Kinetics

What is Kinetics? Rate Laws Effect of Temperature on Rate Constant Mechanisms Catalysts Rate Determining Step Practice Problems Practice Problem Answers Citations

 What is Kinetics?

**Kinetics** is an area of chemistry that explores the rates at which chemical reactions occur and it is largely an experimental science. The speed, or **rate**, of a process is defined as the change in a given quantity over a specific period of time. For chemical reactions, rate is usually associated with the rate at which a reactant is being used up (**rate of disappearance**) or the rate at which a product is being created (**rate of appearance**). This data is often placed on a graph to illustrate the progression of the reaction such as this:



The red line would be the rate of appearance of [HI] and the blue line is the rate of disappearance of [H2] and [I2]. By looking at this graph we can also see that the rate of appearance of [HI] is twice the rate of disappearance of [H2] or [I2] at any given time. In the hypothetical reaction  aA + bB → cC + dD we could measure the rate of disappearance of A by using this equation:

The **instantaneous rate** is another way to measure the rate of a reaction. This is the rate at a particular instance in time, as opposed to a time interval. The technique to determine the instantaneous rate is to draw a tangent to the curved line of the graph. 

How do you set up an expression for a reaction where the stoichiometric relationships are not one-to-one? The reaction rates for each substance are determined using the following equalities in this equation:   Rate Laws

Reactions with a single reactant have a rate expression: Rate=k[A]^m (m) represents the order of the reaction.
 * the order of the reaction is determined experimentally, //not with the coefficients from the balanced equation.//

Reactions with more than one reactant (aA + bB → products) have a rate expression: Rate= k[A]^m[B]^n

where (n) is the reaction order with respect to [B].

 The relationship between concentration and time is expressed differently for each order. The directly proportional relationship is known as **first-order relationship**. If changing the concentration of a reactant had no effect on the reaction rate, the relationship is **zero-order relationship**. A **second-order relationship** is exponential, meaning that doubling the concentration of a reactant will increase the rate by 4 (2^2). The symbols shown above can be used to create a general expression to describe the relationships between the concentration of the reactants and the reaction rate, which is called a **rate law**. The rate law can only be derived experimentally and is not affected by changes in concentration. The variable k above represents the **rate constant**. The total order of the reaction is the sum of the exponents in the rate law. To see how to calculate the rate law from the initial rates of reaction look below at our sample free response questions. Calculating the units of rate constants is very important when studying Kinetics. By rearranging the equation to solve for k and then substituting units into the expression, it is easy to obtain the units for the rate constant. Below is a chart of the rate expression, the integrated rate law, and what needs to be plotted for a straight-line to occur on a graph for each order. These are important to memorize! <span style="color: rgb(0, 0, 0); font-family: Times New Roman;"> <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;"> The amount of time needed to decompose half of a reactant is known as half-life. Half-life equations are also different depending on the order of reactants. Radioactive material is always a first-order process. Some more important equations to memorize are the half-life expressions shown in the table below. These equations are simple plug-and-chug equations.
 * <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">**<span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Order ** || <span style="font-size: 120%; font-family: 'Times New Roman',Times,serif;">**<span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Rate Expression ** || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">**<span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Integrated Rate Law ** || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">**Plot for a straight line** ||
 * **<span style="font-size: 120%; font-family: 'Times New Roman',Times,serif;">0 ** || <span style="font-size: 120%; font-family: 'Times New Roman',Times,serif;">rate = k || [A]-[A]o = -kt || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">[A] vs t ||
 * **<span style="font-size: 120%; font-family: 'Times New Roman',Times,serif;">1st ** || <span style="font-size: 120%; font-family: 'Times New Roman',Times,serif;">rate = k[A] || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">ln[A]-ln[A]o = -kt || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">ln[A] vs t ||
 * **<span style="font-size: 120%; font-family: 'Times New Roman',Times,serif;">2nd ** || <span style="font-size: 120%; font-family: 'Times New Roman',Times,serif;">rate = k[A]^2 || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">1/[A] - 1/[A]o = kt || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">1/[A] vs t ||


