how to calculate rate constant from experimental data

12.3 Rate Laws

Learning objectives.

By the end of this section, you will be able to:

• Explain the form and function of a rate law
• Use rate laws to calculate reaction rates
• Use rate and concentration data to identify reaction orders and derive rate laws

As described in the previous module, the rate of a reaction is often affected by the concentrations of reactants. Rate laws (sometimes called differential rate laws ) or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. As an example, consider the reaction described by the chemical equation

where a and b are stoichiometric coefficients. The rate law for this reaction is written as:

in which [ A ] and [ B ] represent the molar concentrations of reactants, and k is the rate constant , which is specific for a particular reaction at a particular temperature. The exponents m and n are the reaction orders and are typically positive integers, though they can be fractions, negative, or zero. The rate constant k and the reaction orders m and n must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant k is independent of the reactant concentrations, but it does vary with temperature.

The reaction orders in a rate law describe the mathematical dependence of the rate on reactant concentrations. Referring to the generic rate law above, the reaction is m order with respect to A and n order with respect to B . For example, if m = 1 and n = 2, the reaction is first order in A and second order in B . The overall reaction order is simply the sum of orders for each reactant. For the example rate law here, the reaction is third order overall (1 + 2 = 3). A few specific examples are shown below to further illustrate this concept.

The rate law:

describes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:

describes a reaction that is second order in C 4 H 6 and second order overall. The rate law:

describes a reaction that is first order in H + , first order in OH − , and second order overall.

Example 12.3

Writing rate laws from reaction orders.

is second order in NO 2 and zero order in CO at 100 °C. What is the rate law for the reaction?

The reaction is second order in NO 2 ; thus m = 2. The reaction is zero order in CO; thus n = 0. The rate law is:

Remember that a number raised to the zero power is equal to 1, thus [CO] 0 = 1, which is why the CO concentration term may be omitted from the rate law: the rate of reaction is solely dependent on the concentration of NO 2 . A later chapter section on reaction mechanisms will explain how a reactant’s concentration can have no effect on a reaction rate despite being involved in the reaction.

Check Your Learning

has been determined to be rate = k [NO] 2 [H 2 ]. What are the orders with respect to each reactant, and what is the overall order of the reaction?

order in NO = 2; order in H 2 = 1; overall order = 3

The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, determined to be:

What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?

order in CH 3 OH = 1; order in CH 3 CH 2 OCOCH 3 = 0; overall order = 1

A common experimental approach to the determination of rate laws is the method of initial rates . This method involves measuring reaction rates for multiple experimental trials carried out using different initial reactant concentrations. Comparing the measured rates for these trials permits determination of the reaction orders and, subsequently, the rate constant, which together are used to formulate a rate law. This approach is illustrated in the next two example exercises.

Example 12.4

Determining a rate law from initial rates.

This reaction has been studied in the laboratory, and the following rate data were determined at 25 °C.

Determine the rate law and the rate constant for the reaction at 25 °C.

Determine the values of m , n , and k from the experimental data using the following three-part process:

Determine the value of m from the data in which [NO] varies and [O 3 ] is constant. In the last three experiments, [NO] varies while [O 3 ] remains constant. When [NO] doubles from trial 3 to 4, the rate doubles, and when [NO] triples from trial 3 to 5, the rate also triples. Thus, the rate is also directly proportional to [NO], and m in the rate law is equal to 1.

Determine the value of n from data in which [O 3 ] varies and [NO] is constant. In the first three experiments, [NO] is constant and [O 3 ] varies. The reaction rate changes in direct proportion to the change in [O 3 ]. When [O 3 ] doubles from trial 1 to 2, the rate doubles; when [O 3 ] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O 3 ], and n is equal to 1.The rate law is thus:

Determine the value of k from one set of concentrations and the corresponding rate . The data from trial 1 are used below:

Determine the rate law and the rate constant for the reaction from the following experimental data:

rate = k [ CH 3 CHO ] 2 rate = k [ CH 3 CHO ] 2 with k = 6.73 × × 10 −6 L/mol/s

Example 12.5

Determining rate laws from initial rates.

As in Example 12.4 , approach this problem in a stepwise fashion, determining the values of m and n from the experimental data and then using these values to determine the value of k . In this example, however, an explicit algebraic approach (vs. the implicit approach of the previous example) will be used to determine the values of m and n :

Determine the value of m from the data in which [NO] varies and [Cl 2 ] is constant . Write the ratios with the subscripts x and y to indicate data from two different trials:

Using the third trial and the first trial, in which [Cl 2 ] does not vary, gives:

Canceling equivalent terms in the numerator and denominator leaves:

which simplifies to:

Use logarithms to determine the value of the exponent m :

Confirm the result

Determine the value of n from data in which [Cl 2 ] varies and [NO] is constant.

Cancelation gives:

Thus n must be 1, and the form of the rate law is:

Determine the numerical value of the rate constant k with appropriate units. The units for the rate of a reaction are mol/L/s. The units for k are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol 3 /L 3 . The units for k should be mol −2 L 2 /s so that the rate is in terms of mol/L/s.

To determine the value of k once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for k :

Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.

rate 2 rate 3 = 0.00092 0.00046 = k ( 0.0020 ) x ( 0.0040 ) y k ( 0.0020 ) x ( 0.0020 ) y rate 2 rate 3 = 0.00092 0.00046 = k ( 0.0020 ) x ( 0.0040 ) y k ( 0.0020 ) x ( 0.0020 ) y 2.00 = 2.00 y y = 1 rate 1 rate 2 = 0.00184 0.00092 = k ( 0.0040 ) x ( 0.0020 ) y k ( 0.0020 ) x ( 0.0040 ) y rate 1 rate 2 = 0.00184 0.00092 = k ( 0.0040 ) x ( 0.0020 ) y k ( 0.0020 ) x ( 0.0040 ) y 2.00 = 2 x 2 y 2.00 = 2 x 2 1 4.00 = 2 x x = 2 2.00 = 2 x 2 y 2.00 = 2 x 2 1 4.00 = 2 x x = 2 Substituting the concentration data from trial 1 and solving for k yields: rate = k [ OCl − ] 2 [ I − ] 1 0.00184 = k (0.0040) 2 (0.0020) 1 k = 5.75 × 10 4 mol − 2 L 2 s − 1 rate = k [ OCl − ] 2 [ I − ] 1 0.00184 = k (0.0040) 2 (0.0020) 1 k = 5.75 × 10 4 mol − 2 L 2 s − 1

Reaction Order and Rate Constant Units

In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case.

