Ch.13 - Chemical KineticsWorksheetSee 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 enzyme urease catalyzes the reaction of urea, (NH2CONH2), with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of 4.15 10-5-1 at 100 oC. In the presence of the enzyme in water, the reaction proceeds with a rate constant of 3.4 104-1 at 21 oC.If enzymatic reactions could actually be carried out at high temperatures, what would you expect for the rate of the catalyzed reaction at 100 oC as compared to that at 21 oC?
A reaction has a rate constant of 1.19×10−2 /s at 400. K and 0.690 /s at 450. K.Determine the activation barrier for the reaction.
A reaction has a rate constant of 1.19×10−2 /s at 400. K and 0.690 /s at 450. K.What is the value of the rate constant at 425 K?
If a temperature increase from 12.0 oC to 22.0 oC doubles the rate constant for a reaction, what is the value of the activation barrier for the reaction?
At 28 oC, raw milk sours in 4 h but takes 50 h to sour in a refrigerator at 5 oC.Estimate the activation energy in kJ/mol for the reaction that leads to the souring of milk.
The following is a quote from an article in the August 18, 1998, issue of The New York Times about the breakdown of cellulose and starch: "A drop of 18 degrees Fahrenheit [from 77 oF to 59 oF] lowers the reaction rate six times; a 36-degree drop [from 77 oF to 41 oF] produces a fortyfold decrease in the rate."Calculate activation energy for the breakdown process based on the second estimate of the effect of temperature on rate.
The enzyme urease catalyzes the reaction of urea, (NH2CONH2), with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of 4.15 10-5-1 at 100 oC. In the presence of the enzyme in water, the reaction proceeds with a rate constant of 3.4 104-1 at 21 oC.If the rate of the catalyzed reaction were the same at 100 oC as it is at 21 oC, what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions?
Calculate the fraction of atoms in a sample of argon gas at 401 K that have an energy of 11.6 kJ/mol or greater.
The activation energy for the isomerization of methyl isonitrile is 160 kJ/mol.Calculate the fraction of methyl isonitrile molecules that have an energy of 160.0 kJ or greater at 501 K .
Suppose that, in the absence of a catalyst, a certain biochemical reaction occurs x times per second at normal body temperature (37 oC). In order to be physiologically useful, the reaction needs to occur 5500 times faster than when it is uncatalyzed.By how many kJ/mol must an enzyme lower the activation energy of the reaction to make it useful?
The following data show the rate constant of a reaction measured at several different temperatures. Temperature (K) Rate Constant (1/s) 300 2.15×10−2 310 6.55×10−2 320 0.186 330 0.495 340 1.24 Use an Arrhenius plot to determine the activation barrier for the reaction.
The following data show the rate constant of a reaction measured at several different temperatures. Temperature (K) Rate Constant (1/s) 300 2.15×10−2 310 6.55×10−2 320 0.186 330 0.495 340 1.24 Use an Arrhenius plot to determine frequency factor for the reaction.
The following data show the rate constant of a reaction measured at several different temperatures. Temperature (K) Rate Constant (1/s) 310 0.190 320 0.544 330 1.46 340 3.70 350 8.89 Use an Arrhenius plot to determine the activation barrier for the reaction.
The accompanying graph shows plots of ln k versus 1/T for two different reactions. The plots have been extrapolated to the y-intercepts. Which reaction has the larger value for Ea?
The accompanying graph shows plots of ln k versus 1/T for two different reactions. The plots have been extrapolated to the y-intercepts. Which reaction has the larger value for the frequency factor, A?
What is an Arrhenius plot? Explain the significance of the slope and intercept of an Arrhenius plot.
Which reaction do you expect to have the smallest orientation factor?
Reaction A and reaction B have identical frequency factors. However, reaction B has a higher activation energy than reaction A.Which reaction has a greater rate constant at room temperature?
