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 following table.
Given the information in the table above, which of the following is the experimental rate law? | Rate=k[X][Y2]
WHY?
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. |
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 following table.
Based on the information above, determine the initial rate of disappearance of X in experiment 1. | 64 M/s
WHY?
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)=2×32M/s=64M/s. The negative sign indicates that the concentration of X is decreasing. |
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 following table.
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 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-
WHY?
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. |
C12H22O11(aq)+H2O(l)→2C6H12O6(aq)
The chemical equation shown above 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? | 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.
WHY?
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. |
2NO(g)+2H2(g)→N2(g)+2H2O(g)
The information in the data table above represents two different trials for an experiment to study the rate of the reaction between NO(g) and H2(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 frequency of collisions between reactant molecules is greater in trial 2 than it is in trial 1.
WHY?
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. |
Zn(s)+2 HCl(aq)→ZnCl2(aq)+H2(g)
Zn(s)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 0.10MHCl(aq). The rate of reaction between Zn(s) and HCl(aq) is determined by measuring the volume of H2(g) produced over time. In trial 2 of the experiment, 5.0g of powdered Zn(s) is added to 100mL of 0.10MHCl(aq). Which trial will have a faster initial rate of reaction and why? | Trial 2, because the sample of Zn(s) has a greater surface area for the reaction to take place.
WHY?
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. |
CH3I+NaOH→CH3OH+NaI
The rate of the reaction represented by the chemical equation shown above is expressed as rate=k[CH3I][NaOH]. Based on this information, which of the following claims is correct? | The rate of the reaction will double if the concentration of CH31 is doubled while keeping the concentration
of NaOH constant.
WHY?
Based on the information given, doubling the concentration of CH3I while keeping the concentration of NaOHconstant 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. |
2N2O5(g)→4NO2(g)+O2(g)
For the reaction represented by the equation above, the concentration of N2O5 was measured over time. The following graphs were created using the data. | First Order
WHY?
For a first-order reaction, graphing ln[N2O5] versus time produces a straight line. |
Bi 83214→Po 84214+e−1 0
Bismuth-214 undergoes first-order radioactive decay to polonium-214 by the release of a beta particle, as represented by the nuclear equation above.
Which of the following quantities plotted versus time will produce a straight line? | ln[Bi]
WHY?
Because the reaction is a first-order radioactive decay, a plot of nBi versus time would produce a straight line. |
2NH3(g)−→−−−−catalystN2(g)+3H2(g)
The catalyzed decomposition of NH3(g) at high temperature is represented by the equation above. The rate of disappearance of NH3(g) was measured over time for two different initial concentrations of NH3(g) at a constant temperature. The data are plotted in the graph below.
On the basis of the data in the graph, which of the following best represents the rate law for the catalyzed decomposition of NH3(g)? | rate=k
WHY?
The plots of [NH3] versus time for the two different initial concentrations of ammonia are straight lines with identical slopes. The slope is equal to Δ[NH3]Δt, which is constant and equal to the rate of reaction. In other words, the rate is independent of [NH3], and the reaction is zeroth order with rate=k. |
Step 1: H2O2+I−→IO−+H2O
Step 2: H2O2+IO−→H2O+O2+I−
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 H2O2→2 H2O+O2 , and the rate law for elementary step 1 is rate=k[H2O2][I−] .
WHY?
Once all the elementary steps are added and simplified, the resulting equation is 2 H2O2→2 H2O+O2. Because a rate law for an elementary step can be based on the stoichiometric coefficients of the reactants in that step, the rate law for elementary step 1 is rate=k[H2O2][I−], and the rate law for elementary step 2 is rate=k[H2O2][IO−]. |
Step 1: N2O5→NO2+NO3 (slow)
Step 2: NO2+NO3→NO2+NO+O2 (fast)
Step 3: NO+N2O5→3 NO2 (fast)
A proposed reaction mechanism for the decomposition of N2O5(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 2N2O5(g)→4NO2(g)+O2(g) , and the rate law is rate=k[N2O5] .
WHY?