Rate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:

It is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.

The units for a rate constant will vary as appropriate to accommodate the overall order of the reaction. The unit of the rate constant for the second-order reaction described in Example 12.4 was determined to be L mol −1 s −1 . L mol −1 s −1 . For the third-order reaction described in Example 12.5 , the unit for k was derived to be L 2 mol −2 s −1 . L 2 mol −2 s −1 . Dimensional analysis requires the rate constant unit for a reaction whose overall order is x to be L x − 1 mol 1 − x s −1 . L x − 1 mol 1 − x s −1 . Table 12.1 summarizes the rate constant units for common reaction orders.

Note that the units in this table were derived using specific units for concentration (mol/L) and time (s), though any valid units for these two properties may be used.

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Determining the Rate Law from Experimental Data

Learning objectives.

• Use experimental data to determine the rate law for a reaction.

Timer. From Miles Kaufmann/ Wikimedia .

Determining the amount of time a process requires calls for a timer. These devices can be simple kitchen timers (not very precise) or complex systems that can measure to a fraction of a second. Accurate time measurement is essential in kinetics studies for assessing rates of chemical reactions.

In order to experimentally determine a rate law, a series of experiments must be performed with various starting concentrations of reactants. The initial rate law is then measured for each of the reactions. Consider the reaction between nitrogen monoxide gas and hydrogen gas to form nitrogen gas and water vapor.

The following data were collected for this reaction at 1280°C.

Notice that the starting concentrations of NO and H 2 were varied in a specific way. In order to compare the rates of reaction and determine the order with respect to each reactant, the initial concentration of each reactant must be changed while the other is held constant.

The overall rate law then includes both of these results.

Notice that the rate law for the reaction does not relate to the balanced equation for the overall reaction. The coefficients of NO and H 2 are both 2, while the order of the reaction with respect to the H 2 is only one. The units for the specific rate constant vary with the order of the reaction. So far, we have seen reactions that are first or second order with respect to a given reactant. Occasionally, the rate of a reaction may not depend on the concentration of one of the reactants at all. In this case, the reaction is said to be zero-order with respect to that reactant.

• The process of using experimental data to determine a rate law is described.

Use the site below to practice determination of rate constant with experimental data.

http://ibchem.com/IB/ibnotes/full/kin_htm/order_calculation.htm

• How do you carry out experiments for determining rate constants?
• Why is the reaction order with regard to NO a value of 2?
• Why is the reaction order with regard to hydrogen value of 1?
• Chemistry Concepts Intermediate. Authored by : Calbreath, Baxter, et al.. Provided by : CK12.org. Located at : http://www.ck12.org/book/CK-12-Chemistry-Concepts-Intermediate/ . License : CC BY-NC: Attribution-NonCommercial

Reaction Rates & How to Determine Rate Law

Core Concepts

Some reactions will go fast, and some will go slow – the speed of the reaction is it’s reaction rate , which is dictated by a rate law . In this article, we will learn about reaction rates , rate laws, the rate constant, and the reaction order.

The rate of a chemical reaction is determined—and altered—by many factors, including the nature (of reactivity) of reactants, surface area, temperature, concentration, and catalysts. For each unique chemical reaction, rate laws can be written at a rate law equation to show how the concentrations of reactants affect the rate of the reaction. It is important to note that you can only determine rate law experimentally !

Topics Covered in Other Articles

• Activation Energy
• Equilibrium Constant
• Chemical Reaction Types
• Influences on Reaction Rate
• Steady State Approximation
• Concentration Units: Normality and Parts Per Million

What is a Rate Law? The Rate Law Equation

The reaction rate can depend on how concentrated our reactants are. A chemical reaction’s rate law is an equation that describes the relationship between the concentrations of reactants in the reaction and the reaction rate. In the standard form, the rate law equation is written as:

The integrated form of the rate law equation is also important to chemists studying kinetics. Check out this article to learn more about integrated rate laws.

Let’s break down each of these components.

Reaction Rate

Specific rate constant.

Furthermore, the units of the specific rate constant are dependent on the orders of the reaction. This will be discussed more in detail later.

Molar Concentrations of Reactants

The rate law uses the molar concentrations of reactants to determine the reaction rate. Typically, increased concentrations of reactants increases the speed of the reaction, because there are more molecules colliding and reacting with each other.

Orders of Reactants & of the Reaction

How to Determine the Rate Law

There are 2 main questions you’ll see when asked to determine the rate law. The first type asks you to find the rate law from elementary steps. The second type asks you to find the rate law from a table listing different experiments with different reactant concentrations and reaction rates.

Sometimes, you’ll have to find a rate law for a reaction with an intermediate. For that, you’ll need to find the rate determining step.

From Elementary Steps

In many reactions, the chemical equation oversimplifies the reaction process. Usually, there are many intermediate reactions, or elementary steps, that occur to get from the reactants to the products.

The NO 3 and NO 2 cancel out on both sides of the equation, so you’re left with the original equation.

In these problems, you will usually be given the elementary steps and the rates of each of the steps. For example in the equation provided above, step 1 is the slow step, and step 2 is faster. The slower step is used as the rate-determining step —because the rate of reaction can only go as fast as the slowest step. You would use the rate-determining step to write the rate law by using its reactants.

