q's

Lakukan tugas rumah & ujian kamu dengan baik sekarang menggunakan Quizwiz!

The diagram above represents the equilibrium between the two isomers of C2H2Cl2, and the table provides the data collected in an experiment to determine its equilibrium constant, Kc, at 490K. In a second experiment done at the same temperature, [Z]eq≈1.0 M. Which of the following is the approximate equilibrium concentration of Y in the second experiment, and why? A [Y]eq≈1.4 M because [Y]eq−[Z]eq should be the same for the same reaction. B [Y]eq≈1.5 M because the ratio [Z]eq[Y]eq should remain constant when the reaction is done at the same temperature. C [Y]eq≈1.6 M because the ratio [Y]initial[Y]eq should remain constant when the reaction is done at the same temperature. D [Y]eq≈2.0 M because ([Y]initial−[Y]eq)=([Z]initial−[Z]eq) should be the same for the same reaction.

B [Y]eq≈1.5 M because the ratio [Z]eq[Y]eq should remain constant when the reaction is done at the same temperature.

HCl(aq)+H2O(l)⇄H3O+(aq)+Cl−(aq) In 1.0MHCl(aq), HCl is nearly 100 percent dissociated, as represented by the equation above. Which of the following best helps to explain why, in 0.01MHCN(aq), less than 1 percent of HCNis dissociated? A The CN− ion is not very soluble in water, and a solid precipitate would form if more of the HCN dissociated. B Compared to the HCl(aq) solution, the concentration of the HCN(aq) solution is much too dilute to achieve 100 percent dissociation. C The equilibrium constant for the dissociation of HCN(aq) is much smaller than that for the dissociation of HCl(aq). D HCN(aq) reacts with water to form a basic solution, and the high concentration of OH−(aq) interferes with the dissociation process.

C The equilibrium constant for the dissociation of HCN(aq) is much smaller than that for the dissociation of HCl(aq).

2HI(g)⇄H2(g)+I2(g) Kp=PH2PI2P2HI=0.0016 The decomposition of HI(g) at 298K is represented by the equilibrium equation above. When 100.torr of HI(g) is added to a previously evacuated, rigid container and allowed to reach equilibrium, the partial pressure of I2(g) is approximately 3.7torr. If the initial pressure of HI(g) is increased to 200.torr and the process is repeated at the same temperature, which of the following correctly predicts the equilibrium partial pressure of I2(g), and why? A PI2≈14 torr, because it is directly proportional to the square of the initial pressure of HI. B PI2≈0.073 torr, because it is inversely proportional to the square of the initial pressure of HI. C PI2≈7.4 torr, because it is directly proportional to the initial pressure of HI. D PI2≈1.9 torr, because it is inversely proportional to the initial pressure of HI.

C. PI2≈7.4 torr, because it is directly proportional to the initial pressure of HI.

2NO(g)+Cl2(g)⇄2NOCl(g)Kc=2000 A mixture of NO(g) and Cl2(g) is placed in a previously evacuated container and allowed to reach equilibrium according to the chemical equation shown above. When the system reaches equilibrium, the reactants and products have the concentrations listed in the following table. SpeciesConcentration (M) NO(g)0.050 Cl2(g)0.050 NOCl(g)0.50 Which of the following is true if the volume of the container is decreased by one half? A Q=100, and the reaction will proceed toward reactants. B Q=100, and the reaction will proceed toward products. C Q=1000, and the reaction will proceed toward reactants. D Q=1000, and the reaction will proceed toward products.

D Q=1000, and the reaction will proceed toward products.

Reaction 1: NO3(g)+NO2(g)⇄N2O5(g) K=2.6×10−11 In the atmosphere, small water droplets are suspended in the air, forming an aerosol. N2O5(g) can form HNO3(aq) under these conditions, as shown in reaction 2, represented below. Reaction 2:N2O5(g)+H2O(l)→2HNO3(aq) Which of the following predicts the effect that the formation of HNO3(aq) will have on the equilibrium shown in reaction 1, and why? A The equilibrium of reaction 1 will shift toward the formation of more product, because N2O5(g) is removed when it reacts to form HNO3(aq). B The equilibrium of reaction 1 will shift toward the formation of more product, because H2O(l) acts as a catalyst for reaction 1. C The equilibrium of reaction 1 will shift toward the formation of more reactants, because the product N2O5(g) is removed when it reacts to form HNO3(aq). D The equilibrium of reaction 1 is not affected, because neither H2O(l) nor HNO3(aq) are gases.

A The equilibrium of reaction 1 will shift toward the formation of more product, because N2O5(g) is removed when it reacts to form HNO3(aq).

