Ch.13 - Chemical KineticsSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds
Jules Bruno

Heat or thermal energy play a pivotal role in the rate of a chemical reaction. The Arrhenius equation represents the formula that shows the dependence of the rate and rate constant k on the absolute temperature, as well as other variables such as activation energy, and the pre-exponential factor. 



Rate of Reaction 

Generally, as we increase the temperature of a chemical reaction, molecules will absorb this new source of thermal energy. The result is that molecules will move more quickly, striking one another more frequently and with greater force. If molecule A can hit molecule B with enough energy and the correct orientation then they will successful combine and form a new molecule AB. 

Rate-TemperatureRate and Temperature

From the chart provided, as the temperature increases the rate of formation for molecule AB increases. The rule of thumb is that a rise of 10 °C will double the rate of reaction. 

Arrhenius Equation

Although it was well known that increases in temperature led to increases in rate, it wasn’t until the Arrhenius Equation developed by Svante Arrhenius that this temperature-rate relationship could be accurately explained. 

Arrhenius-Equation-rate-coefficientsArrhenius Equation

When examining the variables of the Arrhenius Equation: k is the rate constant and, Ea is the activation energy normally in J/mol or kJ/mol, T is the absolute Temperature in Kelvin, R is the universal gas constant equal to 8.314 J/mol K, and A is the pre-exponential factor, frequency factor or Arrhenius constant. 

Arrhenius-Equation-Components-collision-rates-calculator-vant-HoffArrhenius Equation Components

Since the pre-exponential factor is a constant based on the given chemical reaction and the universal gas constant is always the same value, the way to affect k is based on changing Ea or T. Based on the Arrhenius equation, increasing the rate constant k can be accomplished by either increasing the temperature or decreasing the activation energy, usually through the use of a catalyst.

Let’s approach a question using the Arrhenius equation.

PRACTICE: The rate constant of a reaction at 33.0 °C was measured to be 5.2 × 10−2 s−1. If the frequency factor is 1.2 × 1013 s−1, what is the activation barrier?

STEP 1: Identify the given variables. 

Arrhenius-Equation-Identifying-VariablesArrhenius Equation (Identifying Variables)

STEP 2: Convert the temperature given in Celsius to Kelvin. 

Arrhenius-Equation-Temperature-ConversionArrhenius Equation (Temperature Conversion)

STEP 3: Multiply both sides of the equation by the natural logarithm, ln

Arrhenius-Equation-Natural-LogArrhenius Equation & Natural Log

When multiplying the Arrhenius equation by ln we obtain: 

Derivation-Arrhenius-EquationDerivation of Arrhenius Equation

Rearranging this new formula gives us the straight-line form: 

Arrhenius-Equation-Straight-Line-FormArrhenius Equation (Straight-Line Form)

STEP 4: Plug in the given values into straight-line form of the Arrhenius Equation. 

Arrhenius-Equation-Plugging-ValuesArrhenius Equation (Plugging in values)

Solving for some of the portions simplifies it to: 

Arrhenius-Equation-SimplifyingArrhenius Equation (Simplification 1)

STEP 5: Subtract 30.1158 from both sides. 

Arrhenius-Equation-SubtractingArrhenius Equation (Subtracting)

This simplifies the equation to: 

Arrhenius-Equation-Simplification-2Arrhenius Equation (Simplification 2)


STEP 6: Multiply both sides by 8.314. 

Arrhenius-Equation-Multiplying-by-8.314Arrhenius Equation (Multiplying by 8.314)

The equation is now: 

Arrhenius-Equation-Simplication-3Arrhenius Equation (Simplification 3)

STEP 6: Divide both sides by 0.003266. 

Arrhenius-Equation-Divide-by-0.003266Arrhenius Equation (Divide by 0.003266)

This gives the value of Ea as:

Arrhenius Equation (Isolating Ea)

Arrhenius Equation: Plot-wise approach 

As we’ve shown before, the Arrhenius equation can be written in a non-exponential form or straight-line form.  

Arrhenius-Equation-Non-Exponential-FormArrhenius Equation (Non-Exponential Form)

This form of the Arrhenius equation allows us to interpret our values graphically. 

lnk-1/T-plot-modelLn k vs. 1/T Plot

By setting our y-axis as ln k and our x-axis as the inverse temperature we can correlate the non-exponential form of the Arrhenius equation to a straight-line. 