The rate law does not include CO (the second reactant in the original chemical equation). This is because CO is not used in the slower, rate-determining step, so it does not affect the reaction rate.

From a Table

To determine the rate law from a table, you must mathematically calculate how differences in molar concentrations of reactants affect the reaction rate to figure out the order of each reactant. Then, plug in values of the reaction rate and reactant concentrations to find the specific rate constant. Finally, rewrite the rate law by plugging in the specific rate constant and the orders for the reactants.

Reading the Table

In the experiment, hydrogen iodide HI is the reactant, and H 2 and I 2 are the products. From the table, you can tell that 3 experiments of the same reaction were run, with varying concentrations of HI. In each experiment, there reaction rate was different, as a result of the different concentrations of HI.

Determine Rate Law Using the Table

Finding the order of reactants.

Mathematically, you can use the same process to find reactant orders by plugging values into the following equation:

Finding the Specific Rate Constant

Now that you know the order of reactant HI, you can start to write the rate law. First, plug in the order into the rate law equation.

Units for the Specific Rate Constant

As mentioned earlier, the units for the specific rate constant depend on the order of the reaction. Keep in mind:

• The order of the reactant changes the units on the right side of the equation

As you can see, the order of each reactant affects the units of the specific rate constant.

For More Help, Watch our Interactive Video on Rate Laws!

Example questions for determining rate law.

Write the rate law for the following reaction given the reaction mechanism elementary steps:

Explanation: Since step 1 is the slower step, it is the rate-determining step for this reaction. Write the rate law by plugging in the reactants into the rate law equation.

Explanation:

Finding the Orders of Reactants

Let’s start by finding the order of Reactant A. As you can see in the table, between experiments 1 and 2, the concentration of B changed, but the concentration of A did not—this would not be useful in finding the order of A. However, between experiments 1 and 3, the concentration of A changed, while B did not—this is perfect for finding the order of A because A is the only thing that changed, and therefore is the only variable that could have affected the reaction rate.

Plugging in the values from the table, you get:

*As a side note, you could also do this by comparing values on the table, without using the equation. Between experiments 1 and 2, as B was halved, the reaction rate was also halved. Thus, you know that the concentration of B had a directly proportional effect on the reaction rate, and the order of B was 1.

Finally, find the value of k by plugging in the values from any of the experiments. If we choose to use experiment 1, we get:

Determine Rate Law Equation

Our Latest Awesome Chemistry Video

Rate law practice problems.

In an overall second-order reaction involving two reactants (Reactant X and Reactant Y), you observe the reaction rate double when you double the concentration of X. In another experiment, you halve the concentration of Y. How would you expect the reaction rate to change?

What is the rate law for this reaction?

Rate Law Practice Problem Solutions

1: The rate would halve.

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Chapter 13: Chemical Kinetics

Back to chapter, determining order of reaction, previous video 13.3: concentration and rate law, next video 13.5: the integrated rate law: the dependence of concentration on time.

For a chemical reaction, the rate law expresses the relationship between the reaction rate and reactant concentration. The exponents of the reactant concentrations influence the reaction rate individually and are called reaction orders.

The reaction order is determined experimentally by employing the method of initial rates, where a chemical reaction is repeated multiple times with varying reactant concentrations to measure the initial reaction rates.

An increase in the reactant concentration, producing a linearly-proportional increase in the reaction rate, characterizes a first-order reaction. If doubling the reactant concentration quadruples the initial rate, then a second-order reaction is observed.

However, if the change of reactant concentrations does not affect the initial rate values, then a zero-order reaction is observed.

When the obtained data of initial rates present an indiscernible relationship between the change in initial reactant concentration and the corresponding rate, a ratio of rate laws is calculated.

Here, any two concentration values of the reactant and their corresponding reaction rates are used to determine the reaction order.

But what about reactions with multiple reactants? First, using the method of initial rates, the reaction order of each reactant is determined individually.

Next, the individual reaction orders are expressed as exponents to their respective reactant concentrations to formulate the rate law. Lastly, the summation of individual exponents from the rate law determines the overall reaction order.

While the reaction order depicts the reaction rate’s dependence on the reactant concentration, a direct measure of the relative reaction speed is indicated by the rate constant.

The rate constant k , is the proportionality coefficient relating the reaction rate to the product of reactant concentrations.

The unit of a rate constant depends on the overall reaction order and can be determined by rearranging the rate law to solve for the rate constant. For a zero-order reaction, k has the unit molarity per seconds. The unit for a first-order reaction is 1/s, and for a second-order reaction it is 1/M·s.

A large rate constant indicates a fast reaction. Conversely, a smaller rate constant indicates a slow reaction. A value of zero signifies the absence of any chemical reaction.

Rate laws describe the relationship between the rate of a chemical reaction and the concentration of its reactants. In a rate law, the rate constant k and the reaction orders are determined experimentally by observing how the rate of reaction changes as the concentrations of the reactants are changed. A common experimental approach to the determination of rate laws is the method of initial rates. This method involves measuring reaction rates for multiple experimental trials carried out using different initial reactant concentrations. Comparing the measured rates for these trials permits determination of the reaction orders and, subsequently, the rate constant, which together are used to formulate a rate law.

The rate of a reaction, for example, involving nitric oxide with ozone [NO ( g ) + O 3 ( g ) ⟶ NO 2 ( g ) + O 2 ( g )] can be determined from the experimental data of method of initial rates,  in the laboratory.

From the rate data, a generic rate law; rate = k [NO] m [O 3 ] n is formulated. The values of the reaction orders m and n , and rate constant k are determined from the experimental data using a three-part process:

In step 1, the value of m is determined from the data in which [NO] varies, and [O 3 ] is constant. In trials 3, 4 and 5, [NO] varies while [O 3 ] remains constant. When [NO] doubles from trial 3 to 4, the rate doubles, and when [NO] triples from trial 3 to 5, the rate also triples. Thus, the rate is also directly proportional to [NO], and m in the rate law is equal to 1.

