Kinetics

A chemist is studying the reaction between the gaseous chemical species X and Y₂, represented by the equation below. Initial rates of reaction are measured at various concentrations of reactants. The results are recorded in the following table.

2X      +     Y₂     -->     X₂Y₂

Given the information in the table, which of the following is the experimental rate law?

 

• Rate = k [X] [Y₂]

• Rate = k [X]² [Y₂]

• Rate = k [X] [Y₂]²

• Rate = k [X]² [Y₂]²

2X      +     Y₂     -->    X₂Y₂

A chemist is studying the reaction between the gaseous chemical species X and Y₂, represented by the equation above. Initial rates of reaction are measured at various concentrations of reactants. The results are recorded in the table on the left.

Based on the information, determine the initial rate of disappearance of X in experiment 1.

• 16 M / s

• 32 M / s

• 64 M / s

• 128 M / s

2X      +     Y₂    -->     X₂Y₂

A chemist is studying the reaction between the gaseous chemical species X and Y₂, represented by the equation above. Initial rates of reaction are measured at various concentrations of reactants. The results are recorded in the table on the left.

A second chemist repeated the three experiments and observed that the reaction rates were considerably greater than those measured by the first chemist, even though the concentrations of the reactants and the temperature in the laboratory were the same as they were for the first chemist. Which of the following is the best pairing of a claim about a most likely cause for the greater rates measured by the second chemist and a valid justification for that claim?

• 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.

The chemical equation shown below represents the hydrolysis of sucrose. Under certain conditions, the rate is directly proportional to the concentration of sucrose. Which statement supports how a change in conditions can increase the rate of this reaction?

C₁₂H₂₂O_11(aq)      +      H₂O_(l)      -->      2 C₆H₁₂O_6(l)

• 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.

• 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 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.

• 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.

2 NO_(g)     +      2 H_2(g)     -->     N_2(g)     +     2 H₂O_(g)

The information in the data table on the left represents two different trials for an experiment to study the rate of the reaction between NO_(g) and H_2(g), as represented by the balanced equation above the table. Which of the following statements provides the correct explanation for why the initial rate of formation of N₂ is greater in trial 2 than in trial 1? Assume that each trial is carried out at the same constant temperature.

• The activation energy of 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.

• The value of the rate constant for 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.

Zn_(s)     +     2 HCl_(aq)     -->      ZnCl_2(aq)      +      H_2(g)

 

Zn_(s) reacts with HCl_(aq) according to the equation shown above. In trial 1 of a kinetics experiment, a 5.0g piece of Zn_(s) is added to 100Ml of 0.10 M HCl_(aq). The rate of reaction between Zn_(s) and HCl_(aq) is determined by measuring the volume of H_2(g) produced over time. In trial 2 of the experiment, 5.0g of powdered Zn_(s) is added to 100mL of 0.10 M HCl_(aq). Which trial will have a faster initial rate of reaction and why?

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

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

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

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

CH₃I     +     NaOH     -->     CH₃OH     +     NaI

The rate of the reaction represented by the chemical equation shown above is expressed as rate = k . Based on this information, which of the following claims is correct?

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

• 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.

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

The equations shown below 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:    O      +      O     -->      O₂

Reaction B:    C₂H₄      +      C₂H₄      -->      C₄H₈

Reaction C:    CO      +      O₂      -->      CO₂     +     O

Reaction D:    CH₃I      +      Br^-      -->      CH₃Br     +     I^-

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

• Reaction D, because the reactant Br^- and the product I^- are negatively charged.

• Reaction C, because both CO and O₂ are linear molecules.

• Reaction B, because each C₂H₄ molecule has a double bond.

The two diagrams above represent collisions that take place at the same temperature between a CO molecule and an NO₂ molecule. The products are CO₂ 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 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.

• 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).

The diagram on the left 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 T₁, because a larger fraction of the molecules have about the same energy.

• At T₃, because at this temperature the rate of consumption for about half of the molecules is determined by their orientation.

• At T₄, because a larger fraction of the molecules have an energy that is equal to or greater than the activation energy.

• At T₂, because at this temperature most of the molecules undergo collisions frequently.