Arrhenius-EquationArrhenius Equation (y = mx + b)

By using the straight-line or non-exponential form of the Arrhenius equation we can determine either the frequency factor or the activation energy of a chemical reaction. 


PRACTICE: The rate constant for a reaction was measured as a function of temperature. Based on the plot of ln k versus 1/T, what is the activation energy for the reaction?

Arrhenius-Equation-Plotted-GraphArrhenius Equation & Plotted Graph

STEP 1: Using the equation for the straight line from the chart provided, apply it to the non-exponential form of the Arrhenius equation. 

Arrhenius-Equation-Given-step-by-stepArrhenius Equation & Given Equation

STEP 2: Isolate the slope of the straight line and set it equal to – Ea/R. 

Slope-activation-energySlope & Activation Energy

Since the universal gas constant R is equal to 8.314 J/mol K, solve for the activation energy of the chemical reaction. 

STEP 3: Multiply both sides by – 8.314 to isolate the activation energy. 

Isolating-Ea-Straight-line-formSlope & Isolating Ea

This gives Ea as: 

Ea-J/mol-kJ/molSolve for Ea (J/mol)

Arrhenius Equation: 2-Point Form 

The 2-point form of the Arrhenius Equation is a way mathematically calculating the changes in the rate constant k as the temperature changes. Recall that an increase in temperature causes an increase in the rate constant. 

Arrhenius-Equation-2-point-form-equilibrium-constantsArrhenius Equation (2-Point Form)

Depending on the placement of T1 and T2 the 2-point form of the Arrhenius equation can be written in two different ways. Algebraically either form can be used for solving. 


From the 2-point form of the Arrhenius Equation: 1 and k2 are the rate constants and, Ea is the activation energy normally given in J/mol or kJ/mol, T1and T2 are the absolute Temperatures in Kelvin, R is the universal gas constant equal to 8.314 J/mol K. 


PRACTICE: The activation energy for a reaction is 37.6 kJ/mol. The rate constant for the reaction is 5.4 x 10-3 s-1 at 45 °C. Calculate the rate constant at 145 °C.

STEP 1: Identify the given variables. 

Arrhenius-2-Point-Form-Identifying-VariablesArrhenius 2 -Point Form (Identifying variables)

STEP 2: Convert the temperatures given in Celsius to Kelvin. 

Arrhenius-2-Point-Form-Temperature-ConversionsArrhenius 2-Point Form (Temperature Conversions)

STEP 3: Since the universal gas constant R uses joules in its units then the activation energy Ea must be converted from kJ/mol to J/mol. 

Ea-Unit-ConversionEa (Unit Conversion)

STEP 4: Plug the given variables into the equation. 

Arrhenius-2-Point-Form-Plugging-VariablesArrhenius 2-Point Form (Plugging variables)

STEP 5: Solve the portions of the equation with values into order to simplify. 

Arrhenius-2-Point-Form-Simplification-1Arrhenius 2-Point Form (Simplification 1)

STEP 6: Combine the integers on the right side by multiplying them together. 

Arrhenius-2-Point-Form-Simplification-2Arrhenius 2-Point Form (Simplification 2)

STEP 7: Take the inverse e of the natural logarithm to both sides of the equation to further isolate k2

Arrhenius-2-Point-Form-Simplification-3Arrhenius 2-Point Form (Simplification 3)

When you take the inverse of the natural logarithm you eliminate the ln on the left side. 

Arrhenius-2-Point-Form-Simplification-4Arrhenius 2-Point Form (Simplification 4)

This simplifies the expression to: 

Arrhenius-2-Point-Form-Simplification-5Arrhenius 2-Point Form (Simplification 5)

STEP 8: Solve for the right side of the equation. 

Arrhenius 2-Point Form (Simplification 6)

STEP 9: Multiply both sides by 5.4 x 10-3 s-1

Arrhenius-2-Point-Form-Multiplying-by-5.4Arrhenius 2-Point Form (Multiplying by 5.4 x 10-3)

STEP 10: Isolate the value for k2

Arrhenius 2-Point Form (Isolating k2)

The Arrhenius equations form an integral part in our study of Chemical Kinetics and give a mathematical view of how molecules successful react to form products. Other important concepts dealing with the rates of chemical reactions include general raterate law, the integrated rate lawsenergy diagrams and reaction mechanisms


Jules Bruno

Jules felt a void in his life after his English degree from Duke, so he started tutoring in 2007 and got a B.S. in Chemistry from FIU. He’s exceptionally skilled at making concepts dead simple and helping students in covalent bonds of knowledge.


