$$CH_3I+NaOH\longrightarrow CH_3OH+NaI$$ The rate of the reaction represented by the chemical equation shown above is expressed as Rate=k[CH 3 I][NaOH]. Based on this information, which of the following claims is correct?

The rate of the reaction will double if the concentration of CH 3 I is doubled while keeping the concentration of NaOH constant.

A larger amount of CH 3 OH will be produced if the concentrations of CH 3 I and NaOH are halved.

The mechanism for a chemical reaction is shown above. Which of the following statements about the overall reaction and rate laws of the elementary reactions is correct?

The chemical equation for the overall reaction is 2 H 2 O 2 →2 H 2 O+O 2 , and the rate law for elementary step 1 is Rate=k[H 2 O 2 ][I − ].

The chemical equation for the overall reaction is 2 H 2 O 2 +I − → 2 H 2 O+O 2 +I − , and the rate law for elementary step 2 is Rate=k[H 2 O 2 ][IO − ].

The chemical equation for the overall reaction is H 2 O 2 +IO − → H 2 O+O 2 +I − , and the rate law for elementary step 2 is Rate=k[H 2 O 2 ]2[IO − ].

The chemical equation for the overall reaction is 2 H 2 O 2 →2 H 2 O+O 2 , and the rate law for elementary step 1 is Rate=k[H 2 O 2 ]2.

A proposed reaction mechanism for the decomposition of N 2 O 5 (g) is shown above. Based on the proposed mechanism, which of the following correctly identifies both the chemical equation and the rate law for the overall reaction?

The chemical equation for the overall reaction is 2N 2 O 5 (g)→4NO 2 (g)+O 2 (g), and the rate law is Rate=k[N 2 O 5 ].

The chemical equation for the overall reaction is NO(g)+N 2 O 5 (g)→3NO 2 (g), and the rate law is Rate=k[NO][N 2 O 5 ].

The chemical equation for the overall reaction is 2N 2 O 5 (g)→4NO 2 (g)+O 2 (g), and the rate law is Rate=k[N 2 O 5 ] 2 .

The chemical equation for the overall reaction is N 2 O 5 (g)+NO(g)+NO 3 (g)→4NO 2 (g)+O 2 (g), and the rate law is Rate=k[N 2 O 5 ][NO][NO 3 ].

The overall reaction represented above is proposed to take place through two elementary steps. Which of the following statements about the chemical equation for step 1 and the rate law for the overall reaction is correct?

The chemical equation for step 1 is NO 2 (g)+F 2 (g)→NO 2 F(g)+F(g), and the rate law for the overall reaction is Rate=k[NO 2 ][F 2 ].

The chemical equation for step 1 is NO 2 (g)+F(g)→NO 2 F(g)+F 2 (g), and the rate law for the overall reaction is Rate=k[NO 2 ][F].

The chemical equation for step 1 is 2NO 2 (g)+F 2( g)→2NO 2 F(g), and the rate law for the overall reaction is Rate=k[NO 2 ][F].

The chemical equation for step 1 is 3NO 2 (g)+F 2 (g)+F(g)→3NO 2 F(g), and the rate law for the overall reaction is Rate=k[NO 2 ]2[F 2 ].

The equations shown above represent four elementary reactions. Which of the following identifies the reaction in which the number of successful collisions and reaction rate are independent of the orientation of the reactants and explains why?

Reaction A, because the electron clouds of the O atoms are distributed symmetrically.

Reaction B, because each C 2 H 4 molecule has a double bond.

Reaction C, because both CO and O 2 are linear molecules.

Reaction D, because the reactant Br − and the product I − are negatively charged.

The two diagrams above represent collisions that take place at the same temperature between a CO molecule and an NO 2 molecule. The products are CO 2 and NO. Which diagram most likely represents an effective collision, and why?

Diagram 1 represents an effective collision because the two molecules have the proper orientation to form a new C−O bond as long as they possess enough energy to overcome the activation energy barrier.