Explain the meaning of each term within the Arrhenius equation: activation energy, frequency factor, and exponential factor. Use these terms and the Arrhenius equation to explain why small changes in temperature can result in large changes in the reaction rate.
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?
If a temperature increase from 21.0 oC to 36.0 oC triples the rate constant for a reaction, what is the value of the activation barrier for the reaction?
Assume that water boils at 100.0°C in Houston (near sea level), and at 90.0°C in Cripple Creek, Colorado (near 9500 ft). If it takes 4.8 min to cook an egg in Cripple Creek and 4.5 min in Houston, what is Ea for this process?
The reaction 2 N2O5 → 2 N2O4 + O2 takes place at around room temperature in solvents such as CCl4. The specific rate constant at 293 K is found to be 2.35 x 10–4 s–1 and at 303 K the specific rate constant is found to be 9.15 x 10–4 s–1.Calculate the frequency factor for the reaction at 303 K.
The following is a quote from an article in the August 18, 1998, issue of  The New York Times about the breakdown of cellulose and starch: "A drop of 18 degrees Fahrenheit [from 77 oF to 59 oF] lowers the reaction rate six times; a 36-degree drop [from 77 oF to 41 oF] produces a fortyfold decrease in the rate."Calculate activation energy for the breakdown process based on the first estimate of the effect of temperature on rate. Are the values consistent?
Ethyl chloride vapor decomposes by the first-order reaction C2H5Cl → C2H4 + HCl. The activation energy is 249 kJ/mol and the frequency factor is 1.61014 s-1.Find the temperature at which the rate of the reaction would be twice as fast.
The rate constant for a reaction at 18.0 oC is 0.010 s-1, and its activation energy is 35.8 kJ. Find the rate constant at 56.0 oC.
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.
The rate constant of a reaction is 4.7 x 10-3 s-1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C?
The rate constant of a reaction is 4.50×10−5 L/mol•s at 195°C and 3.20 x 10-3 L/mol•s at 258°C. What is the activationenergy of the reaction?
Understanding the high-temperature formation and breakdown of the nitrogen oxides is essential for controlling the pollutants generated from power plants and cars. The first-order breakdown of dinitrogen monoxide to its elements has rate constants of 0.76/s at 727°C and 0.87/s at 757°C. What is the activation energy of this reaction?
You may want to reference (Pages 568 - 621) Chapter 14 while completing this problem.The reaction H2O2(aq) → H2O(l) + 1/2 O2(g) is first order. At 300 K the rate constant equals 7.0 x 10–4 s–1. If the activation energy for this reaction is 75 kJ/mol, at what temperature would the reaction rate be doubled?
A reaction has a rate constant of 1.25×10−4 s-1 at 28oC and 0.230 s-1 at 78oC. Determine the activation barrier for the reaction.
A reaction has a rate constant of 1.25×10−4 s-1 at 28oC and 0.230 s-1 at 78oC. What is the value of the rate constant at 17  oC ?
The reaction between nitrogen dioxide and carbon monoxide is NO2(g) + CO(g) NO(g) + CO2(g). The rate constant at 701 K is measured as 2.57 M-1s-1 and that at 895 K is measured as 567 M-1s-1.Use the value of the activation energy (Ea = 1.50 x 102 kJ/mol) and the given rate constant of the reaction at either of the two temperatures to predict the rate constant at 551 K.        
The activation energy for the decomposition of HI(g) to H 2(g) and I2(g) is 186 kJ/mol. The rate constant at 555 K is 3.52 x 10-7 L/mol•s. What is the rate constant at 645 K?
A first-order reaction has rate constants of 4.6 x 10 -2 s-1 and 8.1 x 10-2 s-1 at 0°C and 20.°C, respectively. What is the value of the activation energy?
A certain reaction has an activation energy of 54.0 kJ/mol. As the temperature is increased from 22°C to a higher temperature, the rate constant increases by a factor of 7.00. Calculate the higher temperature.
Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to 35°C?
The rate constant for the decomposition of acetaldehyde, CH3CHO, to methane, CH4, and carbon monoxide, CO, in the gas phase is 1.1 × 10−2 L/mol/s at 703 K and 4.95 L/mol/s at 865 K. Determine the activation energy for this decomposition.
An elevated level of the enzyme alkaline phosphatase (ALP) in human serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 °C to 37 °C. What is the activation energy for the ALP–catalyzed conversion of PNPP to PNP and phosphate?
The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:
A certain compound, A, reacts to form products according to the reaction A → P. The amount of A is measured as a function of time under a variety of different conditions and the tabulated results are shown here:Time (s)25.0 oC [A] (M)35.0 oC [A] (M)45.0 oC [A] (M)01.0001.0001.000100.7790.6620.561200.5910.4610.312300.4530.3060.177400.3380.2080.100500.2590.1360.057600.2000.0930.032What is the activation energy for this reaction?
The hydrolysis of the sugar sucrose to the sugars glucose and fructose, C12H22O11 + H2O ⟶ C6H12O6 + C6H12O6 follows a first-order rate equation for the disappearance of sucrose: Rate = k[C12H22O11] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.) (a) In neutral solution, k = 2.1 × 10−11 s −1 at 27 °C and 8.5 × 10−11 s −1 at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
Consider the hypothetical reaction A 2 (g) + B2 (g) → 2AB (g), where the rate law is:The value of the rate constant at 302°C is 2.45 x 10 -4 L/mol•s, and at 508°C the rate constant is 0.891 L/mol•s. What is the activation energy for this reaction? What is the value of the rate constant for this reaction at 375°C?
A slightly bruised apple will rot extensively in about 4 days at room temperature (20°C). If it is kept in the refrigerator at 0°C, the same extent of rotting takes about 16 days. What is the activation energy for the rotting reaction?
At body temperature (37°C), the rate constant of an enzyme-catalyzed decomposition is 2.3 x 1014 times that of the uncatalyzed reaction. If the frequency factor, A, is the same for both processes, by how much does the enzyme lower the Ea?
The activation energy of a reaction is 56.6 kJ/mol and the frequency factor is 1.5 x 1011 /s. Calculate the rate constant of the reaction at 25 oC.
The rate constant of a reaction at 33 oC was measured to be 5.6 x 10−2 s-1 . If the frequency factor is 1.2 x 1013, what is the activation barrier?
The temperature dependence of the rate constant for the reaction is tabulated as follows:Temperature (K)k (M-1s-1)6000.0286500.227001.37506.080023Calculate A.
The activation energy for the isomerization of methyl isonitrile  is 160 kJ/mol.Calculate the fraction of methyl isonitrile molecules that has an energy equal to or greater than the activation energy for a temperature of 521 K. 
The activation energy for the isomerization of methyl isonitrile  is 160 kJ/mol.What is the ratio of the fraction of methyl isonitrile molecules that has an energy equal to or greater than the activation energy at 521 K to that at 501 K?
Suppose that a catalyst lowers the activation barrier of a reaction from 126 kJ/mol to 56 kJ/mol. By what factor would you expect the reaction rate to increase at 25 oC? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.)
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. If an enzyme increases the rate of the hydrolysis reaction by a factor of 1 million, how much lower does the activation barrier have to be when sucrose is in the active site of the enzyme? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical and that the temperature is 298 K.)
One mechanism for the destruction of ozone in the upper atmosphere isc. Ea for the uncatalyzed reactionO3 (g) + O (g) → 2O2 (g)is 14.0 kJ. Ea for the same reaction when catalyzed is 11.9 kJ. What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at 25°C? Assume that the frequency factor A is the same for each reaction.