Additional Problems
Which of the following reactions would you predict to have the largest orientation factor? A) NOCl(g) + NOCl(g) → 2NO(g) + Cl 2(g) B) Br2(g) + H2C=CH2(g)  →  BrH2C-CH2Br(g) C) NH3(g) + BCl3(g)  →  H3N-BCl3(g) D) H(g) + F(g) →  HF(g) E) All of these reactions should have nearly identical orientation factors.
When a reaction is run at 150.0°C, data shows that k = 28.68 s  -1 and Ea = 36.25 kJ/mol. What temperature would the reaction have to be run in order for k to be 1521 s-1?  a. 234.8 °C b. 274.1 °C c. 414.9 °C d. 173.7 °C e. -1.11 °C
A first-order reaction proceeds with a rate constant of 1.5x10  -5 s-1 at 25°C and 2.7x10-3 s-1 at 75°C. According to the Arrhenius equation, what is the activation energy of this reaction in kJ/mol? A. 89.5 kJ/mol B. 1.30 kJ/mol C. 16.2 kJ/mol D. 95.8 kJ/mol
Each line in the graph below represents a different reaction. List the reactions in order of increasing activation energy. A) A< B < C B) C < A < B C) B < A < C D) C < B < A E) this graph is only for first order rxns
For reaction, what generally happens if the temperature is increased? A. a decrease in k occurs, which results in a faster rate. B. a decrease in k occurs, which results in a slower rate. C. an increase in k occurs, which results in a faster rate. D. an increase in k occurs, which results in a slower rate. E. there is no change with k or the rate.
The first-order rearrangement of CH3NC is measured to have a rate constant of 3.61 x 10-15 s-1 at 298 K and a rate constant of 8.66 x 10-7 s-1 at 425 K. Determine the activation energy for this reaction. A. 160. kJ/mol B. 240 kJ/mol C. 417 kJ/mol D. 127 kJ/mol E. 338 kJ/mol
A plot of k versus 1/T has a slope of -7,445. What is E a, for this reaction? A, 90.79 kJ/mol B. 895.5 J/mol C. 304.7 kJ/mol D. 61.90 kJ/mol E. 610.5 J/mol
The reaction CO2(g) + H2(g) → CO(g) + H2O(g) has an activation energy of 62 kJ/mol. If a catalyst is used the activation energy is lowered to 29 kJ/mol. By how much does the reaction rate at 600°C increase in the presence of the catalyst? (Assume the frequency factors for the catalyzed and uncatalyzed reaction are the same.) a) By a factor of 9.23x104 b) By a factor of 94.3 c) By a factor of 1.28x106 d) By a factor of 746 e) By a factor of 1.005  
A reaction is found to have a rate constant of 12.5 s -1 at 25.0 Celsius. When you heat the reaction up by ten degrees Celsius, the rate of the reaction exactly doubles. What is the activation energy for this reaction in kJ/mol? A. 52.9 B. 59.6 C. 62.7 D. 140 E. 523
The rate constant for a reaction was measured as a function of temperature. A plot of ln k versus 1/T is linear and gave a slope equal to -7210. What is the activation energy for the reaction?  
The addition of a catalyst to a chemical reaction helps to lower its activation energy, which then increases the rate of the reaction. If a catalyst lowers the activation barrier from 150 kJ/mol to 42 kJ/mol, by what factor would the reaction rate increase at a temperature of 30°C? Assume that the frequency factor will remain constant. 
At 298 K, the decomposition of ammonia is catalyzed by tungsten. This activation energy is modified by the use of a catalyst, so as to make the reaction run faster. More precisely, the catalyst brings the activation energy down from 335 kJ/mol to 163 kJ/mol. Assuming that the Arrhenius entropic prefactor is the same for the catalyzed and uncatalyzed reactions, which of the following statements is true with reference to the rate constant of the catalyzed and uncatalyzed reactions, at 298 K? a) The rate constant for the catalyzed reaction is 1.07 times larger than the rate constant for the uncatalyzed reaction. b) The rate constant for the catalyzed reaction is 0.933 times the rate constant for the uncatalyzed reaction. c) The rate constant for the catalyzed reaction is 7×10−31 times larger than the rate constant for the uncatalyzed reaction. d) The rate constant for the catalyzed reaction is equal to the rate constant for the uncatalyzed reaction. e) The rate constant for the catalyzed reaction is 1.4×1030 times larger than the rate constant for the uncatalyzed reaction.
A reaction is found to have a rate constant of 12.5 s -1 at 25.0 degrees Celsius. When you heat the reaction up by ten degrees Celsius, the rate of the reaction exactly doubles. What is the activation energy for this reaction in kJ/mol? A. 52.9 B. 59.6 C. 62.7 D. 140 E. 523