Reaction 1: HOCl(aq)+H2O(l)⇄H3O+(aq)+OCl−(aq)K1=[H3O+][OCl−][HOCl]Reaction 2: 2H2O(l)⇄H3O+(aq)+OH−(aq)K2=[H3O+][OH−]Reaction 3: OCl−(aq)+H2O(l)⇄HOCl(aq)+OH−(aq)K3=? Based on the equilibrium constants given above, which of the following gives the correct expression for the equilibrium constant for reaction 3? A K3=K2K1 B K3=K1K2 C K3=K1K2 D K3=1K1K2

A. K3= k2/k1

Fe3+(aq)Colorless+ SCN−(aq)Colorless⇄ FeSCN2+(aq)Red When colorless solutions containing Fe3+(aq) ions and SCN−(aq) ions are combined, a deep-red complex ion, FeSCN2+(aq) quickly forms, as shown in the net ionic equation above. Which of the following explains the observation that adding a few additional crystals of KSCN(s) results in the red color of the solution becoming deeper? A The added KSCN(s) dissolves, disturbing the charge balance in the solution, causing Fe(SCN)3 to precipitate as a red solid. B The added KSCN(s) dissolves, causing the solution to become saturated in SCN− ions, which appear red at high concentrations. C The added KSCN(s) dissolves, causing the reaction system to respond by producing more product to partially consume SCN−(aq) and reduce its concentration. D The added KSCN(s) dissolves, causing the reaction system to respond by forming more Fe3+ ions, which have a deep orange color at high concentrations.

C The added KSCN(s) dissolves, causing the reaction system to respond by producing more product to partially consume SCN−(aq) and reduce its concentration.

A cylinder with a moveable piston is completely filled with a small amount (100 millimoles) of liquid water at a pressure of 1.0atm and a temperature of 80∘C. All the air in the cylinder is excluded. The cylinder is placed in a water bath held at 80∘C. The piston is slowly moved out to expand the volume of the cylinder to 20L as the pressure inside the cylinder is monitored. A plot of the pressure versus volume for the system is shown in the figure above. Which of the following statements most closely indicates, with justification, the region of the curve where the equilibrium represented below occurs? H2O(l)⇄H2O(g) A Region A, because the initial pressure inside the cylinder is equal to the pressure outside the cylinder, so there is no net force on the piston. B Region B, because the pressure inside the cylinder is equal to the vapor pressure of water at 80∘C when both liquid and gas phases are present. C Region C, because the water vapor is behaving according to the ideal gas law as expansion occurs. D Region D, because the pressure inside the cylinder has leveled off.

B Region B, because the pressure inside the cylinder is equal to the vapor pressure of water at 80∘C when both liquid and gas phases are present.

MgCO3(s)⇄Mg2+(aq)+CO32−(aq) A saturated solution of MgCO3 at equilibrium is represented by the equation above. Four different saturated solutions were prepared and kept at the same temperature. A given amount of HCl was added to each solution and data were collected to calculate the molar solubility of MgCO3 as shown in the table above. Which of the following can be concluded from the data? A The molar solubility of MgCO3 decreases with increasing acidity (lower pH). B The molar solubility of MgCO3 increases with increasing acidity (lower pH). C The pH or acidity of the solution has a negligible effect on the molar solubility of MgCO3. D From the data it is not possible to determine the effect of changes in acidity on the molar solubility of MgCO3.

B The molar solubility of MgCO3 increases with increasing acidity (lower pH).

2A(g)+B(g)⇄2C(g) A(g) and B(g) react to form C(g), according to the balanced equation above. In an experiment, a previously evacuated rigid vessel is charged with A(g), B(g), and C(g), each with a concentration of 0.0100M. The following table shows the concentrations of the gases at equilibrium at a particular temperature. [A]eq[B]eq[C]eq0.01800.01400.0020 If the experiment is repeated at a higher temperature at which Kc is larger, which of the following best describes the effect of the temperature change on the concentrations of the gases at equilibrium? A [A]eq, [B]eq, and [C]eq will all increase because Kc increased. B [A]eq and [B]eq will remain constant, but [C]eq will increase because Kcincreased. C There will be a decrease in [A]eq that will be two times the decrease in [B]eqbecause A and B react in a 2-to-1 ratio. D There will be an increase in [A]eq that will be two times the increase in [B]eqbecause A and B react in a 2-to-1 ratio.

C There will be a decrease in [A]eq that will be two times the decrease in [B]eqbecause A and B react in a 2-to-1 ratio.