Diagram 1 represents an effective collision because the molecules have about the same size and the same energy, which leads to a larger rate constant (k).

Diagram 2 represents an effective collision because the molecules are aligned head-to-head and that maximizes the overlap between their atomic orbitals, forming more products.

Diagram 2 represents an effective collision because the oxygen atoms are the most electronegative and will minimize their repulsions when oriented away from each other.

The diagram above shows the distribution of molecular collision energies for equimolar samples of a reactant at different temperatures. Based on the diagram, at which temperature will the reactant be consumed at the fastest rate, and why?

At T 4, because a larger fraction of the molecules have an energy that is equal to or greater than the activation energy.

At T 1, because a larger fraction of the molecules have about the same energy.

At T 2 , because at this temperature most of the molecules undergo collisions frequently.

At T 3 , because at this temperature the rate of consumption for about half of the molecules is determined by their orientation.

Which of the following best describes the elementary step(s) in the reaction mechanism represented in the diagram above?

Two steps: Step 1: 2X(g)→X 2 (g) Step 2: X 2 (g)+Y(g)→X 2 Y(g)

One step: 2X(g)+Y(g)→X 2 Y(g)

One step: X 2 (g)+Y(g)→X 2 Y(g)

Two steps: Step 1: X(g)+Y(g)→XY(g) Step 2: X(g)+XY(g→X 2 Y(g)

A proposed mechanism for the reaction H 2 +2IBr→I 2 +2HBr is shown above. Two experiments were performed at the same temperature but with different initial concentrations. Based on this information, which of the following statements is correct?

The rate of the reaction will undergo a 4-fold increase in the experiment in which the initial concentrations of both H 2 and IBr were doubled.

The rate of the reaction will undergo a 4-fold increase in the experiment in which the initial concentrations of both HI and IBr were doubled.

The rate of the reaction will undergo a 2-fold increase in the experiment in which the initial concentrations of both HI and IBr were doubled.

The rate of the reaction will undergo a 8-fold increase in the experiment in which the initial concentrations of both H 2 and IBr were doubled.

The elementary steps in a proposed mechanism for the decomposition of HCOOH are represented above. Which of the following identifies the catalyst in the overall reaction and correctly justifies the choice?

H 2 SO 4 , because it is consumed in the first step of the mechanism and regenerated in a later step.

HCOOH, because increasing the concentration of HCOOH increases the reaction rate.

HCOOH 2 + , because it is produced in the first step of the mechanism and consumed in a later step.

CO, because it is produced in the final step of the reaction mechanism.

The diagram above shows the reaction energy profiles for a reaction with and without a catalyst. Which of the following identifies the reaction energy profile for the catalyzed reaction, and why?

Profile Y, because it introduces a different reaction path that reduces the activation energy.

Profile X, because a catalyst minimizes the number of elementary steps required for a reaction to proceed.

Profile X, because the activation energy for the reverse reaction is greater than for the forward reaction, which increases its rate.

Profile Y, because an increase in the number of transition states increases the rate of the reaction.

The diagram above illustrates how the reaction N 2 (g) + 3H 2 (g) -> 2NH 3 (g) occurs on the surface of Ru(s) The rate of this reaction is determined by the amount of energy required to break the bond in N 2 This bond is weakened when N 2 is adsorbed on Ru(s) Based on this information, which of the following provides the best reason for the use of Ru(s) for the synthesis of NH 3 ?

It provides a reaction path with a lower activation energy.

It promotes the proper orientation of N and H atoms to form new N - H bonds.

It forms a bond between Ru and N that is stronger than the bond in N 2

It decreases the frequency of collisions between N 2 (g) and H 2 (g)

The following questions relate to the below information. $$XY_2\rightarrow\ X+Y_2$$ The equation above represents the decomposition of a compound XY 2 . The diagram below shows two reaction profiles (path one and path two) for the decomposition of XY 2 .