The rate constant for the decomposition of C 4H4 to C2H2 is 7.2x104 s−1 at 75°C. If the activation energy is 87kJ/mol, what temperature is required to increase the rate constant by a factor of three.   a) 528 K b) 472 K c) 744 K d) 633 K e) 361 K  
A series of experiments were performed to determine the rate constant for the reaction  2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g) at different temperatures. A plot of ln k against 1/T was a straight line with a slope of −1.5275x104 and intercept of 23.6. What is the frequency factor? a) 196.2 b) 23.6 c) 1.5275x104 d) 5.63x10−11 e) 1.78x1010
If a reaction has a frequency factor of 1.2x10 11 s−1 and an activation energy of 62.3 kJ/mol, what is the rate constant at 67°C? a) 32.2 s−1 b) 1.17x10−11 s−1 c) 3.21x10−38  s−1 d) 1.07x10−11  s−1 e) 2.68x10−10 s−1  
Given that the rate constant for a reaction is 1.5×10 10 at 25°C and the activation energy is 38 kJ/mol, what is its rate constant at 37°C?   a) 1.5×1010 b) 2.7×1010 c) 1.6×1010 d) 2.1×10 36 e) 8.3×10 9
Using the graph given below, determine the activation energy (E a). R = 8.314 J/mol K a) 166.3 kJ/mol b) 11.6 kJ/mol c) 83.1 kJ/mol d) 130.2 J/mol
Fill in the blanks below: a. For a given reaction, as temperature increases, k ________ , E a _________ and the rate ________. (increases, decreases, stays the same). b. If you know the rate constant at two different temperatures, how can you determine the activation energy for a reaction?
A certain reaction has an activation energy of 35.87 kJ/mol. If the rate constant of this reaction is 8.53 x 10-3 s-1 at 356.8oC, determine the value of the rate constant at 402.9oC A. 8.53 x 10-3 s-1 B. 1.36 x 10-2 s-1 C. 9.81 x 10-3 s-1 D. 3.40 x 10-2 s-1 E. 8.94 x 10-3 s-1
The Arrhenius equation describes the relationship between the rate constant,  k, and the energy of activation, Ea. k = Ae–Ea / RT In this equation, A is an empirical constant, R, is the ideal-gas constant, e is the base of natural logarithms, and T is the absolute temperature. According to the Arrhenius equation, (A) at constant temperature, reactions with lower activation energies proceed more rapidly. (B) at constant temperature, reactions with lower activation energies proceed less rapidly. (C) at constant energy of activation, reactions at lower temperatures proceed more rapidly. (D) at constant energy of activation, reactions with smaller values of A proceed more rapidly.
The following reaction has an activation energy of 262 kJ/mol.C4H8(g) → 2C2H4(g)At 600.0 K the rate constant is 6.1 x 10-8 s-1. What is the value  of the rate constant at 725.0 K?
The rate constant of a reaction at 33°C was measured to be 5.2 × 10 −2 /s . If the frequency factor is 1.2 × 1013/s, what is the activation barrier?
The rate constant of a first-order reaction is 4.60 x 10 -4 s-1 at 350 °C. If the activation energy is 104 kJ/mol, calculate the temperature at which its rate constant is 8.80 x 10-4 s-1.
The following reaction has an activation energy of 262 kJ/mol.C4H8(g) → 2C2H4(g)At 600.0 K, the rate constant is 6.1 x 10 -8 s-1.What is the value of the rate constant at 720.0 K.K= ? answer in s-1 
An experiment was run at a temperature of T= 496 K and the following initial rates were observed, after which the reaction was then run at a temperature of T = 298 K. From the choices below, what is the calculated Ea?
The activation energy for a reaction is 37.6 kJ/mol. The rate constant for the reaction is 5.4 x 10 −3 s −1 at 45 °C. Calculate the rate constant at 145 °C. A) 5.5 × 10 −3 s −1 B) 0.56 s −1 C) 0.16 s −1 D) 8.4 × 10 −3 s −1 E) 0.38 s −1
A reaction has a rate constant of 1.24 × 10 −4/s at 29°C and 0.229/s at 75°C.a. Determine the activation barrier for the reaction.b. What is the value of the rate constant at 18°C?
The following data show the rate constant of a reaction measured at several different temperatures.                    Temperature(K)                              Rate Constant (1/s)                            300                                                 3.37×10-3                            310                                                 1.08×10-2                            320                                                 3.21×10-2                            330                                                 8.96×10-2                            340                                                 0.235 a. Use an Arrhenius plot to determine the activation barrier for the reaction. b. Use an Arrhenius plot to determine frequency factor for the reaction.
