Kinetics and Equilibrium

A high pressure gaseous synthesis of hydrazine (H4N2) is governed by the following equilibrium: 2 NH3 <--> H4N2 + H2, Keq = 10-7 If a 1 L system is charged with 10 mol of NH3, how much hydrazine will be present at equilibrium? Question 11 Answer Choices A. 0.003 mol B. 0.03 mol C. 0.01 mol D. 0.001 mol

2 NH3 <--> H4N2 + H2, Keq = 10-7 When a 1 L system is charged with 10 mol of NH3, 0.003 mol of H4N2 will be present at equilibrium. At equilibrium we can say that K = [H4N2][H2]/[NH3]2 = 10-7. Allowing [H4N2] = [H2] = x while [NH3] = 10 - 2x, we get x2/(10 - 2x)2 = 10-7. As Keq is small we can safely disregard 2x in the denominator, giving x2 = 10-5, or x = 10-2.5. This value is between 10-2 and 10-3, meaning 0.003 is the only possible answer.

A high pressure gaseous synthesis of hydrazine (H4N2) at moderate temperatures is governed by the following equilibrium: 2 NH3 <--> H4N2 + H2, Keq = 10-2 If 2 mol of each of the three species is added to the flask and the progression of the reaction is monitored, which of the following will be observed? Question 12 Answer Choices A. An increase in the concentration of NH3, and a decrease in the concentrations of H4N2 and H2 B. An increase in the concentration of NH3, and an increase in the concentrations of H4N2 and H2 C. An increase in the concentration of H4N2, and a decrease in the concentrations of NH3 and H2 D. A decrease in the concentration of NH3, and an increase in the concentrations of H4N2 and H2

A. A high pressure gaseous synthesis of hydrazine (H4N2) at moderate temperatures is governed by the following equilibrium: 2 NH3 <--> H4N2 + H2, Keq = 10-2 If 2 mol of each of the three species is added to the flask and the progression of the reaction is monitored, an increase in the concentration of NH3, and a decrease in the concentrations of H4N2 and H2 will be observed as the reaction proceeds to equilibrium. The reaction quotient (Q = [H4N2][H2]/[NH3]2) for the reaction in question at the initial conditions is equal to 1. As this value is greater than Keq, the reaction must move to the left to attain equilibrium. Answer choices indicating that both sides of the reaction move in the same direction, can be easily eliminated, as can choices in which species on the same side of the equilibrium equation move in opposite directions.

A sealed container contains NO2, a brownish-red gas, and N2O4, a colorless gas, at equilibrium at 0°C according to the following reaction 2NO2 (g) <--> N2O4 (g) : . How will the color of the gas mixture change if the container is placed in dry ice and acetone at -78°C? Question 5 Answer Choices A. The gas mixture will become lighter in color. B. The gas mixture will become darker in color. C. Cannot be determined from the information provided. D. The gas mixture's color will not change because it is in equilibrium.

A. A sealed container contains NO2, a brownish-red gas, and N2O4, a colorless gas, at equilibrium at 0°C according to the following reaction: . The gas mixture will become lighter in color if the container is placed in dry ice and acetone at -78°C. The system is at equilibrium at 0°C, so ΔG must equal 0. We also know that during the forward reaction 2 moles of gas are converted to 1 mole of gas, which results in a decrease in entropy (ΔS < 0). From this, the sign of ΔH can be determined because ΔG = 0 = ΔH - TΔS. Solving for ΔH gives us: ΔH = TΔS. ΔH must be negative for the forward reaction because ΔS is also negative. Since the forward reaction is exothermic, it will be thermodynamically favored by a decrease in temperature. As temperature falls, the reaction shifts to the right. N2O4 (colorless) increases and NO2 (brownish-red) decreases, leading to an overall lighter color.

Addition of which of the following salts will most likely result in a precipitate when mixed with an aqueous solution of AgNO3? Question 2 Answer Choices A. NaCl B. Mg(ClO4)2 C. Hg(NO3)2 D. NaCH3COO

A. Addition of NaCl will most likely result in a precipitate when mixed with an aqueous solution of AgNO3. Most Ag+ compounds have low solubility in water. According to solubility rules, exceptions should include nitrates, acetates, and perchlorates. The addition of NaCl would result in the formation of AgCl, a water insoluble compound that would immediately precipitate. The choices NaCH3COO and Mg(ClO4)2 would result in the compounds AgCH3COO and AgClO3, both of which have markedly greater solubility than AgCl. Although Hg(NO3)2 could shift the dissociation equilibrium of AgNO3 to the left, precipitation is much less likely than what would occur with AgCl formation.

If a sample of barium hydroxide (Ba(OH)2) is partially dissolved in pure water at room temperature. If the equilibrium concentrations of Ba2+ and OH- ions are measured, which of the following expressions could be used to find the solubility product for barium hydroxide? Question 19 Answer Choices A. Ksp = [Ba][OH]2 B. Ksp = 2[Ba][OH]2 C. Ksp = 2[Ba][OH] D. Ksp = [Ba][OH]

A. Barium hydroxide (Ba(OH)2) is dissolved in pure water at room temperature. If the equilibrium concentrations of Ba+ and OH- ions are measured, Ksp = [Ba][OH]2 could be used to find the solubility product for barium hydroxide. Barium hydroxide dissolves reversibly via the reaction Ba(OH)2(s) <--> Ba2+(aq) + 2 OH-(aq) Therefore, the equilibrium expression is Ksp = [Ba2+][OH-]2.

Consider the reaction . At equilibrium, a 10 L vessel contains 1.5 moles of HI, 0.1 moles of I2 and 0.3 moles of H2. Which direction will the reaction proceed if 2 moles of HI, and 0.5 moles of both I2 and H2 are mixed in a separate 10 L vessel? Question 3 Answer Choices A. The reaction will proceed forward. Correct Answer (Blank) B. The mixture will be at equilibrium. C. This cannot be determined with the given information. D. The reaction will proceed in reverse.

A. Consider the reaction . At equilibrium, a 10 L vessel contains 1.5 moles of HI, 0.1 moles of I2 and 0.3 moles of H2. The reaction will proceed forward if 2 moles of HI, and 0.5 moles of both I2 and H2 are mixed in a separate 10 L vessel. All of the information needed to answer this question is given, so the answer choice "this cannot be determined with the given information" can be eliminated. First, Keq can be determined with the equilibrium moles of each compound despite not knowing the total pressure of the vessel. The derivation is shown here for clarity but could be skipped as the expression reduces down to the ratio of moles products to moles reactants at equilibrium raised to the power of their coefficients in the balanced equation: Next, calculate the reaction quotient, Q, for the second vessel in a similar manner: Since Q < K, the forward reaction is favored.

The [Pb2+] of a solution is 1.25 × 10-3 M. Calculate the [SO42-] that must be exceeded before PbSO4 can precipitate. (The solubility product of PbSO4 at 25°C is 1.6 × 10-8.) Question 17 Answer Choices A. 1.3 × 10-5 M B. 1.3 × 10-4 M C. 1.6 × 10-8 M D. 6.4 × 10-6 M

A. The [Pb2+] of a solution is 1.25 × 10-3 M. The [SO42-] that must be exceeded before PbSO4 can precipitate is 1.3 × 10-5 M. (The solubility product of PbSO4 at 25°C is 1.6 × 10-8.) The concentrations of the dissolved ions must exceed the Ksp value for precipitate to form. Given the dissolution of PbSO4, PbSO4(s) <--> Pb2+(aq) + SO42-(aq) we have Ksp = [Pb2+][SO42-]. Therefore, the minimum concentration of SO42- needed to form precipitate is [SO42-] = Ksp/[Pb2+] = (1.6 × 10-8)/(1.25 × 10-3) = 1.3 × 10-5 M.

The plot below shows the relationship between rate and concentration of reagent X in the reaction X(g) Y(g). From this data what is the rate law of the reaction? Question 5 Answer Choices A. Rate = k B. Rate = k[X] C. Rate = k[X]2 D. Cannot be determined from the information given.