In step 2, the value of n is determined from data in which [O 3 ] varies, and [NO] is constant. In trials 1,2 and 3, [NO] is constant and [O 3 ] varies. The reaction rate changes in direct proportion to the change in [O 3 ]. When [O 3 ] doubles from trial 1 to 2, the rate doubles; when [O 3 ] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O 3 ], and n is equal to 1. The rate law is thus: rate = k [NO] 1 [O 3 ] 1 = k [NO][O 3 ]

In step 3, the value of k is determined from one set of concentrations (for instance, the data from trial 1) and its corresponding rate.

In reactions where the data from the method of initial rates does not directly imply the relation between initial concentrations and initial rates, a calculation involving the ratio of rate laws can be employed to calculate the reaction order and rate constant.

For example, the general rate law for the reaction 2 NO ( g ) + Cl 2 ( g ) ⟶ 2 NOCl ( g ) is expressed as: rate = k [NO] m [Cl 2 ] n .

The data from the method of initial rates are:

The values of m and n can be determined from the experimental data using an algebraic approach, following which the value of k is determined.

In step 1, the value of m is determined from the data in which [NO] varies and [Cl 2 ] is constant. A ratio of rate laws is expressed by substituting data from two different trials (for instance trial 3 and trial 1).

In step 2, the value of n is determined from data in which [Cl 2 ] varies, and [NO] is constant.

Using the computed values of m and n the rate law is expressed as rate = k [NO] 2 [Cl 2 ].

In step 3, the numerical value of the rate constant k is determined with appropriate units. The units for the rate of a reaction are mol/L·s. The units for k is concluded by substituting the units of all other parameters in the rate law. In this example, the concentration units are mol 3 /L 3 . The units for k should be L 2 /mol 2 ·s so that the rate is in terms of mol/L·s. The value of k is determined once the rate law expression has been solved, by simply substituting the values from any of the experimental trials (for instance trial 1).

Reaction Order and Rate Constant Units

In some reactions, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case. Rate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. Rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.

The units for a rate constant will vary as appropriate to accommodate the overall order of the reaction. The unit of the rate constant for a zero-order reaction is mol/L·s (or M/s) and that for a first-order reaction is 1/s. The unit of the rate constant for a second-order reaction is L/mol·s (or 1/M·s) and that for a third-order reaction is L 2 /mol 2 ·s. Although the specific units for concentration and time are indicated as (mol/L) and (s), any other valid units can be used to represent the properties of concentration and time.

This text is adapted from Openstax, Chemistry 2e, Section 12.3: Rate Laws.

• Determining n, m, and p from reaction orders
• Determining n, m, and p from initial rate data
• Determining the rate constant

Experimentally Determining Rate Law — Overview & Examples - Expii

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14.4 Using Graphs to Determine Rate Laws, Rate Constants, and Reaction Orders

Learning objective.

• To use graphs to analyze the kinetics of a reaction.

In Section 14.3 "Methods of Determining Reaction Order" , you learned that the integrated rate law for each common type of reaction (zeroth, first, or second order in a single reactant) can be plotted as a straight line. Using these plots offers an alternative to the methods described for showing how reactant concentration changes with time and determining reaction order.

We will illustrate the use of these graphs by considering the thermal decomposition of NO 2 gas at elevated temperatures, which occurs according to the following reaction:

Equation 14.26

Experimental data for this reaction at 330°C are listed in Table 14.5 "Concentration of NO" ; they are provided as [NO 2 ], ln[NO 2 ], and 1/[NO 2 ] versus time to correspond to the integrated rate laws for zeroth-, first-, and second-order reactions, respectively. The actual concentrations of NO 2 are plotted versus time in part (a) in Figure 14.15 "The Decomposition of NO" . Because the plot of [NO 2 ] versus t is not a straight line, we know the reaction is not zeroth order in NO 2 . A plot of ln[NO 2 ] versus t (part (b) in Figure 14.15 "The Decomposition of NO" ) shows us that the reaction is not first order in NO 2 because a first-order reaction would give a straight line. Having eliminated zeroth-order and first-order behavior, we construct a plot of 1/[NO 2 ] versus t (part (c) in Figure 14.15 "The Decomposition of NO" ). This plot is a straight line, indicating that the reaction is second order in NO 2 .

Table 14.5 Concentration of NO 2 as a Function of Time at 330°C

Figure 14.15 The Decomposition of NO 2

These plots show the decomposition of a sample of NO 2 at 330°C as (a) the concentration of NO 2 versus t , (b) the natural logarithm of [NO 2 ] versus t , and (c) 1/[NO 2 ] versus t .

We have just determined the reaction order using data from a single experiment by plotting the concentration of the reactant as a function of time. Because of the characteristic shapes of the lines shown in Figure 14.16 "Properties of Reactions That Obey Zeroth-, First-, and Second-Order Rate Laws" , the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method described in Section 14.3 "Methods of Determining Reaction Order" required multiple experiments at different NO 2 concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions.

Figure 14.16 Properties of Reactions That Obey Zeroth-, First-, and Second-Order Rate Laws

Dinitrogen pentoxide (N 2 O 5 ) decomposes to NO 2 and O 2 at relatively low temperatures in the following reaction:

This reaction is carried out in a CCl 4 solution at 45°C. The concentrations of N 2 O 5 as a function of time are listed in the following table, together with the natural logarithms and reciprocal N 2 O 5 concentrations. Plot a graph of the concentration versus t , ln concentration versus t , and 1/concentration versus t and then determine the rate law and calculate the rate constant.

Given: balanced chemical equation, reaction times, and concentrations

Asked for: graph of data, rate law, and rate constant

A Use the data in the table to separately plot concentration, the natural logarithm of the concentration, and the reciprocal of the concentration (the vertical axis) versus time (the horizontal axis). Compare the graphs with those in Figure 14.16 "Properties of Reactions That Obey Zeroth-, First-, and Second-Order Rate Laws" to determine the reaction order.