At what temperature will the rate constant for the reaction H2 (g) + I2 (g) -->HI (g)  have the value 5.3x10-3 M-1 s-1? (Assume k = 5.4 times 10-4 M-1 s-1 at 599 K and k = 2.8x10-2 M-1 s-1 at 683 K.) 
The rate of a certain reaction was studied at various temperatures. The table shows temperature (T) and rate constant (k) data collected during the experiments. Plot the data, and then answer the following questions. What is value of the activation energy, E a, for this reaction? What is the value of the pre-exponential factor (sometimes frequency factor), A, for this reaction?
The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperature, the chemical reactions associated with the cooking of food take place at a greater rate. (a) Some food cooks fully in 6.00 min in a pressure cooker at 120.0°C and in 48.0 minutes in an open pot at 100.0°C. Calculate the average activation energy for the reactions associated with the cooking of this food. kJ mol-1 (b) How long will the same food take to cook in an open pot of boiling water at an altitude of 8500 feet, where the boiling point of water is 91.4°C?
The following data show the rate constant of a reaction measured at several different temperatures. Part A Use an Arrhenius plot to determine the activation barrier for the reaction. Express your answer using three significant figures. Part B Use an Arrhenius plot to determine the frequency factor for the reaction. Express your answer using two significant figures.
The rate constant of a first-order reaction is 4.35 times 10-4 s- at 350. °C. If the activation energy is 137 kJ/mol, calculate the temperature at which its rate constant is 9.30x10-4 S-1.
The rate at which tree crickets chirp is 2.10x102 per minute at 28 ° C but only 36.4 per minute at 0 °C. From these data, calculate the activation energy for the chirping process.
The rate constant (k) for a reaction was measured as a function of temperature. A plot of ln k versus 1/T (in K) is linear and has a slope of -1.16 x 104 K. You may want to reference Section 14.5 while completing this problem. Calculate the activation energy for the reaction. Express your answer using three significant figures with the appropriate units.
A certain reaction has an activation energy of 55.54 kJ/mol. At what Kelvin temperature will the reaction proceed 7.50 times faster than it did at 321 K?
The activation energy of an uncatalyzed reaction is 90 kJ/mol. The addition of a catalyst lowers the activation energy to 52 kJ/mol. Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 15°C? Express your answer using two significant figures.Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 137°C?Express your answer using two significant figures.
A certain reaction has an activation energy of 32.89 kJ/mol. At what Kelvin temperature will the reaction proceed 7.00 times faster than it did at 287 K?  
A certain reaction has an activation energy of 26.98 kJ/mol. At what Kelvin temperature will the reaction proceed 4.50 times faster than it did at 321 K?
If Kc = 0.400 at 40°C and Kc = 0.635 at 90°C, what is ΔH° for the reaction? X ⇌ Y
Enter your answer in the provided box. The rate constant of a first-order reaction is 2.20 x 10-4 s-1 at 350°C. If the activation energy is 137 kJ/mol, calculate the temperature at which its rate constant 19.90 x 10-4 s-1.
Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the following reaction equation. 2H2O2 (aq) → 2H2O (l) + O2 (g) The activation energy for this reaction is 75 kJ mol-1. The enzyme catalase (found in blood) lowers the activation energy to 8.0 kJ mol-1. At what temperature would the non-catalyzed reaction need to be run to have a rate equal to that of the enzyme-catalyzed reaction at 25 °C?
Consider the reaction O (g) + N2 (g) → NO (g) + N (g) At 800°C, the rate constant is 9.7 x 1010 L/(mol s). The activation energy is 315 kJ/mol. Calculate the rate constant for this reaction at 500°C.