A. The plot below shows the relationship between rate and concentration of reagent X in the reaction X(g) Y(g). From this data, the rate law of the reaction is Rate = k. As the concentration of X changes, the reaction rate remains the same. Thus, the rate is zero order with respect to X, and the rate law is simply rate = k. Note that the graph gives experimental information which is sufficient to produce a rate law, eliminating the choice, "cannot be determined from the information given." Rate = k[X] and Rate = k[X]2 are first and second order, respectively, with respect to X, which would lead to changing rate with changing [X].

The reaction SbCl5 <--> Cl2(g) + SbCl3 has a Keq of 3.2 × 10-6. Yet despite the small Keq when SbCl5 is dissolved and stirred in organic solvents for extended periods of time, unexpectedly high levels of solvated SbCl3 are measured in solution. Which of the following best explains this phenomenon? Question 10 Answer Choices A. Cl2(g) escapes from solution, driving the equilibrium toward SbCl3. B. The forward reaction has a large negative entropy, which drives the reaction to the right. C. When large amounts of SbCl5 are added to solution, Keq for the reaction gets larger. D. SbCl3 is an unstable species in organic solvents.

A. The reaction SbCl5 <--> Cl2(g) + SbCl3 has a Keq of 3.2 × 10-6. When SbCl5 is dissolved and stirred in organic solvents for extended periods of time, unexpectedly high levels of solvated SbCl3 are measured in solution because Cl2(g) escapes from solution, driving the equilibrium toward SbCl3. The phenomenon above is a description of Le Chatelier's Principle. The escape of non-solvated gas from solution would force the accumulation of SbCl3 in order to maintain Keq. As Le Chatelier's Principle would predict, if Cl2(g) is removed from equilibrium, the reaction will shift toward the right side of the equation. The value of Keq is unaffected by the quantities of reactants in the mixture and only changes when the temperature does. The forward reaction produces a gas, which would indicate a positive entropy.

The following data were collected for the reaction A + B + C D + E: What is the rate law for this reaction? Question 15 Answer Choices A. Rate = k[A]2[B]2[C] B. Rate = k[A][B]2[C]2 C. Rate = k[A]2[B][C]2 D. Rate = k[A][B][C]

A. Using the information below, we know that the rate law for this reaction is rate = k[A]2[B]2[C]. The following data were collected for the reaction A + B + C D + E: To obtain the rate law, we must compare trials and determine how changes in the initial concentrations of the reactants affect the initial rate of the reaction. Comparing Trials 1 and 4, we see that [A] increased by a factor of 2 and the reaction rate increased by a factor of 4 = 22; thus, the reaction is second order with respect to A. Eliminate Rate = k[A][B][C] and Rate = k[A][B]2[C]2. Comparing Trials 1 and 3, we see that [B] increased by a factor of 2 and the reaction rate increased by a factor of 4 = 22; thus, the reaction is also second order with respect to B. By process of elimination you can choose Rate = k[A]2[B]2[C] as the correct answer without analyzing the rate law with respect to C. Doing so, however, we find that by comparing Trials 1 and 2, we see that [C] increased by a factor of 2 and so did the reaction rate; thus, the reaction is first order with respect to C. Therefore, the rate law for this reaction is given by Rate = k[A]2[B]2[C].

When CaF2 is added to a 0.02 M solution of NaF, the solubility of CaF2 is: Question 18 Answer Choices A. low, because the concentration of F- already present in solution prevents dissolution of CaF2. B. high, due to the common ion effect. C. low, because less water is available for solvation due to the presence of NaF. D. unaffected by the presence of sodium fluoride.

A. When CaF2 is added to a 0.02 M solution of NaF, the solubility of CaF2 is low, because the concentration of F- already present in solution prevents dissolution of CaF2. The addition of dissolved F- (by adding NaF) decreases the solubility of CaF2. This is the common ion effect (Le Châtelier's principle).

If a reaction has a ΔH° < 0 and a ΔS° > 0, the Keq is Question 32 Answer Choices A. greater than 1, and the products are favored. B. greater than 1, and the reactants are favored. C. less than 1, and the products are favored. D. less than 1, and the reactants are favored.

A. A negative ΔH° and positive ΔS° will always give a negative ΔG°. In the context of equilibrium, this means that the products are favored (eliminate choices B and D). This corresponds to a Keq greater than 1 since K is a ratio of [products]/[reactants], making choice A the correct answer.

If the molar solubility of O2 in water is 2.59 x 10-4M at sea level and 25°C, what is the molar solubility of O2 at an elevation of 6,225 ft where atmospheric pressure is 0.805 atm at 25°C? Question 19 Answer Choices A. 2.08 x 10-4M B. 2.59 x 10-4M C. 3.66 x 10-4M D. 4.06 x 10-4M

A. At higher elevation, atmospheric pressure is lowered, which decreases the solubility of a gas in solution. Choice A is the only option with a lower molar solubility than the one given in the question.

Which factors are unaffected by the addition of a catalyst? Question 4 Answer Choices A. Enthalpy of the reaction B. Activation energy of the reverse reaction C. Activation energy of the forward reaction D. Reaction rate

A. Enthalpy of the reaction

Step 1: CO2(g) + H2O(l) ? H2CO3(aq) Step 2: H2CO3(aq) + NH4OH(aq) → NH4HCO3(aq) + H2O(l) Step 3: NH4HCO3(aq) + NaCl(aq) → NaHCO3(s) + NH4Cl(aq) In the mechanism above, ammonium bicarbonate acts as which of the following? Question 11 Answer Choices A. An intermediate B. A catalyst C. A transition state D. An insoluble salt

A. Intermediates in reaction mechanisms are generated and then consumed (they are not seen in the initial reactants or final products - choice A is correct). Catalysts, beyond their role in increasing reaction rate, are not consumed during the reaction (choice B is wrong). Given that ammonium bicarbonate is a stable salt, it is unlikely to be a transition state compound (choice C is wrong) and ammonium salts are soluble (choice D is wrong).

Addition of reactant to a previously equilibrated system will result in which of the following? Question 12 Answer Choices A. K > Q and the forward reaction predominates B. K > Q and the reverse reaction predominates C. K < Q and the forward reaction predominates D. K < Q and the reverse reaction predominates

A. Since a K or Q expression is a ratio of [products]/[reactants], addition of a reactant will decrease the reaction quotient and have no impact on the equilibrium constant which depends only on temperature, thus K > Q (choices C and D are incorrect). When K > Q, the reaction proceeds forward in order to use reactants, make products, and reestablish equilibrium where Q = K (choice B is incorrect and A is correct).

Equal volumes of equimolar solutions of Pb(CH3CO2)2 and HI are mixed, forming a yellow precipitate. Which of the following represents the reaction described? Question 14 Answer Choices A. Pb2+(aq) + 2 I-(aq) → PbI2(s) B. H+(aq) + CH3CO2-(aq) → CH3CO2H (s) C. 2 I-(aq) → I2(s) D. CH3CO2-(aq) + I-(aq) → CH3I (s) + CO2(g)

A. Since both starting compounds are electrolytes and dissociate in solution, the cation of one compound will pair with the anion of the other compound in a double displacement reaction, yielding both of the products suggested in choices A and B. Choice A is the better option, however, because CH3CO2H (acetic acid) is a liquid, not a solid. Most Pb compounds are insoluble and are likely precipitates in such reactions. Choice C can be eliminated since two negatively charged iodide ions cannot pair up to make a neutral I2 molecule unless electrons are removed (an oxidation), which is not indicated here. Choice D can also be eliminated for similar reasons. Two anions cannot form two neutral compounds unless electrons are removed from the system. In addition, methyl iodide should be expected to be a liquid as it has a relatively low molar mass and a low polarity, giving it weak intermolecular forces.

The van der Waals equation for real gases: contains two constants which have unique values for each gas. When comparing these constants for neon and argon, the constant "b" is larger for argon because: Question 34 Answer Choices A. argon atoms have greater real volume than neon. B. argon has more protons than neon. C. argon has stronger intermolecular forces than neon. D. argon has a greater density than neon.