B Write the rate law for the reaction. Using the appropriate data from the table and the linear graph corresponding to the rate law for the reaction, calculate the slope of the plotted line to obtain the rate constant for the reaction.

A Here are plots of [N 2 O 5 ] versus t , ln[N 2 O 5 ] versus t , and 1/[N 2 O 5 ] versus t :

The plot of ln[N 2 O 5 ] versus t gives a straight line, whereas the plots of [N 2 O 5 ] versus t and 1/[N 2 O 5 ] versus t do not. This means that the decomposition of N 2 O 5 is first order in [N 2 O 5 ].

B The rate law for the reaction is therefore

Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a first-order reaction is − k . We can calculate the slope using any two points that lie on the line in the plot of ln[N 2 O 5 ] versus t . Using the points for t = 0 and 3000 s,

Thus k = 4.820 × 10 −4 s −1 .

1,3-Butadiene (CH 2 =CH—CH=CH 2; C 4 H 6 ) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C 4 H 6 as a function of time at 326°C are listed in the following table along with ln[C 4 H 6 ] and the reciprocal concentrations. Graph the data as concentration versus t , ln concentration versus t , and 1/concentration versus t . Then determine the reaction order in C 4 H 6 , the rate law, and the rate constant for the reaction.

second order in C 4 H 6 ; rate = k [C 4 H 6 ] 2 ; k = 1.3 × 10 −2 M −1 ·s −1

For a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of − k . For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of − k . For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k .

Key Takeaway

• Plotting the concentration of a reactant as a function of time produces a graph with a characteristic shape that can be used to identify the reaction order in that reactant.

Conceptual Problems

Compare first-order differential and integrated rate laws with respect to the following. Is there any information that can be obtained from the integrated rate law that cannot be obtained from the differential rate law?

• the magnitude of the rate constant
• the information needed to determine the order
• the shape of the graphs

In the single-step, second-order reaction 2A → products, how would a graph of [A] versus time compare to a plot of 1/[A] versus time? Which of these would be the most similar to the same set of graphs for A during the single-step, second-order reaction A + B → products? Explain.

For reactions of the same order, what is the relationship between the magnitude of the rate constant and the reaction rate? If you were comparing reactions with different orders, could the same arguments be made? Why?

• For a given reaction under particular conditions, the magnitude of the first-order rate constant does not depend on whether a differential rate law or an integrated rate law is used.
• The differential rate law requires multiple experiments to determine reactant order; the integrated rate law needs only one experiment.
• Using the differential rate law, a graph of concentration versus time is a curve with a slope that becomes less negative with time, whereas for the integrated rate law, a graph of ln[reactant] versus time gives a straight line with slope = − k . The integrated rate law allows you to calculate the concentration of a reactant at any time during the reaction; the differential rate law does not.

The reaction rate increases as the rate constant increases. We cannot directly compare reaction rates and rate constants for reactions of different orders because they are not mathematically equivalent.

Numerical Problems

One method of using graphs to determine reaction order is to use relative rate information. Plotting the log of the relative rate versus log of relative concentration provides information about the reaction. Here is an example of data from a zeroth-order reaction:

Varying [A] does not alter the reaction rate. Using the relative rates in the table, generate plots of log(rate) versus log(concentration) for zeroth-, first- and second-order reactions. What does the slope of each line represent?

The table below follows the decomposition of N 2 O 5 gas by examining the partial pressure of the gas as a function of time at 45°C. What is the reaction order? What is the rate constant? How long would it take for the pressure to reach 105 mmHg at 45°C?

How To Calculate

How to calculate rate constant..

Rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances.

By finding out how fast products are made and what causes reactions to slow down we can develop methods to improve production.

Formula to calculate rate constant.

• r is the reaction rate.
• k(T) is the reaction rate constant that depends on temperature.
• [A] is the molar concentrations of substances A in moles per unit volume of solution.
• [B] is the molar concentrations of substances B in moles per unit volume of solution.
• m,n are the partial orders of reaction.

Suppose you are given the following figures that were extracted from the reaction of a certain substance.

• k(T) 1.0 x 10^2
• [A] 0.002 M
• [B] 0.001 M

So all we need to do here is plug in the value.

Therefore, the rate constant rate is 4 x 10^-4 M/s.

Related Articles

How to Calculate Molarity.

How to Calculate Osmolarity.

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Introduction

In many chemical reactions, it is crucial to know the rate constant (k), a measure of how quickly a reaction occurs. The rate constant helps scientists understand the specific conditions under which reactions take place and aids in predicting their outcomes. In this article, we will discuss what a rate constant is, how it is calculated, and the factors that affect its value.

Understanding the Rate Constant

The rate constant is an essential factor in determining the reaction rate, which measures the speed at which reactants are consumed or converted into products. It is mathematically expressed as k in our equations and can have various units depending on the reaction order.

The higher the value of k, the faster the reaction proceeds.

Calculating the Rate Constant Using Experimental Data

To calculate a rate constant, you need experimental data on the concentration of reactants and products at different time points throughout a reaction. Here are the steps to determine k for a given chemical reaction:

1. Determine the reaction order: First-order and second-order reactions are most common, but there can be higher-order reactions as well. The order affects how we calculate k. You can determine the order by analyzing how concentration changes with time.

2. Write a suitable rate equation: For example, if you have a first-order reaction A → B, your equation would be -d[A]/dt = k * [A].

3- Rearrange and integrate: Manipulate your equation to isolate either reactant or product concentration on one side; then integrate both sides over time t.

4. Solve for k: With your integrated equation, input experimental data at two points in time to solve for your unknown variable – rate constant k.

Considerations Affecting Rate Constants

Several factors can influence rate constants:

1- Temperature: Generally, higher temperature means faster reactions due to increased molecular motion and more frequent collisions between particles.

