NEW
Font size
WorksheetsAP Kinetics
Total questions: 25
Worksheet time: 4hrs 19mins
How are the exponents n and m determined?
step 2: O3 + O → 2 O2
The oxygen atom (O) is considered to be a(n)...
(CH3)3CBr(aq) + OH-(aq) → (CH3)3COH(aq) + Br-(aq)
What will the initial rate (in mol/L*s) be in Experiment 4?
rate = k[CO]2[O2]2
rate = k[CO]2[O2]
rate = k[CO][O2]2
rate = k[CO][O2]1/2
Mechanisms in which one elementary step is followed by another are very common.
step 1: A + B → Q
step 2: B + Q → C
net reaction: A + 2B → C
Which of the following is false?
The graph shown depicts the relationship between concentration and time. What is the slope of this line?
-k
-1/k
k
ln[A]o
What is the rate law for this reaction?
rate = k[X]0
rate = k[X]
rate = k[X]2
rate = -k[X]2
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 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.
Zn(s)+2 HCl(aq)→ZnCl2(aq)+H2(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 0.10MHCl(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 1, because there is a higher concentration of Zn(s)
Zn(s) in the reaction mixture.
Trial 1, because the sample of Zn(s)
Zn(s) has less surface area for the reaction to take place.
Trial 2, because there is a higher concentration of HCl(aq)
HCl(aq) in the reaction mixture.
Trial 2, because the sample of Zn(s)
Zn(s) has a greater 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[CH3I][NaOH]. 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 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.
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.
zeroth order
first order
second order
third order
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+I−→ 2 H2O+O2+I− , and the rate law for elementary step 2 is rate=k[H2O2][IO−] .
The chemical equation for the overall reaction is H2O2+IO−→ H2O+O2+I− , and the rate law for elementary step 2 is rate=k[H2O2]2[IO−] .
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−] .
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]2 .
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 NO(g)+N2O5(g)→3NO2(g) , and the rate law is rate=k[NO][N2O5] .
The chemical equation for the overall reaction is 2N2O5(g)→4NO2(g)+O2(g) , and the rate law is rate=k[N2O5] .
The chemical equation for the overall reaction is 2N2O5(g)→4NO2(g)+O2(g) , and the rate law is rate=k[N2O5]2 .
The chemical equation for the overall reaction is N2O5(g)+NO(g)+NO3(g)→4NO2(g)+O2(g) , and the rate law is rate=k[N2O5][NO][NO3] .
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 molecules have about the same size and the same energy, which leads to a larger rate constant (k).
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.
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.
At T3 , because at this temperature the rate of consumption for about half of the molecules is determined by their orientation.
At T2 , because at this temperature most of the molecules undergo collisions frequently.
At T1 , because a larger fraction of the molecules have about the same energy.