A. The constant "b" is a correction for the real, finite volume of gas molecules. The larger the value of "b", the larger the gas in question. As such, choice A is correct. The number of protons does not have a consistent relationship to molecular volume (eliminate choice B). Density (which is mass/volume), has no correlation to the van der Waals equation (eliminate choice D). The correction for intermolecular forces is contained in the constant "a" (eliminate choice B).

Which of the following would violate assumptions made by the kinetic theory of gases? Question 8 Answer Choices A. Condensation B. Intermolecular collisions C. Random molecular trajectories D. Elastic collisions with container walls

A. The kinetic molecular theory of gases makes the following assumptions: the total volume of gas molecules is negligible, the molecules are in constant, random motion (choice C is incorrect), intermolecular forces are absent, and collisions are elastic (choice B and D are incorrect). Condensation is the transition of a gas to liquid and results from intermolecular forces acting on the molecules. Choice A is a violation of the assumptions and is therefore correct.

Which of the following properties would be observed at the conditions indicated by point X on the phase diagram above? Question 5 Answer Choices A. The substance exhibits characteristics of both a liquid and a gas. B. The substance exhibits characteristics of both a liquid and a solid. C. The substance exhibits characteristics of both a solid and a gas. D. The substance spontaneously undergoes decomposition.

A. The point on the phase diagram indicates that the substance is beyond its critical point and will behave as a supercritical fluid. Supercritical compounds, those at conditions beyond their critical points, behave with properties typical of both liquids and gases (choice A is correct, choices B and C are incorrect). While heating a sample sufficiently may result in decomposition, we have no evidence that side reactions would begin to occur and A is the better answer, so choice D can be eliminated.

Four hundred grams of dry ice (solid carbon dioxide) is added to water, causing the water to freeze. Assuming no heat exchange with the surroundings and the initial water temperature to be 0.1°C, what mass of water could be frozen? [Note: ΔHsub (CO2) = 0.571 kJ/g; ΔHfus (H2O) = 334 kJ/g] Question 9 Answer Choices A. 0.69 g B. 1.3 g C. 190 g D. 334 g

A. The sublimation of four hundred grams of dry ice would result in (400 g)(0.571 kJ/g) = ~225 kJ of energy being absorbed from the water. Using ΔHfus (H2O), we can then approximate that (225 kJ)/(334 kJ/g) = ~0.69 g water could be frozen, making choice A correct.

A student in an organic chemistry lab heats a white crystalline substance in a beaker under a vacuum and observes the solid vanish without melting. Which of the following best explains this phenomenon? Question 21 Answer Choices A. Sublimation B. Deposition C. Fusion D. Vaporization

A. The transition from solid to gas is sublimation (choice A is correct). These transitions are most likely to occur when solid substances are held under low pressures and heated. As no liquid was observed, choices C and D can be eliminated. Deposition is the transition from a gas to a solid and was not observed here (choice B is incorrect).

2 SO2(g) + O2(g) 2 SO3(g) A chemist adds a reagent to the following equilibrated reaction and immediately observes the forward and reverse reaction rates to increase by 2.14 and 2.29, respectively. Which of the following was most likely added to the reaction? Question 9 Answer Choices A. SO3(g) B. V2O5(s) C. SO2(g) D. O2(g)

B. 2 SO2(g) + O2(g) 2 SO3(g) A chemist adds V2O5(s) to the following equilibrated reaction and immediately observes the forward and reverse reaction rates to increase by 2.14 and 2.29, respectively. The situation here describes the addition of a catalyst to an equilibrated reaction. Catalysts increase the rates of both the forward and reverse reaction by decreasing the activation energy. Addition of a reactant (SO2(g) and O2(g)) would initially increase the rate of the forward reaction alone while adding a product (SO3(g)) would initially increase only the rate of the reverse reaction. Thus V2O5(s) must serve as a catalyst in this reaction and results in the observed rate changes.

A catalyst would change which of the following? Question 18 Answer Choices A. ΔH B. Ea C. Σ(ΔHproducts) D. Σ(ΔHreactants)

B. A catalyst would change Ea. A catalyst lowers the activation energy, Ea, of a reaction. It does not affect the enthalpy of the reaction at all.

A researcher finds that a unimolecular reaction gives rise to the reaction diagram below. Overall, the reaction is: Question 4 Answer Choices A. first order. B. second order. C. third order. D. of undetermined order based on the data provided.

B. A researcher finds that a unimolecular reaction gives rise to the reaction diagram below. Overall, the reaction is second order. Since the reaction is unimolecular the rate law is: Rate = k[A]x. The order can be determined two ways. First, it can be noticed that doubling the concentration causes the rate to quadruple. Second, the parabolic shape of the curve shows a squared relationship. Either method proves the reaction to be second order.

A sample of solid Mg(OH)2 is added to water and reaches equilibrium with its dissociated ions. Addition of the strong base NaOH would most notably increase the concentration of: Question 20 Answer Choices A. Mg2+ . B. undissociated magnesium hydroxide. C. undissociated sodium hydroxide. D. H3O+.

B. A sample of the salt Mg(OH)2 is dissolved in water and reaches equilibrium with its dissociated ions. Addition of the strong base NaOH would most notably increase the concentration of undissociated magnesium hydroxide. The addition of hydroxide (from NaOH) would decrease the solubility of magnesium hydroxide due to the common ion effect. Therefore, the amount of undissociated Mg(OH)2 would increase.

N2(g) + 3 H2(g) <--> 2 NH3(g), ΔH = -22 kcal/mol What is the effect of decreasing the volume of the system? Question 24 Answer Choices A. An increase in the reverse reaction due to increased pressure B. An increase in the forward reaction due to increased pressure C. An increase in the reverse reaction due to decreased pressure D. An increase in the forward reaction due to decreased pressure

B. Decreasing the volume of the system will result in an increase in the forward reaction due to increased pressure. Le Châtelier's Principle implies that increasing the pressure by decreasing the volume of a container will shift the reaction to the side with the fewer moles of gas. Since the reactant side has 1 + 3 = 4 moles of gas while the product side has just 2, the forward reaction will be favored.

Given the average rate for the following reaction is 2.25 × 10-2 M/min, how long will it take to generate 0.45 moles of NO2(g) in one liter of solution? 2 N2O5(g) → 4 NO2(g) + O2(g) Question 8 Answer Choices A. 15 minutes B. 5 minutes Correct Answer (Blank) C. 10 minutes D. 20 minutes

B. Given the average rate for the following reaction is 2.25 × 10-2 M/min, it would take 5 minutes to generate 0.45 moles of NO2(g) in one liter of solution. The average rate expression for this reaction is as follows: Thus the change in nitrogen dioxide concentration over time is 0.09 M/min (four times that of the average rate) and 0.45 moles would form in 5 minutes (0.45 moles / 0.09 M/min).

If Keq for a given reaction is large in standard state conditions, which of the following is necessarily true about the thermodynamic parameters of the reaction? Question 15 Answer Choices A. ΔG is negative B. ΔG˚ is negative C. ΔH is negative D. ΔH˚ is negative

B. If Keq for a given reaction is large in standard state conditions, we know ΔGo is negative. This can be determined from the relationship ΔG° = -RTlnKeq. The value of ΔH° may be positive or negative, depending on the sign of the standard state entropy change. The non-standard state parameters can be ruled out, as we have no information which could allow us to predict their sign.

If the solubility of AgCl in water is 1.3 × 10-4 mol/L, calculate the solubility product of AgCl. Question 16 Answer Choices A. 6.5 × 10-5 B. 1.7 × 10-8 C. 1 × 10-2 D. 1.3 × 10-4

B. If the solubility of AgCl in water is 1.3 × 10-4 mol/L, the solubility product of AgCl is 1.7 × 10-8. The equilibrium dissociation of AgCl is given by the equation AgCl(s) Ag+(aq) + Cl-(aq), so Ksp = [Ag+][Cl-]. If the concentration of dissolved AgCl is 1.3 × 10-4 M, then [Ag+] = [Cl-] = 1.3 × 10-4 M also. Therefore, Ksp = (1.3 × 10-4)2 = 1.7 × 10-8.