2- Reactant concentrations: The initial concentration of reactants can impact the reaction rate. Typically, a higher concentration results in faster reactions.

3- Catalysts: Catalysts increase the rate of reactions without being consumed, and their presence can significantly impact the value of k.

4- Solvent properties: A solvent’s viscosity, polarity, or acidity can affect the interactions between reactants and influence the rate constant.

Calculating a rate constant allows chemists to predict the speed at which reactions will occur and understand the specific conditions under which they progress. By determining the rate constant, we gain insight into how factors such as temperature, reactant concentration, catalysts, and solvent properties affect the overall reaction kinetics.

How to Calculate a Rate: A Comprehensive ...

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Chemistry Steps

General Chemistry

Chemical kinetics.

We mentioned in the previous post that the order of a reaction can be determined only by experiment. Most often, this experiment consists of measuring the initial rate of the reaction by changing the concentration of the reactant and monitoring how it affects the rate.

For example, the rate law for a hypothetical reaction where molecule A transforms into products can be written as:

A → Products

Rate = k [A] n

where k is the rate constant and n is the reaction order .

Our objective is to determine the reaction order by calculating the n from a set of experiments. Keep in mind that:

• If n = 0 , the reaction is zero-order, and the rate is independent of the concentration of A.
• If n = 1 , the reaction is first-order , and the rate is directly proportional to the concentration of A.
• If n = 2 , the reaction is second-order , and the rate is proportional to the square of the concentration of A.

Now, suppose we run three experiments, and the following data is obtained for the concentration-rate correlation:

In every experiment, the concentration of A is doubled, and what we see is that the rate of the reaction doubles as well. Therefore, the initial rate is directly proportional to the initial concentration, and thus, we have a first-order reaction :

Rate = k [A] 1

If it was a zero-order reaction , the following data for the concentration-rate relationship would have been obtained:

The data for a zero-order reaction indicates that the rate does not depend on the concentration of reactants.

For a second-order reaction , doubling the concertation quadrupoles the reaction rate, and therefore, we would expect the following data:

If the numbers are not obvious for determining how the rate changes with concentration, you can pick the data from any set of two experiments, write the rate law, and divide them to see how the rate changes.

For example, going back to the data for a first-order reaction , we can divide the rate of experiments 1 and 2:

\[\frac{{Rate\;2}}{{Rate\;1}}\;{\rm{ = }}\;\frac{{k{{[{{\rm{A}}_2}]}^{\rm{n}}}}}{{k{{[{{\rm{A}}_1}]}^{\rm{n}}}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.050}}\;M{\rm{/s}}}}{{{\rm{0}}{\rm{.025}}\,M{\rm{/s}}}}\]

\[\frac{{Rate\;2}}{{Rate\;1}}\;{\rm{ = }}\;\frac{{\cancel{k}{{{\rm{(0}}{\rm{.20}}\,\cancel{M}{\rm{)}}}^{\rm{n}}}}}{{\cancel{k}{{{\rm{(0}}{\rm{.10}}\,\cancel{M}{\rm{)}}}^{\rm{n}}}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.050}}\;\cancel{{M{\rm{/s}}}}}}{{{\rm{0}}{\rm{.025}}\,\cancel{{M{\rm{/s}}}}}}\]

\[\frac{{{{{\rm{(0}}{\rm{.20}}\,{\rm{)}}}^{\rm{n}}}}}{{{{{\rm{(0}}{\rm{.10}}\,{\rm{)}}}^{\rm{n}}}}}\;{\rm{ = }}\;2\]

2 n = 2, therefore,

Determining the Value of Rate Constant

To determine the value of the rate constant, write the rate law expression:

Rate = k [A]

Now, you can pick data from any experiment and plug the numbers into a rate law experiment. Let’s use the data from e xperiment 1 .

Rate 1 = k [A 1 ]

\[k\, = \,\frac{{{\rm{rat}}{{\rm{e}}_{\rm{1}}}}}{{\left[ {{{\rm{A}}_{\rm{1}}}} \right]}}\; = \;\frac{{0.025\,M/s}}{{0.10\,M}}\; = \;0.25\,{s^{ – 1}}\]

Let’s now do another example with a real reaction between carbon dioxide and hydrogen and determine the reaction order with respect to each reactant , the overall order , and the value of the rate constant .

Carbon dioxide, CO 2 , reacts with hydrogen to give methanol (CH 3 OH), and water.

CO 2 ( g ) + 3H 2 ( g ) ⇆ CH 3 OH( g ) + H 2 O( g )

In a series of experiments, the following initial rates of disappearance of CO 2 were obtained:

Determine the rate law and calculate the value of the rate constant for this reaction.

To determine the overall reaction order, we need to determine it with respect to both reactants. Let’s first determine the order in CO 2 . Find two experiments where the concentration of H 2 is kept constant while the concentration of CO 2 is changed . In experiments 1 and 2 , the concentration of CO 2 is doubled from 0.640 M to 1.28 M while the concentration of H 2  is kept at 0.220 M . We see from the table, that doubling the concentration of CO 2 quadruples the rate of the reaction (1.08 x 10 -2 ÷ 2.7 x 10 -3 = 4). Therefore, the reaction is   second-order in CO 2 .

Now, let’s find two experiments where the concentration of CO 2 is kept constant while that of H 2  is changed . In experiments 1 and 3 , the concentration of CO 2 is kept at 0640 M while the concentration of H 2  is doubled from 0.220 M to 0.440 M . We see from the table, that doubling the concentration of H 2 had doubled the reaction rate (5.4 x 10 -3 ÷ 2.7 x 10 -3 = 2 ). Therefore, the reaction is  first-order in H 2 .

The rate law, therefore, is:

Rate = k [CO 2 ] 2 [H 2 ]

And the overall order of the reaction is 2+1 = 3 – it is a third-order reaction .