When the elementary steps are added and simplified, the net chemical equation for the overall reaction is 2N2O5(g)→4NO2(g)+O2(g). Because step 1 is the slowest (rate-determining) step, rate=k[N2O5]. |
Step 1:?? (slow)
Step 2: NO2(g)+F(g)→NO2F(g) (fast)
Overall: 2 NO2(g)+F2(g)→2NO2F(g)
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 NO2(g)+F2(g)→NO2F(g)+F(g), and the rate law for the overall reaction is rate=k[NO2][F2].
WHY?
The chemical equations for the two elementary steps should add up to give the net equation for the overall reaction. Based on the chemical equations for step 2 and the overall reaction, the chemical equation for step 1 is NO2(g)+F2(g)→NO2F(g)+F(g). Because step 1 is the slow (rate-determining) step, the rate law for the overall reaction is set by the stoichiometry of step 1 and is equal to rate=k[NO2][F2]. |
Reaction A: O+O→O2
Reaction B: C2H4+C2H4→C4H8
Reaction C: CO+O2→CO2+O
Reaction D: CH3I+Br−→CH3Br+I−
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.
WHY?
The electron cloud is distributed symmetrically around the O atoms; thus the orientation of the two reacting Oatoms during a collision is not a factor in determining the success of the collision and the reaction rate. |
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.
term-11
WHY?
In order for this reaction to form CO2 and NO, the reactants CO and NO2 must be oriented in a way that allows for a bond between N and O to be broken while a new bond between C and O forms. Diagram 1 represents the best orientation for an effective collision as long as the molecules have the minimum energy required to overcome the activation energy barrier. |
The diagram above shows the distribution of molecular 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 T4 , because a larger fraction of the molecules have an energy that is equal to or greater than the activation energy.
WHY?
The rate of consumption of a reactant depends on temperature, and for most reactions an increase in temperature increases the rate. This is a result of an increase in the number of molecules whose energy is equal to or greater than the minimum energy required to overcome the activation energy (Ea) barrier. This is shown in the energy distribution corresponding to T4 because a larger portion of the curve is located at energies greater than or equal to Ea. |
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)→X2(g)
Step 2: X2(g)+Y(g)→X2Y(g)
WHY?
From the first diagram to the second diagram, two X atoms joined to form the intermediate X2. From the second diagram to the third diagram, the X2 joined with a Y to form the final product X2Y. |
The diagram above shows the progress of the chemical reaction for the synthesis of ammonia from its elements. The adsorption of the N2 molecules on the surface of Ru weakens the triple bond between the two N atoms. Based on the diagram, what is the role of Ru in this process? | Ru is a Catalyst
WHY?
The amount of Ru atoms does not change. Ru is a catalyst that provides an alternate path to the process of breaking the triple bond between the two N atoms, which is the rate-limiting step in the synthesis of NH3. |
The proposed mechanism for a reaction involves the three elementary steps represented by the particle models shown above. On the basis of this information, which of the following models represents an intermediate in the overall reaction? | Model representing NNO
WHY?
N2O is a reaction intermediate. N2O is generated in step 2 and is consumed in step 3. Intermediates, such as N2O and N2O2, do not appear in the overall chemical equation of the reaction, which is 2NO+2H2→N2+2H2O. |
O3(g)+O(g)→2O2(g)
The decomposition of O3 occurs according to the balanced equation above. In the presence of NO, the decomposition proceeds in two elementary steps, as represented by the following mechanism.
Step 1: O3+NO→NO2+O2 (slow)
Step 2: NO2+O→NO+O2 (fast)
Based on the information, which of the following is the rate law? | Rate=k[O3][NO]
WHY?
Rate=k[O3][NO] is consistent with step 1 being the rate-determining step. When the first step of a multistep reaction is the slowest step, the rate of reaction depends on the concentration and stoichiometry of the reacting species in this elementary reaction. |
Step 1: 2NO2(g)→NO3(g)+NO(g) (slow)
Step 2: NO3(g)+CO(g)→NO2(g)+CO2(g) (fast)
A proposed two-step mechanism for the chemical reaction NO2(g)+CO(g)→NO(g)+CO2(g) is shown above. Which of the following equations is a correct rate law that is consistent with the elementary steps in the mechanism? | Rate=k[NO2]^2
WHY?