The addition of a catalyst to a chemical reaction will bring about: Question 1 Answer Choices A. an increase in activation energy and an increase in the rate of the reverse reaction. B. a decrease in activation energy and an increase in the rate of the forward reaction C. an increase in activation energy and a decrease in the rate of the forward reaction. D. a decrease in activation energy and a decrease in the rate of the reverse reaction.

B. The addition of a catalyst to a chemical reaction will bring about a decrease in activation energy and an increase in the rate of the forward reaction. A catalyst increases the rate of both the forward and reverse reactions, and it does this by lowering the activation energy.

The following reaction is allowed to reach equilibrium: Cl2O(g) + 2 OH-(aq) <--> 2 OCl-(aq) + H2O(l) What will be the effect of then doubling the hydroxide concentration? Question 22 Answer Choices A. The reaction quotient will be four times the equilibrium constant, favoring the reverse reaction. B. The reaction quotient will be one-fourth the equilibrium constant, favoring the forward reaction. Correct Answer (Blank) C. The reaction quotient will be one-half the equilibrium constant, favoring the forward reaction. D. The reaction quotient will be twice the equilibrium constant, favoring the forward reaction.

B. The following reaction is allowed to reach equilibrium: Cl2O(g) + 2 OH-(aq) <--> 2 OCl-(aq) + H2O(l) The effect of then doubling the hydroxide concentration will be that the reaction quotient will be one-fourth the equilibrium constant, favoring the forward reaction. First, eliminate "the reaction quotient will be four times the equilibrium constant, favoring the reverse reaction"; adding a reactant favors the forward reaction. Since the equilibrium constant, K, and the reaction quotient, Q, are inversely proportional to [OH-]2, doubling [OH-] will cause K to decrease by a factor of 22 = 4. Thus, the reaction quotient, Q, will be one-fourth of K, favoring the forward reaction.

When a catalyst is added to a chemical reaction, what will be its effect on the energy of the activated complex and on the rate of the reaction? Question 16 Answer Choices A. The energy of the activated complex will increase, and the reaction rate will decrease. B. The energy of the activated complex will decrease, and the reaction rate will increase. C. The energy of the activated complex and the reaction rate will decrease. D. The energy of the activated complex and the reaction rate will increase.

B. When a catalyst is added to a chemical reaction, the energy of the activated complex will decrease, and the reaction rate will increase. A catalyst lowers the activation energy of a reaction (the energy of the activated complex), causing an increase in the reaction rate.

Which compound is a weak electrolyte? Question 2 Answer Choices A. CaCl2 B. H2CO3 C. PbSO4 D. CH3OH

B. All ionic compounds are strong electrolytes, regardless of how soluble they are. This means that both choices A and C can be eliminated. Both carbonic acid and methanol are molecular compounds, but since carbonic acid dissociates to a small extent in solution (it's a weak acid) while methanol doesn't, B is the best answer.

Which procedure below can be used to prepare a 0.75 M solution of KC2H3O2(aq) (MW = 98 g/mol)? Question 10 Answer Choices A. Dissolve 98.0 g of KC2H3O2 in 750 g of water. B. Dissolve 146 g of KC2H3O2 in enough water to make 2.0 L of solution. C. Dissolve 73.5 g of KC2H3O2 in enough water to make 750 mL of solution. D. Dissolve 73.5 g of KC2H3O2 in 1000 g of water.

B. Molarity is defined as the number of moles of solute per liter of solution, so a correct procedure must entail diluting with water to a specific volume of solution and not mixing with a specified amount of water (eliminate choices A and D). In choice B, the molarity can be calculated as follows, indicating it is the correct answer: while the calculation for choice C shows why it is incorrect:

Given the following experimental data, which of the following would allow for the determination of the rate law for the reaction in question? A. The reaction order for nitrogen trioxide B. The rate constant C. The activation energy for the reaction D. Additional trials with constant carbon monoxide levels

B. The isolation method can be used to determine the reaction order for reagents in question to help formulate a rate law. The data provided here can be used to determine the reaction order of nitrogen trioxide (choices A and D are wrong) but is insufficient to ascertain the reaction order of carbon monoxide. If we were provided with the rate constant, the reaction order for carbon monoxide could be calculated and a complete rate law could be determined (choice B is correct). While the activation energy is a component of the rate constant and provides valuable information about the reaction, it would not aid us in determining the rate law (choice C is wrong).

Some high-Pb commercial bronzes consist of 40% Cu, 34% Zn and 26% Pb (wt. %). In such a material, what is the mole fraction of Pb? Question 33 Answer Choices A. 0.05 B. 0.10 C. 0.25 D. 0.37

B. The mole fraction of Pb in bronze alloy consisting of 40 wt.% Cu, 34 wt.% Zn, and 26 wt.% Pb is 0.10. Assuming 100 g of bronze, 40 g will be Cu (63.5 g/mol), 34 g will be Zn (65 g/mol) and 26 g will be Pb (209 g/mol).40 g / (63.5 g/mol) = ~0.6 mol Cu34 g / (65 g/mol) = ~0.5 mol Zn26 g / (209 g/mol) = ~0.12 mol PbTherefore, there are ~0.12 mol of Pb in a total of ~1.2 mol of bronze. Since 0.12 / 1.2 = 0.1, answer choice B is the best.

A chemist adds two moles of nitrogen to the following mixture: 3 moles CO2, 2 moles O2, and 6 moles N2. She seals the cylinder containing the gaseous mixture and reads a final pressure of 1250 kPa. What is the final partial pressure of carbon dioxide? Question 16 Answer Choices A. 175 kPa B. 288 kPa C. 340 kPa D. 576 kPa

B. The partial pressure of a gas can be determined using Dalton's Law which states that total pressure of a mixture of gases is equal to the sum of the partial pressures of each constituent. Each partial pressure can then be calculated by multiplying the total pressure by the mole fraction of the gas in question. The final gaseous mixture contains a total of 13 moles of gas, resulting in a mole fraction of 3/13 = 0.23 for carbon dioxide. The partial pressure of carbon dioxide can then be determined by multiplying (0.23)(1250 kPa) = 288 kPa, making choice B correct.

A graduate student fills a piston with helium but inadvertently adds too much of the gas. To determine the quantity of helium that entered the cylinder, he carefully adjusts the pressure to 1 atm and temperature to 0°C and reads the volume to be 2.2 L. How much helium is present in the piston? (Note: R = 0.082 L?atm/mol?K) Question 28 Answer Choices A. 0.01 mol He B. 0.1 mol He C. 0.2 mol He D. 2.2 mol He

B. The quantity of gas can be calculated using the ideal gas law: n = PV/RT = (1 atm)(2.2 L) / (0.082 L?atm/mol?K)(273 K) = 0.1 mole of helium. Alternatively, and since the value of the gas constant wasn't given, the volume of one mole of gas at STP is 22.4 L. Here we observe 2.2 L of gas at STP which would correspond with ~0.1 moles of the gas, which makes choice B correct.

Given the reaction: 2 SO3 + Cl2 → 2 SOCl2 + O2, which of the following represents the best rate expression for the reaction? Question 6 Answer Choices A. rate = k [SO3]2[Cl2] B. The rate expression cannot be determined without more information C. rate = k [SO3][Cl2] D. rate = k [SO3][Cl2]2

B. The rate expression cannot be determined without more information

A chemist heats a 0.18 g sample of water initially at -25°C to 45°C. How much energy was required for this process? [Note: ΔHfus = 6.0 kJ/mol; cice = 38.0 J/mol?K; cliquid water = 74.5 J/mol?K] Question 17 Answer Choices A. 10 J B. 0.1 kJ C. 10 kJ D. 100 kJ

B. The total amount of energy required to perform this process is the sum of energy required to heat the sample to its melting point, to melt the sample, and then to heat it to 45°C. Using q = mcΔT, the amount of energy required to heat the sample to its melting point is (0.01 mol)(38.0 J/mol?K)(25 K) = 9.5 J. Melting the ice requires only 60 J and heating it to 45°C requires (0.01 mol)(74.5 J/mol?K)(45 K) = 33.5 J. Thus the total amount of energy required is 9.5 J + 60 J + 33.5 J ≈ 100 J (choice B is correct). Note that the answer choices are very widely spaced, requiring only an approximation of the correct answer.