An equimolar mixture of X(g) and Y(g) is placed inside a rigid container at constant temperature. The particle diagram above represents the changes that occur over time. Based on the particle diagram, which of the following best predicts whether or not the system has reached equilibrium by 300s? A It is not possible to determine that the system has reached equilibrium by 300s because the stoichiometry of the reaction is not known. B It is not possible to determine that the system has reached equilibrium by 300s because the amounts of X, Y, and XY have continued to change. C The system has reached equilibrium by 300s because the rate of formation of XY is constant. D The system has reached equilibrium by 300s because the rates of consumption of X and Y are equal.

B. It is not possible to determine that the system has reached equilibrium by 300s because the amounts of X, Y, and XY have continued to change.

X(g)+Y(g)⇄XY(g) In an experiment, X(g) and Y(g) were combined in a rigid container at constant temperature and allowed to react as shown in the equation above. The table provides the data collected during the experiment. Based on the data, which of the following claims is most likely correct? A The reaction was about to reach equilibrium 15 minutes after the reactants were combined because the concentrations of X and XY were almost the same. B The reaction reached equilibrium between 75 minutes and 155 minutes after the reactants were combined because the concentrations of X and XYremained constant. C The reaction did not reach equilibrium because only 86% of the initial concentration of X was consumed. D The reaction did not reach equilibrium because initially there was no XYinside the container.

B. The reaction reached equilibrium between 75 minutes and 155 minutes after the reactants were combined because the concentrations of X and XYremained constant.

Reaction 1: NO3(g)+NO2(g)⇄N2O5(g) K=2.6×10−11 During the day, solar radiation is absorbed by NO3(g), resulting in its decomposition. Which of the following best explains whether the equilibrium concentration of N2O5(g) in the atmosphere in the daytime is different from that in the nighttime, and why? A [N2O5] will be higher during the day, because the decomposition of NO3(g)results in an increase in the rate of production of N2O5(g). B [N2O5] will be higher during the day, because NO2(g) will be in excess, leading to an increase in the rate of production of N2O5(g). C [N2O5] will be higher at night, because the decomposition of NO3(g) in the daytime will result in an increase in the rate of consumption of N2O5(g) to reform NO3(g). D [N2O5] will be about the same at nighttime and daytime, because the amount of NO2(g) will not be changed and the equilibrium will not be affected.

C [N2O5] will be higher at night, because the decomposition of NO3(g) in the daytime will result in an increase in the rate of consumption of N2O5(g) to reform NO3(g).

X(g)+Y(g)⇄XY(g) The particle diagram above illustrates the changes that take place when X(g) and Y(g) are placed inside a rigid container at constant temperature. Which of the following is a characteristic of a system at equilibrium that is best represented by the particle diagram? A The particle diagram shows that initially the reaction proceeds to the right to form products, which is a characteristic of a system at equilibrium. B The particle diagram shows that after 200s the rate of the reverse reaction is slower than the rate of the forward reaction, which is a characteristic of a system at equilibrium. C The particle diagram shows that after 200s there are no observable changes in the amounts of reactants and products, which is a characteristic of a system at equilibrium. D The particle diagram shows that between 0s and 200s the rates of the forward and reverse reactions are the same, which is a characteristic of a system at equilibrium.

C. The particle diagram shows that after 200s there are no observable changes in the amounts of reactants and products, which is a characteristic of a system at equilibrium.

The table above shows data for two reactions carried out in two separate evacuated 1.0-liter rigid containers at constant temperature of 298K. To each container 0.50mol of the appropriate reactants was added, and the reaction was allowed to reach equilibrium. Based on this information, which of the following correctly compares the relative concentrations of BrCl and NO present inside their respective containers at equilibrium? A [BrCl]eq=[NO]eq because equimolecular mixtures of the reactants were allowed to reach equilibrium at the same constant temperature. B [BrCl]eq>[NO]eq because Br2 and Cl2 are larger molecules that can collide more frequently to form products. C [BrCl]eq>[NO]eq because the much larger Keq for reaction 1 means that a much higher concentration of products will be present at equilibrium for reaction 1 compared with reaction 2. D [BrCl]eq<[NO]eq because the much larger Keq for reaction 1 means that hardly any products will be present at equilibrium compared with reaction 2.

C. [BrCl]eq>[NO]eq because the much larger Keq for reaction 1 means that a much higher concentration of products will be present at equilibrium for reaction 1 compared with reaction 2.