The decomposition of NH3 to N2 and H2 was studied on two surfaces:Without a catalyst, the activation energy is 335 kJ/mol.b. How many times faster is the reaction at 298 K on the W surface compared with the reaction with no catalyst present? Assume that the frequency factor A is the same for each reaction.
Consider the following two reactions:O + N2 → NO + N          Ea = 315 kJ/molCl + H2 → HCl + H          Ea = 23 kJ/molThe frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25 oC.
Ethyl chloride vapor decomposes by the first-order reaction C2H5Cl → C2H4 + HCl. The activation energy is 249 kJ/mol and the frequency factor is 1.61014 s-1.Find the value of the specific rate constant at 720 K .
The gas-phase reaction of NO with F2 to form NOF and F has an activation energy of Ea = 6.3 kj/mol and a frequency factor of A = 6.0 x 108 M-1 s - 1. The reaction is believed to be bimolecular:NO(g) + F2 (g) → NOF (g) + F(g).Calculate the rate constant at 106 oC.
Consider the following reaction: O3(g) → O2(g) + O(g)Using the results of the Arrhenius analysis (Ea= 93.1kJ/mol and A = 4.361011M), predict the rate constant at 325 K.
Human liver enzymes catalyze the degradation of ingested toxins. By what factor is the rate of a detoxification changed if an enzyme lowers the Ea by 5 kJ/mol at 37˚C?
In an experiment, a sample of NaClO3 was 90% decomposed in 48 min. Approximately how long would this decomposition have taken if the sample had been heated 20 °C higher? (Hint: Assume the rate doubles for each 10 °C rise in temperature.)
The rate constant at 325 °C for the decomposition reaction C4H8 ⟶ 2C2H4 is 6.1 × 10−8 s −1, and the activation energy is 261 kJ per mole of C4H8. Determine the frequency factor for the reaction.
A popular chemical demonstration is the “magic genie” procedure, in which hydrogen peroxide decomposes to water and oxygen gas with the aid of a catalyst. The activation energy of this (uncatalyzed) reaction is 70.0 kJ/mol. When the catalyst is added, the activation energy (at 20.°C) is 42.0 kJ/mol. Theoretically, to what temperature (°C) would one have to heat the hydrogen peroxide solution so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction at 20.°C? Assume the frequency factor A is constant, and assume the initial concentrations are the same.
The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If the uncatalyzed reaction takes about 2400 years to occur, about how long will the catalyzed reaction take? Assume the frequency factor A is constant, and assume the initial concentrations are the same.
A certain compound, A, reacts to form products according to the reaction A → P. The amount of A is measured as a function of time under a variety of different conditions and the tabulated results are shown here:Time (s)25.0 oC [A] (M)35.0 oC [A] (M)45.0 oC [A] (M)01.0001.0001.000100.7790.6620.561200.5910.4610.312300.4530.3060.177400.3380.2080.100500.2590.1360.057600.2000.0930.032The same reaction is conducted in the presence of a catalyst, and the following data are obtained:Time (s)25.0 oC [A] (M)35.0 oC [A] (M)45.0 oC [A] (M)01.0001.0001.0000.10.7240.6680.5980.20.5110.4330.3410.30.3750.2910.2020.40.2750.1900.1190.50.1980.1220.0710.60.1410.0800.043What is the activation energy for this reaction in the presence of the catalyst?
Cobra venom helps the snake secure food by binding to acetylcholine receptors on the diaphragm of a bite victim, leading to the loss of function of the diaphragm muscle tissue and eventually death. In order to develop more potent antivenins, scientists have studied what happens to the toxin once it has bound the acetylcholine receptors. They have found that the toxin is released from the receptor in a process that can be described by the rate lawRate = k [acetylcholine receptor–toxin complex]If the activation energy of this reaction at 37.0°C is 26.2 kJ/mol and A = 0.850 s  -1, what is the rate of reaction if you have a 0.200-M solution of receptor–toxin complex at 37.0°C?