The relationship of temperature to the rate constant of a reaction is: Question 1 Answer Choices A. dependent upon the order of the reaction. B. positively correlated, but non-linear. C. inversely proportional. D. linearly proportional.

B. positively correlated, but non-linear.

Based on the information above, what is the rate law of the following reaction? A + 2 B C + 2 D Question 6 Answer Choices A. Rate = k[A][B] B. Rate = k[A]2 C. Rate = k[A]2[B] D. Rate = k[A]2[B]2

C. Based on the information above, the rate law of the following reaction is Rate = k[A]2[B]. A + 2 B C + 2 D Using experiments 1 and 2, where [B] is held constant the concentration of A is tripled and the rate increases by a factor of 9, which makes the order for [A] second order. Using experiments 2 and 3, the concentration of A is held constant and the concentration of B is increased by 5 fold. This causes an increase in rate of 5 fold as well, so the reaction is 1st order in B. This makes the overall rate equation rate = k[A]2[B].

A reaction between two species is experimentally observed to be second order overall. If the concentration of one of the species doubles, what happens to the reaction rate? Question 3 Answer Choices A. The rate doubles B. The rate is halved C. Cannot be determined D. The rate quadruples

C. A reaction between two species is experimentally observed to be second order overall. If the concentration of one of the species doubles, the reaction rate cannot be determined. There are three possibilities for a two species reaction to produce an overall second order rate law: (1) rate = k[A][B], (2) rate = k[A]2, (3) rate = k[B]2. Let's assume that A is the species doubled. In the first case, doubling A will cause the rate to double. In the second case, doubling A will quadruple the rate. Finally, in the third case, doubling A will cause no change in the rate. Therefore, more information is needed to answer the question properly.

If carbon dioxide gas is in equilibrium with an aqueous solution of carbonic acid, then the CO2 solubility can be increased by: Question 8 Answer Choices A. decreasing the pH. B. increasing the temperature. C. increasing the pressure. D. adding carbonic anhydrase, an enzyme which increases the rate of the dissolution reaction of CO2 in water.

C. If carbon dioxide gas is in equilibrium with an aqueous solution of carbonic acid, then the CO2 solubility can be increased by increasing the pressure. The solubility of a gas in a liquid increases with increasing pressure. Solubility of a gas decreases with increasing temperature. Catalysts have no effect on the thermodynamics of processes. They never change an equilibrium, but they can help a system come to equilibrium quicker by increasing the rate of transformations. Decreasing the pH would result in a more acidic solution, which would drive the equilibrium towards CO2.

In the diagram, 1, 2, 3, and 4 represent, respectively: Question 17 Answer Choices A. Ea (forward), the activated complex, ΔH, Ea (reverse) B. the activated complex, ΔH, Ea (forward), Ea (reverse) C. the activated complex, Ea (forward), ΔH, Ea (reverse) D. Ea (reverse), the activated complex, ΔH, Ea (forward)

C. In the diagram, 1, 2, 3, and 4 represent, respectively the activated complex, Ea (forward), ΔH, Ea (reverse). Point 1 is the energy of the activated complex. Line 3 is the difference in energy between the products and reactants; this is ΔH. Thus, the answer must be "the activated complex, Ea (forward), ΔH, Ea (reverse)".

Which of the following answer choices accurately describes the thermodynamics of the following reaction? (Keq = 0.5 at 25°C) Question 4 Answer Choices A. Reactants are favored at equilibrium and ΔG° will be less than zero. B. Products are favored at equilibrium and ΔG° will be less than zero. C. Reactants are favored at equilibrium and ΔG° will be greater than zero. Correct Answer (Blank) D. Products are favored at equilibrium and ΔG° will be greater than zero.

C. Reactants are favored at equilibrium and ΔG° will be greater than zero accurately describes the thermodynamics of the following reaction. (Keq = 0.5 at 25°C) This is a two by two question, so approach each part of the answer choices separately. Given Keq < 1, reactants are favored at equilibrium, eliminating the choices written "products are favored..." If Keq < 1, then ΔG° will be greater than zero (non-spontaneous in the forward direction) according to the equation: ΔGº = -RTlnKeq.

Shvo's Catalyst is an attractive promoter of gas-phase hydrogenation and dehydrogenation in the cyclohexene/cyclohexane system (see the equilibrium below): If a microporous cylinder designed to allow slow effusion is charged with the three species, after an initial rapid equilibration, what changes in the relative amounts of the species, if any, will be observed? Question 13 Answer Choices A. The relative concentration of cyclohexane will increase and the relative concentration of cyclohexene will decrease. B. Shvo's catalyst will enhance the forward rate, pushing the equilibrium to the right. C. The relative concentration of cyclohexene will increase and the relative concentration of cyclohexane will decrease. D. The equilibrated mixture shows no changes through time.

C. Shvo's Catalyst is an attractive promoter of gas-phase hydrogenation and dehydrogenation in the cyclohexene/cyclohexane system (see the equilibrium below): If a microporous cylinder designed to allow slow effusion is charged with the three species, after an initial rapid equilibration, it will be observed that the relative concentration of cyclohexene will increase and the relative concentration of cyclohexane will decrease. Effusion of the three gaseous species from the cylinder does not occur at the same rate. The rate of effusion is governed by average molecular velocity, and hence molecular weight. Since H2 is by far the lightest, if will effuse out of the cylinder much faster than the other species, pushing the equilibrium to the left, and increasing the relative concentration of cyclohexene. Catalysts do not affect equilibrium, so that answer can be safely eliminated.

The formation constant (Kf) for the complex ion [Fe(en)3]2+ is 5.0 × 109, whereas the Kf for [Fe(ox)3]4- is 1.7 × 105 (en = 1,2 diaminoethane, ox = oxalate anion). Based on this information, which of the following statements is true? Question 1 Answer Choices A. The higher the oxidation state on the transition metal, the larger the formation constant. B. Complexes with bidentate ligands always have higher Kf values than the corresponding monodentate complexes. C. 1,2-Diaminoethane is a stronger ligand with iron(II) than the oxalate ion. D. A reaction between oxalate ion and [Fe(en)3]2+ will form [Fe(ox)3]4- completely.

C. The formation constant (Kf) for the complex ion [Fe(en)3]2+ is 5.0 × 109, whereas the Kf for [Fe(ox)3]4- is 1.7 × 105 (en = 1,2 diaminoethane, ox = oxalate anion). Based on this information, 1,2-diaminoethane is a stronger ligand with iron(II) than the oxalate ion. Formation constants are equilibrium constants for the formation of complex ions. The larger the equilibrium constant of a reaction, the larger the ratio of products to reactants at equilibrium will be. The oxalate complex with iron(II) has a lower Kf than the diamine complex. Therefore, any competitive reaction will favor the diamine complex, so [Fe(en)3]2+ is favored to form, not [Fe(ox)3]4-. Since both complexes have iron(II) as the central metal, there isn't any information to support the relationship suggested between oxidation state and the value of the formation constant. When comparing competing ligands, the one with the larger Kf will yield the more stable complex, which is due to the strongest Lewis acid/Lewis base interactions. Therefore, the diamine must be the stronger ligand. Both of the ligands in the question are bidentate, so there is no information to use for comparison between ligand type and value of Kf.

The following experimental data were collected for the chemical reaction A + B C: Which of the following gives the rate law? Question 14 Answer Choices A. Rate = k[B]2 B. Rate = k[A] C. Rate = k[A][B] D. Rate = k[B]

C. Using the information below, we know that Rate = k[A][B] gives the rate law. The following experimental data were collected for the chemical reaction A + B C: To obtain the rate law, we must compare trials and determine how changes in the initial concentrations of the reactants affect the initial rate of the reaction. Comparing Trials 1 and 2, we see that [A] increased by a factor of 2 and so did the reaction rate; thus, the reaction is first order with respect to A. Eliminate Rate = k[B] and Rate = [B]2. Comparing Trials 2 and 3, we see that [B] increased by a factor of 2 and so did the reaction rate; thus, the reaction is also first order with respect to B. Therefore, the rate law for this reaction is given by Rate = k[A][B].