For elementary steps, the stoichiometry of the slowest step is used to establish a proposed rate law. In the slow step, the stoichiometric coefficient of NO2 is 2; therefore, the proposed rate law should be rate=k[NO2]2. |
Step 1: H2+IBr→HI+HBr (slow)
Step 2: HI+IBr→I2+HBr (fast)
A proposed mechanism for the reaction H2+2IBr→I2+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 H2 and IBr were doubled.
WHY?
Based on the proposed mechanism, the rate-determining step is step 1 and the rate law for the reaction should be rate=k[H2][IBr]. As a result, if two experiments were performed at the same temperature but the initial concentrations of H2 and IBr were doubled for one of the experiments, the rate of the reaction would quadruple for that experiment. |
Step 1: 2NO⇄(NO)2 (fast)
Step 2: (NO)2+O2⇄2NO2 (slow)
The elementary steps in a proposed mechanism for the reaction 2 NO(g)+O2(g)→2NO2(g) are represented by the equations above. Which of the following is the rate law for the overall reaction that is consistent with the proposed mechanism? | rate=k[NO]^2[O2]
WHY?
The rate of the forward reaction in step 1 depends on the collision of two NO molecules, so its rate law is rateforward=kf[NO]2. The rate of the reverse reaction in step 1 depends only on the concentration of (NO)2, so its rate law is ratereverse=kr[(NO)2]. Because step 1 is an equilibrium, the forward and reverse rates are equal, meaning that kf[NO]2=kr[(NO)2]. Step 2 is the slow (rate-determining) step, which depends on the collision between (NO)2 and O2. Its rate law is rateStep 2=k2[(NO)2][O2]. From Step 1, kf[NO]2=kr[(NO)2], therefore [(NO)2]=kfkr[NO]2. Substituting this expression into the rate law for step 2 yields the equation rateStep 2=(k2)(kfkr)[NO]2[O2] = k′[NO]2[O2]. If the rate constant for the overall reaction k′ is renamed as k, the final rate law expression is rate=k[NO]2[O2]. |
H2(g)+I2(g)→2HI(g)
For the reaction between H2 and I2, shown above, the following two-step reaction mechanism is proposed.
Step 1: I2⇄2I (fast equilibrium)
Step 2: H2+2I→2HI (slow)
What is the rate law expression for this reaction if the second step is rate determining? | Rate=k[H2][I2]
WHY?
The rate law of a reaction is set by the stoichiometry of the rate-determining step. The rate law for the rate-determining step (step 2) is rate=k2[H2][I]2. However, intermediates should not be included in the rate law. Step 1 is a fast equilibrium reaction, and the rates of the forward and reverse reactions are equal: kf[I2]=kr[I]2. Solving for [I]2gives [I]2=kf[I2]kr. Substituting for [I]2 in the rate law for step 2 gives rate=k2[H2]kf[I2]kr, which is equal to rate=k[H2][I2]. |
Step 1: 2X(g)⇄X2(g) (fast)
Step 2: X2(g)+Y(g)→X2Y(g) (slow)
The rate law for the hypothetical reaction 2X(g)+Y(g)→X2Y(g) is consistent with the mechanism shown above. Which of the following mathematical equations provides a rate law that is consistent with this mechanism? | rate=k[X]^2[Y]
WHY?
Since step 2 is the rate-determining step, rate=k2[X2][Y]. X2 is an intermediate formed in a fast equilibrium step, thus, kforward[X]2=kreverse[X2]. Rearranging to solve for [X2] and substituting in the first expression, rate=k[X]2[Y], where k=k2kforwardkreverse. |
Step 1: HCOOH+H2SO4→HCOOH2++HSO4−
Step 2: HCOOH2+→HCO++H2O
Step 3: HCO++HSO4−→CO+H2SO4
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? | H2SO4, because it is consumed in the first step of the mechanism and regenerated in a later step.
WHY?
The H2SO4 is consumed in the first step of the mechanism and regenerated in a later step. This is the definition of a catalyst. |
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.
WHY?
A catalyst increases the rate of a chemical reaction by either increasing the number of effective collisions or, as in this case, providing a different reaction path with a lower activation energy. |