A sample of N2O4(g) is placed into an evacuated container at 373K and allowed to undergo the reversible reaction N2O4(g)⇄2NO2(g). The concentration of each species is measured over time, and the data are used to make the graph shown above. Which of the following identifies when equilibrium is first reached and provides a correct explanation? A At 14 seconds, because [N2O4] is twice [NO2], which implies that the forward and reverse reaction rates are equal. B At 23 seconds, because [NO2] equals [N2O4], which shows that equal concentrations are present at equilibrium. C At 40 seconds, because [NO2] is twice [N2O4], which matches the stoichiometry of the balanced chemical equation. D At 60 seconds, because [NO2] and [N2O4] remain constant, indicating that the forward and reverse reaction rates are equal.

D. At 60 seconds, because [NO2] and [N2O4] remain constant, indicating that the forward and reverse reaction rates are equal.

A student investigates the effects of pH on the solubility of AgOH(s), which dissolves in water according to the equation AgOH(s)⇄Ag+(aq)+OH−(aq). The value for Ksp for AgOH is 2.0×10−8 at 298K. The student places the same mass of AgOH(s) into 50.0mL of different solutions with specific pH values and measures the concentration of Ag+ ions in each solution after equilibrium is reached. Based on the data in the table, what can be concluded about the solubility of AgOH? A The solubility of AgOH is unaffected by pH because Ksp is a constant value at 298K. B The solubility of AgOH is unaffected by pH because it is the same value for pH 7.00 and pH 8.00. C AgOH is more soluble at higher pH because lower concentrations of OH−(aq) shift the solubility equilibrium toward Ag+(aq) and OH−(aq). D AgOH is less soluble at higher pH because higher concentrations of OH−(aq) shift the solubility equilibrium toward solid AgOH.

D AgOH is less soluble at higher pH because higher concentrations of OH−(aq) shift the solubility equilibrium toward solid AgOH.

Reaction 1: CO(g)+3H2(g)⇄CH4(g)+H2O(g)K1=[CH4][H2O][CO][H2]3Reaction 2: CO2(g)+H2(g)⇄CO(g)+H2O(g)K2=[CO][H2O][CO2][H2]Reaction 3: CH4(g)+2H2O(g)⇄CO2(g)+4H2(g)K3=? The chemical equations and equilibrium expressions for two reactions at the same temperature are given above. Based on the information, which of the following expressions can be used to calculate the value of K3 for reaction 3 at the same temperature? A K3=(−K1)+(−K2) B K3=(−K1)−(−K2) C K3=K1×K2 D K3=1K1×1K2

D K3=1/K1×1/K2

The system represented by the equation above is allowed to establish equilibrium. The initial pressures of the substances are given in the table. Which of the following explains what the system will do as it approaches equilibrium? A Q=(80.)2(10.)(2.0)>Kp and equilibrium will be approached by producing NOBrbecause the forward reaction is faster than the reverse reaction. B Q=(80.)(10.)(2.0)<Kp and equilibrium will be approached by producing NOBrbecause the forward reaction is faster than the reverse reaction. C Q=(80.)2(10.)2(2.0)<Kp and equilibrium will be approached by consuming NOBrbecause the reverse reaction is faster than the forward reaction. D Q=(80.)2(10.)2(2.0)>Kp and equilibrium will be approached by consuming NOBrbecause the reverse reaction is faster than the forward reaction.

D Q=(80.)2(10.)2(2.0)>Kp and equilibrium will be approached by consuming NOBrbecause the reverse reaction is faster than the forward reaction.

Cl2(aq)+2H2O(l)⇄H3O+(aq)+Cl−(aq)+HOCl(aq) Kc=[H3O+][Cl−][HOCl]/[Cl2]=4.8×10−4 The equilibrium reaction between Cl2(aq) and H2O(l) at 25°C is represented by the chemical equation shown above. If a solution at equilibrium at 25°C is diluted with distilled water to twice its original volume, which of the following gives the value for Qc and predicts the response by the system immediately after dilution? A Qc=4Kc, and the rate of the reverse reaction will be greater than the rate of the forward reaction. B Qc=4Kc, and the rate of the forward reaction will be greater than the rate of the reverse reaction. C Qc=Kc/4, and the rate of the reverse reaction will be greater than the rate of the forward reaction. D Qc=Kc/4, and the rate of the forward reaction will be greater than the rate of the reverse reaction.

D Qc=Kc/4, and the rate of the forward reaction will be greater than the rate of the reverse reaction.