 * ** Order ** || ** Half-Life Expression ** ||
 * 0 || t½ = [A]o / 2k ||
 * 1st || t½ = 0.693 / k ||
 * 2nd || t½ = 1 / k [A]o ||

[|Mechanism Graph] (First Order) by Lauren Wilhoit

<span style="display: block; font-size: 12pt; font-family: 'Times New Roman',Times,serif; text-align: left;"><span style="font-size: 140%; color: rgb(0, 16, 255);"> How to determine the rate expression without a calculator : (contributed by Erin Green)

Let's look at the example experiment used in the free response questions.


 * = <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Experiment ||= <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Initial [A] M ||= <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Initial [B] M ||= <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Initial Rate (M s^-1) ||
 * = <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">1 ||= <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.10 ||= <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.10 ||= <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.10 ||
 * = <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">2 ||= <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.20 ||= <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.10 ||= <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.40 ||
 * = <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">3 ||= <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.20 ||= <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.20 ||= <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.40 ||

<span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">First, we want to find the rate of solution A. To do this, we need to compare two experiments where the value of A has changed and the value of B has stayed constant. That means we want to look at experiments <span style="color: rgb(255, 0, 0);">1 and 2. The concentrations of B are held constant, so let's look at what has happened to A. We see that the value has <span style="color: rgb(255, 0, 0);">doubled from .10 to .20. Now look at the rate- the rate <span style="color: rgb(255, 0, 0);">quadrupled ( but let's think of it as changing by 2^2.) When 1 value changes, and the rate changes by the square value, assume it is<span style="color: rgb(255, 0, 0);"> second order.

Next, we want to find the rate of solution B. To do this, we need to compare two experiments where the value of B has changed and the value of A has stayed constant. That means we want to look at experiments <span style="color: rgb(255, 0, 0);">2 and 3. The concentrations of A are held constant, so let's look at what has happened to B. We see that the value has <span style="color: rgb(255, 0, 0);">doubled from .10 to .20. Now look at the rate- it remained <span style="color: rgb(255, 0, 0);">constant. When 1 value changes, and the rate is unaffected by an increase in concentration ( stays constant), assume is is<span style="color: rgb(255, 0, 0);"> zero order.

The rate law for this experiment would then be **rate =k[A]^2[B]^0**

REMEMBER: zero order= 1 concentration changes and the rate does NOT change 1st order= 1 concentration changes and the rate changes the same 2nd order= 1 concentration changes and the rate changes by the squared value (2, 2^2) <span style="font-size: 22pt; color: rgb(0, 0, 255); font-family: 'Comic Sans MS',cursive;"> Effect of Temperature on Rate Constant <span style="font-size: 12pt; line-height: 200%; font-family: 'Times New Roman','serif';"><span style="color: rgb(0, 0, 0);">Temperature has a definite effect on the rate of chemical reactions. An increase in temperature corresponds to an increase in rate, although there are a small number of exceptions. The rate constant is affected by changes in temperature and not changes in concentration. There are 2 main theories that together explain the relationships between temperature and rates. They are called the **collison theory** and the **transition-state theory**. The collion theory proposes that increases in temperature increase reaction rates by increasing the number of collisons that occur between particles and by increasing the kinetic energy that particles possess when they collide. The transition-state theory focuses on the rates at which particles form and are formed from activated complex. There are three main factors that determine if a reaction will occur: <span style="display: block; font-size: 12pt; line-height: 200%; font-family: 'Times New Roman','serif'; text-align: left;">1.<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> The concentration of activated complexes 2.<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> The rate at which the complexes break apart 3. The direction that the complexes fall apart (whether they break apart into the products or fall back apart as reactants)

One factor that each theory shares is the view that two colliding particles will react only if their combined kinetic energies meet or exceed a threshold value known as the **activation energy (//Ea//).** If the activation energy cannot be achieved, then the reaction will not occur. The **activated complex** is a temporary species that forms as a result of the collision beween particles. Its existence is temporary, so it will eventually fall apart. If the potential energy of an exothermic reaction (gives off heat) is plotted against the progress of the reaction, the resulting graph looks like this:



<span style="font-size: 120%; font-family: 'Times New Roman',Times,serif;">Since the potential energy in the graph shows that the kinetic energy must be sufficient to reach the top of the “hill” for the reaction to continue. The potential energy drops below its original level when the products are formed because the products are more stable then the reactants. The excess energy is released mostly as heat. The potential energy of an endothermic reaction (takes in heat) is plotted against the progress of the reaction then it will look like this: <span style="display: block; font-size: 22pt; color: rgb(0, 0, 255); font-family: 'Comic Sans MS',cursive; text-align: left;">



<span style="display: block; font-size: 22pt; color: rgb(0, 0, 255); font-family: 'Comic Sans MS',cursive; text-align: left;"> Mechanisms <span style="display: block; font-size: 120%; color: rgb(0, 0, 0); font-family: 'Times New Roman',Times,serif; text-align: left;">**Rate mechanisms** are a sequence of steps that show what really happen. A reaction mechanism is a method that is used to show the intermediate processes that occur during a reaction. It lists the proposed changes that take place to the reactants as the products are being formed. They usually are made of 2 or 3 different chemical reactions which are called **elementary steps**, which are shown one on top of the other. This is the balanced chemical equation of the reaction of nitrogen dioxide and carbon monoxide: NO2 (g) + CO (g) → NO (g) + CO2 (g) If we really look at this reaction we can see that the substance NO3 is present in the equation, but it is not present in the overall equation as a reactant or product. That is because the reaction occurs in 2 steps called reaction mechanism: Step 1: NO2 + NO2 →NO3 + NO Step 2: NO3 + CO → NO2 + CO2 A reaction mechanism must correctly represent the overall equation in order to be considered correct. In order to determine if it is correct or not you must add the 2 steps together. To do this you must first make sure all reactants are on the left side and all products on the right. Any substances that are found on both sides may be canceled such as this:

Elementary steps: NO2 + NO2 →NO3 + NO NO3 + CO → NO2 + CO2 Sum of steps: NO2 + NO2 + NO3 + CO → NO3 + NO + NO2 + CO2 (cancel any substances found on both sides) Overall equation: NO2 + CO → NO2 + CO2 Any substance that is neither a reactant or product, but is seen during the reaction is called an **intermediate.** Similar to intermediates are **catalysts**. Catalysts are substances that change the rate of a reaction without being consumed. They help the reaction occur at a faster rate and, like intermediates, take part in the reaction but they are not seen in the overall reaction. [|Click here for determining catalysts and intermediates]. <span style="display: block; font-size: 120%; color: rgb(0, 0, 255); font-family: 'Times New Roman',Times,serif; text-align: left;">Video contributed by Sammy Brennan. <span style="display: block; font-size: 22pt; color: rgb(0, 0, 255); font-family: 'Comic Sans MS',cursive; text-align: left;"> Catalysts <span style="display: block; font-size: 120%; color: rgb(0, 0, 0); font-family: 'Times New Roman',Times,serif; text-align: left;">A catalyst is a substance that changes the rate of the reaction without being consumed by lowering the activation energy. There are two different types of **catalysts**: **heterogeneous and homogeneous**. Heterogeneous catalysts are substances that are in a different phase than the reactants, while homogeneous catalysts are substances that are in the same phase as the reactants. An **enzyme** is an organic catalyst that is produced in living organisms. Enzymes are proteins with a high molecular mass. They can only exist at a certain temperature, otherwise their shape will change. Watch this great video on catalysts: <span style="display: block; font-size: 22pt; color: rgb(0, 0, 255); font-family: 'Comic Sans MS',cursive; text-align: left;">media type="file" key="Catalysts.asf" <span style="color: rgb(0, 0, 0); font-family: Times New Roman;"> <span style="color: rgb(0, 0, 0); font-family: Times New Roman;"> <span style="display: block; font-size: 22pt; color: rgb(0, 0, 255); font-family: 'Comic Sans MS',cursive; text-align: left;"> Rate determining step <span style="display: block; font-size: 12pt; color: rgb(0, 0, 0); font-family: 'Times New Roman',Times,serif; text-align: left;">The rate determining step is always the slowest step of all the elementary steps. The slow one is the rate determining step because the reaction can only go as fast as its slowest step. For example, in the elementary steps: NO2 + F2 → NO2F + F (slow) F + NO2 → NO2F (fast) the first equation is the rate determining step because it is considered the slow step. In this case the rate law we can predict is: Rate = k[NO2][F2]. This rate law is just an educated guess. To detirmine if it is correct you must compare it to experimental data. Even if the experimental data supports the predicted rate law, it does not mean our rate law is correct. If it doesn't match, then it allows us to rule out our mechanism, but if it does match the experimental data then we know we have produced the best explanation for the time being.