Which of the following would have no effect on an equilibrium in which one mole of gas is decomposed into two moles of gas? Question 5 Answer Choices A. Reduction of the volume of the container B. Addition of either one of the products C. Addition of an inert gas at constant volume D. Addition of the reactant

C. Addition of an inert gas at constant volume

Why will increasing the temperature of a reaction increase reaction rate? It increases collision frequency. It decreases activation energy. It increases the probability of reaction upon collision. Question 24 Answer Choices A. I only B. II only C. I and III only D. I, II, and III

C. Increasing temperature increases reaction rate by increasing the average kinetic energy of molecules in solution, which increases their probability of collision (greater velocity results in more chances for collision) as well as provides them with additional energy to help overcome the activation energy necessary to achieve the transition state (statements I and III are true; choices A and B are wrong). While decreasing activation energy will increase reaction rate, an increase in temperature is unlikely to have a favorable impact on the stability of the transition state (statement II is not true, choice C is correct).

Given the average rate for an unknown reaction is described by -Δ[H2]/Δt, -Δ[I2]/Δt, and (1/2) Δ[HI]/Δt, what is the most likely reaction? Question 18 Answer Choices A. HI(g) → 1/2 H2(g) + 1/2 I2(g) B. H2(g) + I2(g) → HI(g) C. H2(g) + I2(g) → 2 HI(g) D. H2(g) + I2(g) → 1/2 HI(g)

C. Reaction rate can be characterized by changes in reactants or products over time. In the generic reaction aA → bB, the reaction rate = -(1/a) Δ[A]/Δt = (1/b) Δ[B]/Δt. Thus we can reconstruct our reaction from the above rates and find the reaction rate to be H2(g) + I2(g) → 2 HI(g) (choice C is correct). Note that simply utilizing process of elimination for unbalanced reactions would eliminate answer choices B and D.

Why is temperature constant during any phase change? Question 30 Answer Choices A. All phases have the same energy, so no temperature change is observed during the phase transitions. B. Energy is used to break or form bonds rather than change temperature during a phase change. C. Energy is used to break or form intermolecular forces rather than change temperature during a phase change. D. Added or removed energy either decreases or increases the entropy of the system, respectively, thereby making temperature changes impossible.

C. Solids, liquids, and gases all have different potential energies (they are listed in order of increasing energy) so choice A can be eliminated. While a phase change does affect the entropy of the system, adding energy makes a more disordered system, thereby increasing entropy (eliminate choice D). A phase change is a physical, not chemical process, so no chemical bonds are broken or formed in the process (eliminate choice B). The energy added to a system causing a phase change is used to break the intermolecular forces holding molecules together rather than make molecules move faster (associated with a temperature change).

Catalysts increase reaction rate by stabilizing the transition state. What impact does this have on the reverse reaction? Question 1 Answer Choices A. No impact: a separate catalyst would be required for the reverse reaction B. No impact: the reverse reaction is catalytically resistant C. Increases reverse reaction rate to the same degree as the forward reaction D. Increases reverse reaction rate by changing the reverse reaction intermediates

C. Stabilizing the transition state increases the rate of the forward and reverse reactions to the same degree (choice C is correct). As catalysts have no impact on equilibrium, the forward and reverse reaction rates must remain proportional at equilibrium. Note that kinetics and thermodynamics are separate entities in chemistry, thus a change in a kinetic factor is unlikely to impact a thermodynamic factor. Addition of a catalyst will not alter the intermediates of a mechanism but will stabilize the transition state (choice D is wrong). The forward and reverse reactions are intimately linked and one cannot be independently catalyzed without the other (choices A and B are wrong).

catalyst is discovered that doubles the rate for the second step in the following mechanism. What impact does this have on the overall reaction? Step 1: 2 NO2 → NO + NO3 (slow) Step 2: NO3 + CO → NO2 + CO2 (fast) Question 15 Answer Choices A. Doubles the reaction rate B. Halves the reaction rate C. No change to the reaction rate D. Insufficient information to determine the change in rate

C. The catalyst functions on the second step in the mechanism, but this is the 'fast' step, meaning we cannot increase the overall reaction rate by accelerating this step alone (choice C is correct). To increase the overall reaction rate, we would need to increase the reaction rate of the first step (the rate limiting step). Note that answer choices where you are given the option of having insufficient information should only be selected if you are certain necessary information was omitted.

A block of ice is formed from water at 0°C, removed from the water and while frozen is trimmed to a volume of exactly 10 cm3. The weight of this block is found to be 9.2 g. The block is then carefully submerged into a sample of heavy-isotope enriched water (D218O) equilibrated at 0°C. After 20 min the block was removed, measured and weighed. Which of the following is the density of the ice block after the experiment? Question 27 Answer Choices A. 0.85 g/mL B. 0.92 g/mL C. 0.96 g/mL D. 1.21 g/mL

C. The density of the ice after the experiment was 0.96 g/mL. The initial density of the block was 9.2/10 = 0.92 g/mL (recall 1 cm3 = 1 mL). At 0°C there is an equilibrium between the solid and liquid phases. Whereas the size of the block may not change much, this equilibrium at the surface involves the loss of light water (H216O), and the incorporation of heavy water. As such we could expect a slight increase in the density of the new ice (eliminating choices A and B). Choice D suggests a mass increase of ~30% in the block, which would not be possible even if all the light water was exchanged for heavy water.

I2(g) + Br2(g) <--> 2 IBr(g) What can be concluded about the reaction above if its equilibrium constant is measured to be 1.2 x 102? Question 6 Answer Choices A. Twice as many products as reactants are present at equilibrium. B. The ΔG° for this reaction is positive. C. Products are more stable than reactants. D. The reaction rapidly establishes equilibrium.

C. The equilibrium constant is a ratio of products to reactants, so when K > 1, there are more products than reactants at equilibrium. At equilibrium, the side of a reaction with the greatest stability will be favored (choice C is correct). While products are heavily favored, it is to a much larger degree than twice as many products as reactants present at equilibrium (choice A is incorrect). Given the equilibrium favors the products, the ΔG° will be negative, not positive (choice B is incorrect). Since kinetics and thermodynamics are very distinct topics in chemistry, the equilibrium constant tells us nothing about the rate of a reaction and how quickly it reaches equilibrium (choice D is incorrect).

Which of the following balanced reactions most likely possesses an equilibrium constant where Keq = [HI]2/[H2]? Question 29 Answer Choices A. H2(g) + 2 I3-(aq) ? 2 I2(s) + 2 HI(aq) B. H2(g) + 2 I-(aq) ? 2 HI(g) C. H2(g) + I2(s) ? 2 HI(g) D. 1/2 H2(g) + 1/2 I2(s) ? HI(g)

C. The equilibrium constant is dictated by the concentration of products over reactants, raised to the power of their coefficients. Both pure solids and liquids are omitted, but gases and aqueous substances are included. Thus for the equilibrium expression [HI]2/[H2], we would expect HI as a gaseous or aqueous product with a coefficient of 2 and H2 as a reactant with a coefficient of 1 (choice D is incorrect). Charge must be balanced in a chemical reaction, and any aqueous ions should appear in the equilibrium expression if appropriately part of the balanced reaction (choices A and B are incorrect). Thus the equilibrium expression Keq = [HI]2/[H2] best describes the balanced reaction H2(g) + I2(s) ? 2 HI(g) (choice C is correct).

Two balloons are filled with equal quantities of either helium or nitrogen and a microscopic hole is placed on the surface of each balloon. Which of the following best describes the balloons after several minutes? Question 22 Answer Choices A. The nitrogen-filled balloon will deflate more quickly because nitrogen possesses a greater amount of kinetic energy. B. The nitrogen-filled balloon will deflate more quickly as it will effuse at a greater rate than helium. C. The helium-filled balloon will deflate more quickly given the smaller molar mass of helium gas. D. The helium-filled balloon will deflate more quickly due to helium's lesser density than the surrounding air.