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the rate constant for the reaction increases by a factor of 2.50 x 103 as compared with the uncatalyzed reaction. Assuming the frequency factor A is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.
The rate of a certain reaction doubles for every 10 °C rise in temperature. (a) How much faster does the reaction proceed at 45 °C than at 25 °C?
The rate of a certain reaction doubles for every 10 °C rise in temperature.(b) How much faster does the reaction proceed at 95 °C than at 25 °C?
A friend studies a first-order reaction and obtains the following three graphs for experiments done at two different temperatures.Which graph probably represents the lower temperature? How do you know?
The rate constant (k) for a reaction was measured as a function of temperature. A plot of k versus 1/T(in K) is linear and has a slope of -6499 K. Calculate the activation energy for the reaction.
The rate constant (k) for a reaction was measured as a function of temperature. A plot of k versus 1/T(in K) is linear and has a slope of −1.19 x 104 K. Calculate the activation energy for the reaction.
You may want to reference (Pages 587 - 592)Section 14.5 while completing this problem.Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320˚C? (The activation energy for this reaction is 160 kJ/mol)(a) 8.1 x 10–15 s–1(b) 2.2 x 10–13 s–1(c) 2.7 x 10–9 s–1(d) 2.3 x 10-–1 s–1(e) 9.2 x 103 s–1
The data shown below were collected for the following first-order reaction: N2O(g) → N2(g) + 1/2 O2(g) Temperature (K)  Rate Constant  (1/s)8003.24 x 10 - 59000.0021410000.061411000.955Use an Arrhenius plot to determine the activation barrier for the reaction.
The data shown below were collected for the following first-order reaction: N2O(g) → N2(g) + 1/2 O2(g) Temperature (K)  Rate Constant  (1/s)8003.24 x 10 - 59000.0021410000.061411000.955Use an Arrhenius plot to determine frequency factor for the reaction.
The temperature dependence of the rate constant for the reaction is tabulated as follows:Temperature (K)k (M-1s-1)6000.0286500.227001.37506.080023Calculate Ea.
The data shown below were collected for the following second-order reaction: Cl2(g) + H2(g) → HCl(g) + H(g) Temperature (K)  Rate Constant  (L/mol)900.003571000.07731100.9561207.781Use an Arrhenius plot to determine the activation barrier for the reaction.
The data shown below were collected for the following second-order reaction: Cl2(g) + H2(g) → HCl(g) + H(g) Temperature (K)  Rate Constant  (L/mol)900.003571000.07731100.9561207.781Use an Arrhenius plot to determine the frequency factor for the reaction.
The following data show the rate constant of a reaction measured at several different temperatures. Temperature (K)  Rate Constant (1/s)3100.1903200.5443301.463403.703508.89Use an Arrhenius plot to determine the frequency factor for the reaction.
The reaction(CH3)3CBr + OH- → (CH3)3COH + Br-in a certain solvent is first order with respect to (CH 3)3CBr and zero order with respect to OH 2. In several experiments, the rate constant k was determined at different temperatures. A plot of ln(k) versus 1/T was constructed resulting in a straight line with a slope value of -1.10 x 104 K and y-intercept of 33.5. Assume k has units of s-1.b. Determine the value of the frequency factor A.
The reaction(CH3)3CBr + OH- → (CH3)3COH + Br-in a certain solvent is first order with respect to (CH 3)3CBr and zero order with respect to OH 2. In several experiments, the rate constant k was determined at different temperatures. A plot of ln(k) versus 1/T was constructed resulting in a straight line with a slope value of -1.10 x 104 K and y-intercept of 33.5. Assume k has units of s-1.c. Calculate the value of k at 25°C.
The kinetics of the following second-order reaction were studied as a function of temperature: C2H5Br(aq) + OH–(aq) → C2H5OH(l) + Br–(aq)Temperature (oC)k (L/mol )258.81 x 10–5350.000285450.000854550.00239650.00633Determine the frequency factor for the reaction.