<span style="display: block; font-size: 22pt; color: rgb(0, 0, 255); font-family: 'Comic Sans MS',cursive; text-align: left;"> Practice Problems **Multiple Choice:** <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">1. If a reaction, 2A + 3B → products, is first order for A and second order for B, the rate law for the overall reaction will be written as Rate = A)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> k[A][B] B)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> k[A][B]^2 C)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> k[A]^2[B]^2 D)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> k[A]^2[B]^3 E)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> k[A]^2[B]

2. The rate law for the reaction 2A + B → C was found to be Rate = k[A][B]^2. If the concentration of B is tripled, what will happen to the rate of the reaction? A)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> It will stay the same. B)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> It will increase by two times. C)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> It will increase by three times. D)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> It will increase by six times. E)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> It will increase by nine times.

<span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">
 * Experiment || Initial [NO2] //m// || Initial Reaction Rate M s -1 ||
 * <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">1 || 0.010 || 7.1 x 10-5 ||
 * <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">2 || 0.020 || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">28 x 10-5 M s ||

3. Nitrogen dioxide decomposes according to the following reaction: 2NO2 → 2NO(g) + O2(g) For the initial rate data given above, what is the value of //k// in the rate law? <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">A) 0.010 M B) 7.1 x 10^-3 s^-1 C) 3.5 x 10^-3 s^-1 D) 0.71 M^-1 s^-1 E) 4

2H2O2 (l) → 2H2O(l) + O2(g) 4. Hydrogen peroxide decomposes according to the reaction above, which is first order for H2O2 and has a half-life 18.0 minutes. If an H2O2 solution that was initially 0.80 M is allowed to decompose for 72 minutes, what will the concentration be at that time? A)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> 0.80 M B)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> 0.40 M C)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> 0.20 M D)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> 0.10 M E)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> 0.05 M

5. Which of the following would produce a linear graph for the following reaction? A→ product(s) A)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> mol A vs. time if the reaction is first order for A B)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> 1/ [A] vs. time if the reaction is first order for A C) 1/[A]^2 vs. time if the reaction is second order for A D) ln [A] vs. time if the reaction is first order for A E) ln [A] vs. time if the reaction is second order for A

Rate= k[HCrO4-][HSO3-]^2[H+] 6. In the reaction above, what is the order of the reaction with respect to H+? A)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> 0 B)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> 1 C)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> 2 D)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> 3 E)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> 4

<span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Rate= k[NO]^2[Br2] 7. What is the order for the overall reaction shown above? A)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Zero Order B)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> First Order C)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Second Order D)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Third Order E)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Not enough information

8. Which of the graphs below represents an exothermic reaction? A)



B)



<span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;"><span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">9. A catalyst does not affect which of the following? I.<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Activation energy II.<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Reaction rate III.<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Potential energies reactants or products A)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> I only B)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> II only C)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> III only D)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> I and II E)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> II and III

<span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">2A + B → products 10. The rate law for the reaction above is Rate= k[A]^2. What effect would adding additional B have? (Assume no change in temperature or volume.) A)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Both the value of k and the reaction rate would decrease. B)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Both the value of k and the reaction rate would increase. C)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> The value of k would remain constant, but the rate would decrease. D)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> The value of k would remain constant, but the rate would increase. E)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Neither the value of k nor the reaction rate would change.