C. The escape of molecules through a hole, or effusion, is governed by Graham's law. Using this law, we can predict that the gas with the smaller molecular weight will have a greater rate of effusion (choices A and B can be eliminated) and will deflate more quickly (choice C is correct). While the relative density of helium is important in making a balloon float, it will not play a role in its effusion (choice D is incorrect).

A saturated, aqueous solution of KBr is heated at its boiling point for five minutes, at which point a white solid appears in the flask. Which of the following explains these observations? Question 26 Answer Choices A. KBr is less soluble at high temperatures and precipitates out of the solution upon boiling. B. KBr is unstable at the boiling point of water and decomposes into an insoluble product. C. KBr precipitates as the solution becomes more concentrated due to the evaporation of water. D. KBr can no longer be solvated as the average kinetic energy of the water molecules increases.

C. The solubility of most solids increases with temperature, eliminating choice A. Ionic compounds are generally quite stable in solution as well, ruling out choice B. The average kinetic energy of boiling water does not increase since the temperature of boiling water will not change, making D a false statement. A saturated solution has reached the maximum amount of solute dissolved in the given amount of solvent. Boiling the solution and reducing the amount of solvent present forces some solute to precipitate, keeping a saturated solution.

When chemical equilibrium is reached, which of the following relationships is always true? Question 2 Answer Choices A. ∆H = 0 B. ∆Ho = 0 C. ∆G = 0 D. ∆Go = 0

C. ∆G = 0

A scientist adds reactants and an enzyme to a beaker placed on the benchtop. As the reaction proceeds, which of the following events would have the smallest effect on the rate of the forward reaction? Question 2 Answer Choices A. More reactants are added to the beaker. B. The temperature of the beaker is changed. C. The solution is stirred. D. Products are removed from the beaker

D. A scientist adds reactants and an enzyme to a beaker placed on the benchtop. Removing products from the beaker would have the smallest effect on the rate of the forward reaction. Temperature, reactant concentrations, and activation energy are the three major factors that play a role in a reaction's kinetics. If the temperature of the beaker changes, the rate of the forward reaction will be affected. When more reactants are added or the mixture stirred, the probability of collisions among the molecules increases. Thus, the forward reaction rate will increase. While removing products will shift the equation to the right (favoring the production of more products) by Le Chatelier's Principle, this is an effect on the equilibrium, or thermodynamics, of the reaction and does not affect the rate of the forward reaction.

As the temperature at which a reaction takes place is increased: Question 12 Answer Choices A. the reaction rate will remain constant, but the rate constant will increase. B. the reaction rate and the rate constant will remain constant. C. the reaction rate will increase, but the rate constant will remain constant. D. the reaction rate and the rate constant will increase.

D. As the temperature at which a reaction takes place is increased the reaction rate and the rate constant will increase. An increase in the rate constant always corresponds to a faster reaction rate, so the statements "the reaction rate will increase, but the rate constant will remain constant" and "the reaction rate will remain constant, but the rate constant will increase" are eliminated. Increasing the temperature of a reaction always increases the reaction rate, so the statement "the reaction rate and the rate constant will increase" must be correct.

Which of the following is necessarily true for any reaction at equilibrium? Question 14 Answer Choices A. TΔS < 0 B. TΔS > 0 C. ΔG˚= 0 D. ΔH = TΔS

D. For a reaction at equilibrium it is necessarily true that ΔH = TΔS. This result stems from the fact that at equilibrium ΔG = 0, and since ΔG = ΔH - TΔS, ΔH = TΔS. It is not true that ΔG° must equal 0, as this is only a condition where there is no difference in free energy between the reacting species. Likewise, TΔS can be either positive or negative, depending on the sign of ΔH.

N2(g) + 3 H2(g) <--> 2 NH3(g), ΔH = -22 kcal/mol Increasing the temperature of the reaction at equilibrium will most likely: Question 23 Answer Choices A. increase the forward reaction. B. increase the heat of reaction. C. decrease the heat of reaction. D. decrease the forward reaction.

D. Increasing the temperature of the reaction at equilibrium will most likely decrease the forward reaction. Since the heat of reaction (ΔH) is not a function of temperature, it will not increase or decrease the heat of reaction. Since the reaction is exothermic, Le Châtelier's principle implies that increasing the temperature will decrease the forward reaction.

The equilibrium constant for the reaction is 617 at a given temperature. In a reaction vessel, the partial pressures of H2 and I2 are each 1 atm and the partial pressure of HI is 12 atm. What is the value of the reaction quotient, Q? Question 6 Answer Choices A. 7.0 × 10-3 B. 8.3 ×10-2 C. 12 D. 144

D. The equilibrium constant for the reaction is 617 at a given temperature. In a reaction vessel, the partial pressures of H2 and I2 are each 1 atm and the partial pressure of HI is 12 atm. The value of the reaction quotient, Q is 144. Although calculated in the same manner, Q is independent of Keq. Q can be solved using partial pressures:

The following equilibrium exists in a closed container: N2(g) + O2(g) <--> 2 NO(g) ΔH = +181 kJ/mol Which of the following perturbations would favor the formation of NO(g)? Question 21 Answer Choices A. Increasing the pressure B. Decreasing the pressure C. Decreasing the temperature D. Increasing the temperature

D. The following equilibrium exists in a closed container: N2(g) + O2(g) <--> 2 NO(g) ΔH = +181 kJ/mol Increasing the temperature would favor the formation of NO(g). Since both sides of the reaction have two gas molecules, changes in pressure will have no effect on the equilibrium; this eliminates "increasing the pressure" and "decreasing the pressure". According to Le Châtelier's principle, endothermic reactions are shifted to the right at higher temperatures.

The molar solubility of HgCl2 in water is 0.27 M. If 0.1 mol of NaCl is added to 1 L of a 0.27 M aqueous solution of HgCl2, what will happen? Question 7 Answer Choices A. NaCl will not dissolve and no precipitate will form. B. NaCl will dissolve and no precipitate will form. C. NaCl will not dissolve and HgCl2 will precipitate. D. NaCl will dissolve and HgCl2 will precipitate.

D. The molar solubility of HgCl2 in water is 0.27 M. If 0.1 mol of NaCl is added to 1 L of a 0.27 M aqueous solution of HgCl2, NaCl will dissolve and HgCl2 will precipitate. The solution of HgCl2 is saturated. NaCl is very soluble in water (solubility is over 6 M at 25°C) and will be able to dissolve completely. Because of the common ion effect, as NaCl dissolves, it will decrease the solubility of HgCl2 and cause it to precipitate.

Which of the following mechanisms best describes the overall reaction CH4 + Cl2 → CH3Cl + HCl with an overall rate expression of rate = k [CH4][Cl2]? Question 10 Answer Choices A. CH4 + Cl2 → •CH3 + •Cl + HCl (fast) •CH3 + •Cl → CH3Cl (slow) B. CH4 + 2 Cl2 → •CH3 + HCl + 3Cl• (slow)CH3 + Cl• → CH3Cl (fast) C. CH4 + 2 Cl2 → •CH3 + HCl + 3Cl• (fast)CH3 + Cl• → CH3Cl (slow) D. CH4 + Cl2 → •CH3 + •Cl + HCl (slow) •CH3 + •Cl → CH3Cl (fast)

D. The overall reaction CH4 + Cl2 → CH3Cl + HCl with an overall rate expression of rate = k [CH4][Cl2] is best described by the mechanism: CH4 + Cl2 → •CH3 + •Cl + HCl (slow) •CH3 + •Cl → CH3Cl (fast) A mechanism describes the elementary steps that compose an overall reaction. The steps in a mechanism must add to give the overall reaction and the overall rate expression should be that of the slow step. The only answer choice which satisfies both these requirements is CH4 + Cl2 → •CH3 + •Cl + HCl (slow) •CH3 + •Cl → CH3Cl (fast)

The rate-determining step of a chemical reaction is the step: Question 13 Answer Choices A. that is the fastest. B. involving the most molecular mass. C. involving the least molecular mass. D. that is the slowest.