/Ag2CO3(s)⇄2Ag+(aq)+CO32−(aq) The chemical equation above represents the equilibrium that exists in a saturated solution of Ag2CO3. If S represents the molar solubility of Ag2CO3, which of the following mathematical expressions shows how to calculate S based on Ksp ? A S=Ksp−−−√ B S=Ksp/2−−−√ C S=Ksp/2−−−√3 D S=Ksp/4−−−√3

D S=Ksp/4−−−√3

2X(g)+Y2(g)⇄2XY(g) A reversible reaction is represented by the equation above. The amounts of reactants and products at time 1 are shown in the particle diagram on the left. The particle diagram on the right shows the amounts of reactants and products at time 2. Based on the diagrams, what can be inferred about the relative rates of the forward and reverse reactions between time 1 and time 2 ? A Nothing can be inferred because the total number of X and Y atoms is the same in each diagram. B Nothing can be inferred because the temperature of the system may have been changed. C The rate of the reverse reaction is greater than the rate of the forward reaction. D The rate of the forward reaction is greater than the rate of the reverse reaction.

D The rate of the forward reaction is greater than the rate of the reverse reaction.

Reaction 1: NO3(g)+NO2(g)⇄N2O5(g) K=2.6×10−11 A mixture of NO3 and a ten-fold excess of NO2 are placed inside a rigid container at constant temperature and allowed to reach equilibrium. Which of the following provides a correct comparison of the equilibrium concentrations of these chemical species, and why? A [N2O5]<<[NO3], because a small K value indicates that the consumption of the reactants is favored at equilibrium. B [NO2]<<[N2O5], because a small K value indicates that the consumption of the reactants is favored at equilibrium. C [NO2]<<[NO3], because a small K value indicates that the formation of products is not favored at equilibrium. D [N2O5]<<[NO2], because a small K value indicates that the formation of products is not favored at equilibrium.

D [N2O5]<<[NO2], because a small K value indicates that the formation of products is not favored at equilibrium

CdF2(s)⇄Cd2+(aq)+2F−(aq) A saturated aqueous solution of CdF2 is prepared. The equilibrium in the solution is represented above. In the solution, [Cd2+]eq=0.0585M and [F−]eq=0.117M. Some 0.90MNaF is added to the saturated solution. Which of the following identifies the molar solubility of CdF2 in pure water and explains the effect that the addition of NaF has on this solubility? A The molar solubility of CdF2 in pure water is 0.0585M, and adding NaF decreases this solubility because the equilibrium shifts to favor the precipitation of some CdF2. B The molar solubility of CdF2 in pure water is 0.0585M, and adding NaF has no effect on the solubility because only changes in temperature can increase or decrease the molar solubility of an ionic solid. C The molar solubility of CdF2 in pure water is 0.117M, and adding NaF decreases this solubility because the equilibrium shifts to favor the precipitation of some CdF2. D The molar solubility of CdF2 in pure water is 0.176M, and adding NaF increases this solubility because the Na+ ions displace the Cd2+ ions, causing the equilibrium to shift to favor the products.

A. The molar solubility of CdF2 in pure water is 0.0585M, and adding NaF decreases this solubility because the equilibrium shifts to favor the precipitation of some CdF2.

AgCl(s) ⇄ Ag+(aq)+Cl−(aq)Ksp=1.8×10−10 Shown above is information about the dissolution of AgCl(s) in water at 298K. In a chemistry lab a student wants to determine the value of s, the molar solubility of AgCl, by measuring [Ag+] in a saturated solution prepared by mixing excess AgCl and distilled water. How would the results of the experiment be altered if the student mixed excess AgCl with tap water (in which [Cl−]=0.010M) instead of distilled water and the student did not account for the Cl− in the tap water? A The value obtained for Ksp would be too small because Cl−(aq) ions would be attracted to the Ag+ ions in the AgCl crystals, thus preventing water molecules from reaching the crystals. B The value obtained for Ksp would be too small because less AgCl(s) would dissolve because of the common ion effect due to the Cl−(aq) already in the water. C The value obtained for Ksp would be too large because more AgCl(s) would dissolve because of the attractions between Ag+ ions in the AgCl crystals and the Cl−(aq) ions in the water. D The results of the experiment would not be altered because 0.010M is such a small concentration of Cl−(aq) ions and thus has no effect on the dissolution of AgCl(s).

B The value obtained for Ksp would be too small because less AgCl(s) would dissolve because of the common ion effect due to the Cl−(aq) already in the water.


Set pelajaran terkait

7th Grade Math Midterm Study Guide

View Set

Business Finance Ch6 Quiz - Connect

View Set

Enlisted PPME Block 5: Rules of Engagement, General Principles

View Set

Article 312- CABINETS, CUTOUT BOXES, AND METER SOCKET ENCLOSURES

View Set

Dosage Calculation 3.0 Critical Care Medications

View Set