QUESTION 11The rate constant is constant at all temperatures.            a. True            b. False QUESTION 12Consider the following reaction at equilibrium, 2NH3 (g) ⇌ N2 (g) + 3H2 (g). If the concentration of hydrogen gas is increased,           a. then the equilibrium will shift to the right.           b. then the forward and reverse reaction will speed up.           c. nothing happens.           d. then the equilibrium will shift to the left.
Enter your answer in the provided box.Given the same reactant concentrations the reaction CO (g) + Cl2 (g) → COCl2 (g) at 229°C is 47.1 times as fast as the same reaction at 155°C. Calculate the activation energy for this reaction. Assume that the frequency factor is constant.
Rate constants for the reaction NO2 (g) + CO (g) → NO (g) + CO2 (g) are 1.3 M-1 s-1 at 700 K and 23.0 M-1 s-1 at 800 K. Part AWhat is the value of the activation energy in kJ/mol? Part BWhat is the rate constant at 730 K? Express your answer using two significant figures
A certain reaction has an activation energy of 63.32 kJ/mol. At what Kelvin temperature will the reaction proceed 7.00 times faster than it did at 339 K?
At 552.3 K, the rate constant for the thermal decomposition of SO 2Cl2 is 1.02 x 10-6 s-1. If the activation energy is 210 kJ/mol, calculate the Arrhenius pre-exponential factor and determine the rate constant at 600 K.
The activation energy of a certain reaction is 34.5 Kj/mol. At 21°C, the rate constant is 0.0190s-1  . At what temperature at degrees Celsius would this reaction go twice as fast.Given the initial rate constant is 0.0190s-1 at an initial temperature of 21°C what would the rate constant be at a temperature of 120 °C for the same reaction described in Part A?
The following reaction has an activation energy of 262 kJ/mol. C4H8 (g) → 2C2H4 (g) At 600.0 K the rate constant is 6.1 x 10-8 s-1. What is the value of the rate constant at 735.0 K?
A particular first-order reaction has a rate constant of 1.35 x 102s-1 at 25.0°C. What is the magnitude of k at 41.0°C if Ea = 55.5 kJ/mol.(a) 2.00 x 1047 (b) 154 (c) 4.23 x 102 (d) 1 23 x 104 (e) 1.35 x 102
The following data show the rate constant of a reaction measured at several different temperatures. Part AUse an Arrhenius plot to determine the activation barrier for the reaction. Express your answer using three significant figures. Part BUse an Arrhenius plot to determine the frequency factor for the reaction. Express your answer using two significant figures. 
We know that for a given reaction when the temperature increases from 100 K to 200 K the rate constant doubles. What is the activation energy in kJ/mol?
A reaction has a rate constant of 1.28 x 10-4 s-1 at 28°C and 0.227 s-1 at 76°C. Part ADetermine the activation barrier for the reaction.Express your answer in units of kilojoules per mole. 
Question 1The activation energy for the decomposition of N2O4 is 57.2. kJ/mol. If the rate constant for the reaction is 3.4 x 104s-1 at 25 °C, what is the rate constant at 75 °C? 
The high temperature in an automobile engine cause nitrogen and oxygen to from noxius nitrogen monoxide. Use the following data on temperature dependence of the rate constant to determine the activation energy for this reaction.
A certain reaction has an activation energy of 25.36 kJ/mol. At what Kelvin temperature will the reaction proceed 3.00 times faster than it did at 357 K?
The rate constant of a reaction is 9.25 x 10-5 L/mol • s at 195°C and 7.65 x 10-3 L/mol • s at 258°C. What is the activation energy of the reaction? Enter your answer in scientific notation.
Use the following information to determine the value of Ea for a typical reaction: 
A certain reaction has an activation energy of 74.13 kJ/mol. At what Kelvin temperature will the reaction proceed 4.50 times faster than it did at 333 K? 
Calculate the rate constant ar 200 C for a reaction that has a rate constant of  9.50x10-4 s-1 at 90°C and an activation energy of 52.2 KJ/mol. 
For the first-order reaction N2O5 (g) --> 2NO2 (g) + 1/2 O2 (g) t1/2 = 22.5 h at 20 ºC and 1.5 h at 40 ºC. Calculate the activation energy of this reaction. Ea = kJ/mol If the Arrhenius constant A = 2.05 times 1013 s-1, determine the value of k at 29 ºC. k = s-1
The following reaction has an activation energy of 262 kJ/mol. C4H8(g) → 2C4H4(g) At 600.0K the rate constant is 6.1 x 10-8 s-1. What is the value of the rate constant at 835.0K?