The [presence of a catalyst in path two

A higher temperature in path one

A higher temperature in path two

The presence of a catalyst in path one

$$C_{12}H_{22}O_{11}+H_2O\rightarrow\ 2C_6H_{12}O_6$$ The hydrolysis of sucrose is represented by the chemical equation above. This reaction is extremely slow in aqueous solution. however, when sucrase is added as shown in the diagram, the rate of the reaction is about 6,000,000 times faster. based on this information, which of the following best explains the large increase in the rate of hydrolysis that occurs with the addition of sucrase?

The reaction proceeds through a different reaction path with a lower activation energy in which the sucrase-sucrose complex is formed as an intermediate.

The addition of the sucrase reduces the number of effective collisions between the water and sucrose molecules.

The reaction proceeds through a different reaction that is independent of temperature and concentration.

The addition of the sucrase decreases the number of sucrose molecules that have the minimum energy required to overcome the activation energy barrier.

The role of a catalyst in a chemical reaction is to

lower the activation energy for the reaction

decrease the amount of reactants that must be used

supply the activation energy required for the reaction to proceed

increase the amounts of products formed at equilibrium

increase the entropy change for the reaction

$$X\rightarrow\ products$$ A student studied the kinetics of the reaction represented above by measuring the concentration of the reactant, X, over time. The data are plotted in the graph below. Which of the following procedures will allow the student to determine the rate constant, k, for the reaction?

Plot in ln[X] versus time and determine the magnitude of the slope.

Plot 1/[X] versus time and determine the magnitude of the slope.

Run another trial of the experiment with a different initial concentration, plot the data on the same graph, and see where the curves intersect.

Run another trial of the experiment at a different temperature, plot the data on the same graph, and see where the curves have the same slope.

A kinetics experiment is set up to collect the gas that is generated when a sample of chalk, consisting primarily of solid CaCO 3 , is added to a solution of ethanoic acid, CH 3 COOH. The rate of reaction between CaCO 3 and CH 3 COOH is determined by measuring the volume of gas generated at 22°C and 1 atm as a function of time. Which of the following experimental conditions is most likely to increase the rate of gas production?

Decreasing the particle size of the CaCO3 by grinding it into a fine powder

Decreasing the volume of ethanoic acid solution used in the experiment

Decreasing the concentration of the ethanoic acid solution used in the experiment

Decreasing the temperature at which the experiment is performed

Two samples of Mg(s) of equal mass were placed in equal amounts of HCI(aq) contained in two separate reaction vessels. Particle representations of the mixing of Mg(s) and HCI(aq) in the two reaction vessels are shown in Figure 1 and Figure 2 above. Water molecules are not included in the particle representations. Which of the reactions will initially proceed faster, and why?

The reaction in Figure 2, because more Mg atoms are exposed to HCI(aq) in Figure 2 than in Figure 1

The reaction in Figure 1, because the atoms of Mg are more concentrated than those in Figure 2

The reaction in Figure 1, because the Mg(s) in Figure 1 has a larger mass than the Mg(s) in Figure 2

The reaction in Figure 2, because the Mg(s) in Figure 2 has less surface area than the Mg(s) in Figure 1

$$H_3AsO_4+3I^-+2H3O^+\rightarrow\ H_3AsO_3+I_3^-+H_2O$$ The oxidation of iodide ions by arsenic acid in acidic aqueous solution occurs according to the stoichiometry shown above. The experiment rate law of the reaction is: Rate=k[H 3 AsO 4 ][I - ][H 3 O + ] According to the rate law for the reaction, an increase in the concentration of the hydronium ion has what effect on this reaction?

The rate of the reaction increases

The value of the equilibrium constant increases

The value of the equilibrium constant decreases

Neither the rate nor the value of the equilibrium constant is changed

AP Chemistry Unit 5 Review

Authored by Sarah Callo

Chemistry

12th Grade

NGSS covered

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MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Rate = k[X][Y₂]

Rate = k[X]²[Y₂]

Rate = k[X][Y₂]²

Rate = k[X]²[Y₂]²

Answer explanation

Comparing experiment 1 with experiment 2, a doubling of [Y₂]_i while keeping [X]_i constant resulted in a doubling of the reaction rate. Thus, the reaction is first order with respect to Y₂. A similar comparison of experiment 2 with experiment 3 reveals that the reaction is first order with respect to X. Thus, the exponent of the concentrations of reactants in the rate law are both equal to 1.