//Step 1:// NO(g) + Cl2(g) ↔ NOCl2(g) (fast equilibrium) //Step 2:// NO(g)+ NOCl2(g)→ 2NOCl (g) (slow) <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">11. Which of the following rate laws is consistent with the reaction mechanism shown above? A)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Rate= k[NO][Cl2] B)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Rate= k[NO]^2 C)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Rate= k[NO][NOCl2] D)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Rate= k[NO]^2[Cl] E)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Rate= k[NO]^2[Cl][NOCl2]

<span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">12. The reaction mechanism for the chlorination of trichloromethane is shown below: Cl2 (g)↔ 2Cl(g) (fast) Cl (g) + CHCl3 (g) → HCl (g) + CCl3(g) (slow) Cl (g) + CCl3 (g) → CCl4 (g) (fast) Which of the following represents an intermediate in the reaction? A)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> CHCl3 B)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Cl C)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> Cl2 D)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> CCl4 E)<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"> HCl

<span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">1) A hypothetical reaction takes palce according to the following equation: 2A + B → 3C + 2D The following rate data were obtained from three different experiments, all of which were performed at the same temperature:
 * Free Response:**


 * <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Experiment || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Initial [A]^m || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Initial [B]^n || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Initial rate M s^-1 ||
 * <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">1 || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.10 || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.10 || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.10 ||
 * <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">2 || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.20 || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.10 || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.40 ||
 * <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">3 || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.20 || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.20 || <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">0.40 ||

<span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;"><span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">A) Determine the rate law for the reaction. B) Calculate the rate constant, k. C) Calculate the rate of the reaction when [A] = 0.30 M and [B] = 0.30 M.

2) The proposed mechanism for the reaction in this question is shown below: Step 1: 2NO (g) ↔ N2O2 (g) (fast) Step 2: N2O2 (g) + H2 (g) ↔ H2O2 (g) + N2 (g) (slow) Step 3: H2 (g) + H2O2 (g) ↔ 2H2O (fast)

A) Write the balanced equation for this reaction. B) Identify the rate-determining step. C) Write a rate law that is most consistent with this mechanism. D) Based on your rate law, determine the effect of doubling the concentration of NO (while maintaining constant amounts of other reactants) on the reaction rate.

<span style="color: rgb(0, 0, 0); font-family: Times New Roman;"> <span style="font-size: 22pt; color: rgb(0, 0, 255); font-family: 'Comic Sans MS',cursive;"> Practice Problem Answers <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">Multiple Choice: <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;"> 1) B 2) E 3) D 4) E 5) D 6) B 7) D 8) A 9) C 10) E 11) D 12) B

Free Response: [|Click here for the answer to Free Response 1 Part A using one method] [|Click here for the answer to Free Response 1 Part A using a different method] [|Click here for the answer to Free Response 1 Part B] [|Click here for the answer to Free Response 1 Part C] [|Click here for the answer to Free Response 2 Part A] [|Click here for the answer to Free Response 2 Part B & C] [|Click here for the answer to Free Response 2 Part D]

<span style="font-size: 22pt; color: rgb(0, 0, 255); font-family: 'Comic Sans MS',cursive;"> Citations

<span style="display: block; font-size: 50%; color: rgb(0, 0, 0); font-family: 'Times New Roman',Times,serif; text-align: left;"><span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">[] [] <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">[] [] __Peterson’s Master AP Chemistry__ by Brett Barker (pgs. 379-410) <span style="font-size: 12pt; font-family: 'Times New Roman',Times,serif;">__Chemistry: Matter and its Changes__ by Brady Senese (pgs. 519-561) __Chemistry: Principles and Reactions__ by Masterton and Hurley (pgs. 282-311)