To calculate the value of the rate constant , use the numbers from any experiment for the following equation:

\[k\; = \;\frac{{{\rm{rate}}}}{{{{\left[ {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right]}^2}\left[ {{{\rm{H}}_{\rm{2}}}} \right]}}\]

\[k\; = \;\frac{{{\rm{2}}{\rm{.7  \times  1}}{{\rm{0}}^{{\rm{ – 3}}}}\;\cancel{{\rm{M}}}{\rm{/s}}}}{{{{\left( {{\rm{0}}{\rm{.640}}\;{\rm{M}}} \right)}^{\rm{2}}}\left( {{\rm{0}}{\rm{.220}}\;\cancel{{\rm{M}}}} \right)}}\;{\rm{ = }}\;{\rm{3}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ – 2}}}}\;{{\rm{M}}^{{\rm{ – 2}}}}{{\rm{s}}^{{\rm{ – 1}}}}\;\]

Here is a 77-question, Multiple-Choice  Quiz on Chemical Kinetics:

Chemical Kinetics Quiz

• Reaction Rate
• Rate Law and Reaction Order
• Integrated Rate Law
• The Half-Life of a Reaction
• Determining the Reaction Order Using Graphs
• Units of Rate Constant k
• How Are Integrated Rate Laws Obtained
• Activation Energy
• The Arrhenius Equation
• Chemical Kinetics Practice Problems

Iron(II) ion is oxidized by hydrogen peroxide in an acidic solution.

2Fe 2 + ( aq ) + H 2 O 2 ( aq ) + 2H + ( aq ) → 2Fe 3+ ( aq ) + 2H2O( l )

The rate law for the reaction is determined to be rate = k [H 2 O 2 ][Fe 2 + ]. The rate constant, at certain temperature, is 2.56 x 10 24 / M · s. Calculate the rate of the reaction at this temperature if [H 2 O 2 ] = 0.48 M and [H 2 O 2 ] = 0.070 M .

For the kinetics of the reaction

 2NO( g ) + Cl 2 ( g ) → 2NOCl( g )

The following data were obtained:

a) What is reaction order in Cl 2 and NO?

b) What is the rate law?

c) What is the value of the rate constant?

The date for the initial rate of the following reaction is listed in the table below:

A + B → C + D

(a) What is the order of reaction with respect to A and to B?

(b) What is the overall reaction order?

(c) What is the value of the rate constant, k ?

Consider the reaction

A(g) + B(g) ⇌ C(g)

The following data were obtained at a certain temperature:

Using the data, determine the order of the reaction and calculate the rate constant:

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A + 2B ---> C + 2D Exp. Initial A (mol/L) Initial B (mol/L) Init. Rate of Formation of C (M min -1 ) 1 0.10 0.10 3.0 x 10 -4 2 0.30 0.30 9.0 x 10 -4 3 0.10 0.30 3.0 x 10 -4 4 0.20 0.40 6.0 x 10 -4

1) compare exp. 1 and exp. 3. A remains constant and B is tripled. The rate from 1 to 3 remains constant. Conclusion: B is not in the rate law expression. 2) compare exp 1 to exp 2. The concentration of A triples (and we don't care what happens to B). The rate triples. Conclusion: first order in A. 3) we can also show first order in A by comparing exp 1 to exp 4. The concentration of A doubles and the rate doubles. Remember, B is not part of the rate law, so we don't pay any attention to it at all. rate = k[A]

Trial 1: [A] = 0.50 M; [B] = 1.50 M; Initial rate = 4.2 x 10 -3 M/min Trial 2: [A] = 1.50 M; [B] = 1.50 M; Initial rate = 1.3 x 10 -2 M/min Trial 3: [A] = 3.00 M; [B] = 3.00 M; Initial rate = 5.2 x 10 -2 M/min

Look at trial 1 and trial 2. B is held constant while A triples. The result is that the rate triples. Conclusion: A is first order.

Look at trials 2 and 3. The key to this is that we already know that the order for A is first order. Both concentrations were doubled from 2 to 3 and the rate goes up by a factor of 4. Since A is first order, we know that a doubling of the rate is due to the concentration of A being doubled. So, we look at the concentration change for B (a doubling) and the consequent rate change (another doubling - remember the overall increase was a factor of 4 - think of 4 as being a doubled doubling). Conclusion: the order for B is first order. The rate law is rate = k [A] [B]

Exp. Initial A (mmol/L) Initial B (mmol/L) Init. Rate of Formation of products (mM min -1 ) 1 4.0 6.0 1.60 2 2.0 6.0 0.80 3 4.0 3.0 0.40

1) Look at experiments 2 and 1. From 2 to 1, we see that A is doubled (while B is held constant). A doubling of the rate with a doubling of the concentration shows that the reaction is first order with respect to A. 2) Now compare experiments 1 and 3. The concentration of A is held constant while the concentration of B is cut in half. When B is cut in half, the overall rate is cut by a factor of 4 (which is the square of 2). This shows the reaction is second order in B. 3) The rate law is this: rate = k [A] [B] 2 4) Note that the comparison in (2) can be reversed. Consider that the concentration of B is doubled as you go from exp. 3 to exp. 1. When the concentration is doubled, the rate goes up by a factor of 4 (which is 2 2 ). 5) We can use any set of values to determine the rate constant: rate = k [A] [B] 2 1.60 mM min -1 = k (4.0 mM) (6.0 mM) 2 k = 0.011 mM -2 min -2 the units on k can be rendered in this manner: k = 0.067 L 2 mmol -2 min -1

Exp. Initial A (mol/L) Initial B (mol/L) Init. Rate of Formation of products (M s -1 ) 1 0.040 0.040 9.6 x 10 -6 2 0.080 0.040 1.92 x 10 -5 3 0.080 0.020 9.6 x 10 -6