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NGSS.HS-PS1-5

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

16 M/s

32 M/s

64 M/s

128 M/s

Answer explanation

The initial rate of disappearance of X is equal to twice the initial rate of appearance of X₂Y₂and given by the following: (Δ[X]/Δt)=−2 × (Δ[X₂Y₂]/Δt) = 2 × (initial rate of reaction)= 2× 32 M/s = 64 M/s. The negative sign indicates that the concentration of X is decreasing.

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NGSS.HS-PS1-5

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

The pressures of the gases used by the second chemist must have been lower than those used by the first chemist, thus the collisions between reacting particles were less frequent than they were in the first chemist's experiments.

The pressures of the gases used by the second chemist must have been lower than those used by the first chemist, thus the number of collisions with sufficient energy to cause reaction was lower than it was in the first chemist's experiments.

The second chemist must have added a catalyst for the reaction, thus providing a different reaction pathway for the reactant particles to react with an activation energy that was lower than that of the uncatalyzed reaction in the first chemist's experiments.

The second chemist must have added a catalyst for the reaction, thus providing energy to reactant particles to increase their rate of reaction compared to their rate of reaction in the first chemist's experiments.

Answer explanation

The presence of a catalyst is the most likely cause of the increased rates measured by the second chemist. An alternative reaction pathway with a lower activation energy would explain the greater reaction rates compared to the rates measured by the first chemist.

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NGSS.HS-PS1-5

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis.

Decreasing the temperature will increase the frequency of the collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis.

Increasing the concentration of sucrose will increase the rate of hydrolysis by increasing the frequency of the collisions between the sucrose and the water molecules.

Decreasing the concentration of sucrose will increase the rate of hydrolysis by increasing the frequency of the collisions between the sucrose and the water molecules.

Answer explanation

The rate changes proportionally to the change in the concentration of sucrose and the increase in the rate of hydrolysis can be explained by an increase in the frequency of the collisions between the reactant molecules.

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NGSS.HS-PS1-5

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

The activation energy of the reaction is smaller in trial 2 than it is in trial 1.

The frequency of collisions between reactant molecules is greater in trial 2 than it is in trial 1.

The value of the rate constant for the reaction is smaller in trial 2 than it is in trial 1.

The value of the rate constant for the reaction is greater in trial 2 than it is in trial 1.

Answer explanation

The concentration of H₂ in trial 2 has been doubled. As reactant concentration increases, the frequency of collisions between reactant molecules increases. This leads to an increase in reaction rate.

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NGSS.HS-PS1-5

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Trial 1, because there is a higher concentration of Zn(s) in the reaction mixture.

Trial 1, because the sample of Zn(s) has less surface area for the reaction to take place.

Trial 2, because there is a higher concentration of HCl(aq) in the reaction mixture.

Trial 2, because the sample of Zn(s) has a greater surface area for the reaction to take place.

Answer explanation

The sample of powdered Zn(s) in trial 2 has a greater surface area than the piece of Zn(s) in trial 1; therefore, more Zn atoms are exposed to HCl(aq) in trial 2 than in trial 1, leading to a faster initial reaction rate.

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NGSS.HS-PS1-5

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

The reaction will proceed at a slower rate with increasing temperature.

The rate of the reaction will double when the concentrations of both CH₃I and NaOH are doubled.

The rate of the reaction will double if the concentration of CH₃I is doubled while keeping the concentration of NaOH constant.

A larger amount of CH₃OH will be produced if the concentrations of CH₃I and NaOH are halved.

Answer explanation

Based on the information given, doubling the concentration of CH₃I while keeping the concentration of NaOH constant will double the rate of the reaction. Based on collision theory, the rate is faster as a result of the increased frequency of collisions when the concentration is increased.