D. The rate-determining step of a chemical reaction is the step that is the slowest. Molecular mass is unrelated to the rate of a reaction, so eliminate the statements "involving the most molecular mass" and "involving the least molecular mass".

The equilibrium between methylamine bound, and unbound, to iron pentacarbonyl is found to have a constant Keq of ~30 and an activation energy of 13 kcal/mol. NH2CH3 + Fe(CO)5 <--> (CH3NH2)Fe(CO)5 If the amine is switched to the very bulky diisopropylethylamine, the activation energy of binding is found to increase to 39 kcal/mol. With this data, approximately what is the new value of Keq? Question 9 Answer Choices A. Keq will increase by a factor of 3 B. Keq will decrease by a factor of log 3 C. Keq will decrease by a factor of 3 D. There is no information in the question to determine the new Keq.

D. With the material given in the question, there is no information in the question to determine the new Keq for the reaction with diisopropylethylamine. The question gives information on the new activation energy, which is strictly a kinetic measurement. It says nothing about the thermodynamics of the system, so Keq with diisopropylethylamine is unknown.

CH4(g) + 2 S2(g) <--> CS2(g) + 2 H2S(g) Once the system shown reaches equilibrium, increasing [CH4] will result in each of the following EXCEPT: Question 20 Answer Choices A. a decrease in the reaction quotient. B. an increase in the total quantity of H2S(g) formed at equilibrium. C. a decrease in the total quantity of S2(g) formed at equilibrium. D. an increase in the Keq for the reaction.

D. An increase in reactant concentration will decrease the reaction quotient (choice A is a true statement, so can be eliminated). With an increase in reactant concentration, an equilibrium shift to the products occurs, resulting in increased total quantities of each product being formed (choice B is true, so can be eliminated). When the equilibrium shifts to the products side, it must use up reactants, so the amount of S2 should decrease (choice C is true, so can be eliminated). Changes in reactant or product concentration do not result in a change in Keq as this value is only determined by the temperature of the system. Since temperature remains constant, the value of K does not change, making choice D the correct answer.

Once equilibrium has been established, each of the following are true EXCEPT: Question 3 Answer Choices A. the system has reached an energy minimum. B. no net flux of reactant to product occurs. C. the rates of the forward and reverse reactions are equal. D. equal quantities of products and reactants are present.

D. At equilibrium, the quantities of reactant and product are not necessarily equal, however the proportions will be consistent (choice D is not true of a system at equilibrium and is the answer). By definition, a reaction at equilibrium will have equal rates of forward and reverse reactions, meaning no net flux of reactants or products (choices B and C are true of a system at equilibrium and can be eliminated). At equilibrium, the system has established an energy minimum, which accounts for the spontaneous progression of the reaction toward equilibrium (choice A is true of a system at equilibrium and can be eliminated).

The table above lists the formation constants for some common complex ions with the oxalate ion (ox, C2O42-). Which metal ion would react most completely with 100 mL of a 1.0 M solution of potassium oxalate to form the corresponding complex ion? Question 7 Answer Choices A. Cobalt(II) B. Iron(II) C. Cobalt(III) D. Iron(III)

D. Formation constants, Kf, are just another form of an equilibrium constant. Therefore the complex with the largest formation constant will have the most complete reaction and result in the highest concentration. The last complex in the table, [Fe(ox)3]3-, has the largest Kf, so the metal must be iron (eliminate choices A and C). Since the charge on oxalate is -2, three ligands will contribute -6, and require the metal to have a +3 charge (eliminate choice B).

A researcher finds a significant deviation from her experimental results when compared to her predictions using the ideal gas law. Which of the following would most likely decrease this discrepancy? Question 13 Answer Choices A. Increasing the pressure and temperature of the gas B. Increasing the pressure and decreasing the temperature of the gas C. Decreasing the pressure and temperature of the gas D. Decreasing the pressure and increasing the temperature of the gas

D. Real gases deviate from the assumptions made for ideal gases to differing degrees based on temperature and pressure. Real gases behave more 'ideally' when at high temperatures and low pressures (choice D is correct). This minimizes intermolecular interactions (e.g. condensation) and would decrease the error observed by the researcher.

The following table shows the values of the van der Waals constants for a select group of gases: Based on these data, which of the following shows the gases listed in order by increasing condensation temperature at constant pressure? Question 4 Answer Choices A. Ne < He < Ar < O2 B. O2 < Ar < He < Ne C. O2 < Ar < Ne < He D. He < Ne < Ar < O2

D. The condensation temperature of a gas at constant pressure will depend upon the attractive forces between the particles. The stronger the intermolecular forces of the gas, the higher the condensation temperature. In other words, the gas doesn't need to be cooled much in order to condense if the particles are strongly attracted to each other. The strength of the intermolecular force of a gas is directly proportional to the magnitude of the "a" constant in the van der Waals equation. The smallest value in the table is for He, so it must be first in the sequence. Eliminate choices A, B, and C, and choose choice D. Note that the sequence for choices A and B are related to the values of the "b" constant, which depicts molecular volume, not intermolecular forces.

Which of the following elementary steps is best described by the following rate law: rate = k [NO][ClNO2]? Question 25 Answer Choices A. 2 ClNO(l) + NO(g) <--> NO2(g) + 2 ClNO(g) B. 2 ClNO(g) + NO(g) <--> NO2(g) + 2 ClNO(g) C. ClNO2(l) + NO(g) <--> NO2(g) + ClNO(g) D. ClNO2(g) + NO(g) <--> NO2(g) + ClNO(g)

D. The rate law for each elementary step can be derived by taking each non-liquid/solid reactant and raising it to the power of its coefficient. The rate laws for the answer choice A-D are k [NO], k [NO][ClNO]2, k [NO], and k [NO][ClNO2], respectively (choice D is correct).

A chemist expands four liters of gas initially at STP under constant temperature conditions. Given the final pressure is 200 mm Hg, what is the final volume? Question 23 Answer Choices A. 1.1 L B. 4.2 L C. 8.0 L D. 15.2 L

D. This question requires you to be familiar with standard temperature and pressure (273 K, 760 mm Hg) as well as the ideal gas law. Given that the number of moles of the gas, temperature, and the ideal gas constant remain unchanged, the ideal gas law can be rearranged to P1V1 = P2V2. V2 = P1V1 / P2 = (760 mm Hg)(4 L) / 200 mm Hg = 15.2 L, making choice D correct.

In a chemical equilibrium: Question 3 Answer Choices A. the forward and reverse reaction have stopped each other. B. the rate constants of the forward and reverse reactions are equal. C. the products and reactants are static. D. the concentrations of the products and reactions are static.

D. the concentrations of the products and reactions are static.

The following data have been recorded for the reaction A + B C + D. What is the most likely rate expression? Question 7 Answer Choices A. Rate = k[A][B]2 B. Rate = k[A]2[B] C. Rate = k[A]2[B]2 D. Rate = k[A][B]

The most likely rate expression is Rate = k[A][B]2. It can be seen from the table that if the concentration of A is increased 5 fold (entries 1 and 2), there is a corresponding 5-fold increase in the observed rate; i.e., the rate expression must be first order in A. Furthermore, with a 4-fold increase in B (entries 2 and 3), the rate becomes 16 = 42 times greater. Therefore, the rate expression must be second order in B.

Find the reaction rate law for the reaction A + B C: Question 11 Answer Choices A. Rate = k[A]2[B]2 B. Rate = k[A]2[B] C. Rate = k[A][B]2 D. Rate = k[A][B]

The reaction rate law for the reaction A + B C is Rate = k[A][B]2. To obtain the rate law, we must compare trials and determine how changes in the initial concentrations of the reactants affect the initial rate of the reaction. Comparing Trials 1 and 3, we see that [A] increased by a factor of 3 and so did the reaction rate; thus, the reaction is first order with respect to A. Eliminate Rate = k[A]2[B] and Rate = k[A]2[B]2. Comparing Trials 1 and 2, we see that [B] increased by a factor of 4 and the reaction rate increased by a factor of 16 = 42. Thus, the reaction is second order with respect to B. Therefore, the rate law for this reaction is given by Rate = k[A][B]2.