1) Examine exps. 2 and 3. A remains constant while B is doubled in concentration from 3 to 2. The result of this change is that the rate of the reaction doubles. We conclude that the reaction is first order in B. 2) Now we look at exps 1 and 2. B remains constant while the concentration of A doubles. As a result of the doubled concentration, the rate also doubles. Conclusion: the reaction is first order in A. 3) The rate law for this reaction is: rate = k [A] [B]

rate = k [A] [B] 9.6 x 10 -6 M s -1 = k (0.040 M) (0.040 M) k = 0.0060 M -1 s -1

rate = (0.0060 M -1 s -1 ) (0.12 M) (0.015 M) rate = 1.08 x 10 -5 M s -1

2ClO 2 (aq) + 2OH¯(aq) ---> ClO 3 ¯(aq) + ClO 2 ¯(aq) + H 2 O(l)

Determination #1: [ClO 2 ] o = 1.25 x 10¯ 2 M; [OH¯] o = 1.30 x 10¯ 3 M Initial rate for formation of ClO 3 ¯ = 2.33 x 10¯ 4 M s -1

Determination #2: [ClO 2 ] o = 2.50 x 10¯ 2 M; [OH¯] o = 1.30 x 10¯ 3 M Initial rate for formation of ClO 3 ¯ = 9.34 x 10¯ 4 M s -1

Determination #3: [ClO 2 ] o = 2.50 x 10¯ 2 M; [OH¯] o = 2.60 x 10¯ 3 M Initial rate for formation of ClO 3 ¯ = 1.87 x 10¯ 3 M s -1

1) Compare #1 and #2. The concentration of ClO 2 doubles (hydroxide remains constant) and the rate goes up by a factor of four (think of it as two squared). This means the reaction is second order with respect to ClO 2 . 2) Compare #2 and #3. The concentration of hydroxide is doubled while the [ClO 2 ] remains constant. The rate doubles, showing that the reaction is first order in hydroxide. 3) The rate law is: rate = k [ClO 2 ] 2 [OH¯] 4) Calculation for the rate constant: 1.87 x 10¯ 3 M s -1 = k (2.50 x 10¯ 2 M) 2 (2.60 x 10¯ 3 M) k = 1.15 x 10 3 M¯ 2 s -1 Often the rate constant unit is rendered thusly: L 2 mol -2 s -1 . Note that the overall order of the rate law is third order and that this is reflected in the unit associated with the rate constant.

rate = (1.15 x 10 3 M¯ 2 s -1 ) (8.25 x 10¯ 3 M) 2 (5.35 x 10¯ 2 M) rate = 4.19 x 10¯ 3 M s -1

Exp. [A] [B] Initial rate of formation of C 1 0.6 0.15 6.3 x 10 -3 2 0.2 0.6 2.8 x 10 -3 3 0.2 0.15 7.0 x 10 -4

rate 1 / rate 3 = k 1 [A 1 ] x [B 1 ] y / k 3 [A 3 ] x [B 3 ] y k's will cancel and [B] will cancel rate 1 / rate 3 = [A 1 ] x / [A 3 ] x 0.0063 / 0.0007 = (0.6) x / (0.2) x 9 = 0.6 x / 0.2 x 9 = 3 x rate = k[A] 2

Exp. [CH 3 CHO] [CO] Rate (M s -1) 1 0.30 0.20 0.60 2 0.10 0.30 0.067 3 0.10 0.20 0.067

1) Note experiments 2 and 3. The [CH 3 CHO] remains constant while the [CO] changes. However, no change of the reaction rate is observed. From this, we conclude that CO is not part of the rate law. 2) Examine experiment 3 compared to experiment 1. from 3 to 1, the [CH 3 CHO] triples and the rate goes up by a factor of 9 (which is 3 2 ). Conclusion: the reaction is second order in CH 3 CHO. 3) The rate expression is this: rate = k [CH 3 CHO] 2

Q + X --> products Trial [Q] [X] Rate 1 0.12 M 0.10 M 1.5 x 10 -3 M/min 2 0.24 M 0.10 M 3.0 x 10 -3 M/min 3 0.12 M 0.20 M 1.2 x 10 -2 M/min

1) Examine trials 1 and 2. [X] is held constant while [Q] is doubled (from 1 to 2). As a result, the rate doubles. We conclude that the reaction is first order in Q. 2) Examine trials 3 and 1. The concentration of Q does not change from 3 to 1 but the concentration for X is doubled. When this happens, we observe an eight-fold increase in the rate of the reaction. We conclude that the reaction is third order in X. This arises from the fact that 2 3 = 8. In more detail: rate 3 / rate 1 = k 3 [Q 3 ] x [X 3 ] y / k 1 [Q 1 ] x [X 1 ] y k's will cancel and [Q] will cancel rate 3 / rate 1 = [X 3 ] x / [X 1 ] x 0.012 / 0.0015 = (0.20) x / (0.10) x 8 = 0.20 x / 0.10 x 8 = 2 x x = 3

rate = k [Q] [X] 3 0.012 M min -1 = k (0.12 M) (0.20 M) 3 k = 12.5 L 3 mol -3 min -1

Trial [NO] o [O 2 ] o Initial reaction rate, M/s 1 0.020 0.010 0.028 2 0.020 0.020 0.057 3 0.040 0.020 0.227

1) Doubling O 2 doubles reaction rate (trials 2 and 1). 2) Doubling NO increases reaction rate 8x (trials 2 and 3). 3) Conclusion: first order O 2 , third order NO.

A + B ---> 2C

(a) rate = k [A] 2 [B] (b) rate = k [A] [B] (c) rate = k [A] [B] 2 (d) rate = k [A] 1/2 [B]

R = k[1] 0.5 = 1 R = k[2] 0.5 = 1.4 R = k[4] 0.5 = 2 R = k[8] 0.5 = 2.8 R = k[16] 0.5 = 4

The reaction rate is not doubled when you double the concentration. Notice that when the concentration changed from 2 to 8, the R went from 1.4 to 2.8. So when the concentration is quadrupled, the R doubles.

When the concentration doubles, the rate goes up by a factor which is the square root of two.