Tags

NGSS.HS-PS1-5

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AP Chemistry Unit 5 Review

The initial rate of disappearance of X is equal to twice the initial rate of appearance of X₂Y₂and given by the following: (Δ[X]/Δt)=−2 × (Δ[X₂Y₂]/Δt) = 2 × (initial rate of reaction)= 2× 32 M/s = 64 M/s. The negative sign indicates that the concentration of X is decreasing.

The presence of a catalyst is the most likely cause of the increased rates measured by the second chemist. An alternative reaction pathway with a lower activation energy would explain the greater reaction rates compared to the rates measured by the first chemist.

The rate changes proportionally to the change in the concentration of sucrose and the increase in the rate of hydrolysis can be explained by an increase in the frequency of the collisions between the reactant molecules.

The concentration of H₂ in trial 2 has been doubled. As reactant concentration increases, the frequency of collisions between reactant molecules increases. This leads to an increase in reaction rate.

The sample of powdered Zn(s) in trial 2 has a greater surface area than the piece of Zn(s) in trial 1; therefore, more Zn atoms are exposed to HCl(aq) in trial 2 than in trial 1, leading to a faster initial reaction rate.

Based on the information given, doubling the concentration of CH₃I while keeping the concentration of NaOH constant will double the rate of the reaction. Based on collision theory, the rate is faster as a result of the increased frequency of collisions when the concentration is increased.

For a first-order reaction, graphing ln[N₂O₅] versus time produces a straight line.

Because the reaction is a first-order radioactive decay, a plot of n_Bi versus time would produce a straight line.

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AP Chemistry Unit 5 Review

The initial rate of disappearance of X is equal to twice the initial rate of appearance of X2Y2 and given by the following: (Δ[X]/Δt)=−2 × (Δ[X2Y2]/Δt) = 2 × (initial rate of reaction)= 2× 32 M/s = 64 M/s. The negative sign indicates that the concentration of X is decreasing.

The presence of a catalyst is the most likely cause of the increased rates measured by the second chemist. An alternative reaction pathway with a lower activation energy would explain the greater reaction rates compared to the rates measured by the first chemist.

The rate changes proportionally to the change in the concentration of sucrose and the increase in the rate of hydrolysis can be explained by an increase in the frequency of the collisions between the reactant molecules.

The concentration of H2 in trial 2 has been doubled. As reactant concentration increases, the frequency of collisions between reactant molecules increases. This leads to an increase in reaction rate.

The sample of powdered Zn(s) in trial 2 has a greater surface area than the piece of Zn(s) in trial 1; therefore, more Zn atoms are exposed to HCl(aq) in trial 2 than in trial 1, leading to a faster initial reaction rate.

Based on the information given, doubling the concentration of CH3I while keeping the concentration of NaOH constant will double the rate of the reaction. Based on collision theory, the rate is faster as a result of the increased frequency of collisions when the concentration is increased.

For a first-order reaction, graphing ln[N2O5] versus time produces a straight line.

Because the reaction is a first-order radioactive decay, a plot of nBi versus time would produce a straight line.

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The initial rate of disappearance of X is equal to twice the initial rate of appearance of X₂Y₂and given by the following: (Δ[X]/Δt)=−2 × (Δ[X₂Y₂]/Δt) = 2 × (initial rate of reaction)= 2× 32 M/s = 64 M/s. The negative sign indicates that the concentration of X is decreasing.

The concentration of H₂ in trial 2 has been doubled. As reactant concentration increases, the frequency of collisions between reactant molecules increases. This leads to an increase in reaction rate.

Based on the information given, doubling the concentration of CH₃I while keeping the concentration of NaOH constant will double the rate of the reaction. Based on collision theory, the rate is faster as a result of the increased frequency of collisions when the concentration is increased.

For a first-order reaction, graphing ln[N₂O₅] versus time produces a straight line.

Because the reaction is a first-order radioactive decay, a plot of n_Bi versus time would produce a straight line.