The kinetics of the following second-order reaction were studied as a function of temperature: C2H5Br(aq) + OH–(aq) → C2H5OH(l) + Br–(aq)Temperature (oC)k (L/mol )258.81 x 10–5350.000285450.000854550.00239650.00633Determine the rate constant at 15˚C.
The kinetics of the following second-order reaction were studied as a function of temperature: C2H5Br(aq) + OH–(aq) → C2H5OH(l) + Br–(aq)Temperature (oC)k (L/mol )258.81 x 10–5350.000285450.000854550.00239650.00633Determine the activation energy for the reaction.
The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows:Temperature (K)Rate Constant (s-1)3003.2 x 10 - 113201.0 x 10 - 93403.0 x 10 - 83552.4 x 10 - 7From these data, calculate the activation energy in units of kJ/mol.
The rate constant of a reaction is measured at different temperatures. A plot of the natural log of the rate constant as a function of the inverse of the temperature (in kelvins) yields a straight line with a slope of -8500 K-1 . What is the activation energy (Ea) for the reaction?
The rate constant for the gas-phase decomposition of N 2O5,N2O5 → 2NO2 + 1/2 O2has the following temperature dependence:Make the appropriate graph using these data, and determine the activation energy for this reaction.
The reaction(CH3)3CBr + OH- → (CH3)3COH + Br-in a certain solvent is first order with respect to (CH 3)3CBr and zero order with respect to OH 2. In several experiments, the rate constant k was determined at different temperatures. A plot of ln(k) versus 1/T was constructed resulting in a straight line with a slope value of -1.10 x 104 K and y-intercept of 33.5. Assume k has units of s-1.a. Determine the activation energy for this reaction.
Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, for the reaction was measured at several different temperatures and the data are shown here:What is the value of the activation energy (in kJ/mol) for this reaction?
For the reaction2N2O5 (g) → 4NO2 (g) + O2 (g)the following data were collected, whereRate = - Δ [N2O5] / ΔtCalculate Ea for this reaction.
Experimental values for the temperature dependence of the rate constant for the gas-phase reactionNO + O3 → NO2 + O2are as follows:Make the appropriate graph using these data, and determine the activation energy for this reaction.
Is each of these statements true? If not, explain why.(f) The activation energy of a reaction depends on temperature.
Is each of these statements true? If not, explain why.(k) Temperature has no effect on the frequency factor (A).
Is each of these statements true? If not, explain why.(n) The orientation probability factor (p) is near 1 for reactions between single atoms.
The decomposition of NH3 to N2 and H2 was studied on two surfaces:Without a catalyst, the activation energy is 335 kJ/mol.a. Which surface is the better heterogeneous catalyst for the decomposition of NH 3? Why?b. How many times faster is the reaction at 298 K on the W surface compared with the reaction with no catalyst present? Assume that the frequency factor A is the same for each reaction.c. The decomposition reaction on the two surfaces obeys a rate law of the formRate = k [NH3] / [H2]How can you explain the inverse dependence of the rate on the H 2 concentration?
The activation energy of an uncatalyzed reaction is 93 kJ/mol . The addition of a catalyst lowers the activation energy to 56 kJ/mol .Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 26 oC?
The activation energy of an uncatalyzed reaction is 93 kJ/mol . The addition of a catalyst lowers the activation energy to 56 kJ/mol .Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at 126 oC?
Suppose that a certain biologically important reaction is quite slow at physiological temperature (37 oC) in the absence of a catalyst.Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction in order to achieve a 4×105-fold increase in the reaction rate?
Which factor is most sensitive to changes in temperature—the frequency of collisions, the orientation factor, or the fraction of molecules with energy greater than the activation energy?
You may want to reference (Pages 588 - 592)Section 14.5 while completing this problem.In which of the following reactions would you expect the orientation factor to be least important in leading to reaction:NO + O → NO2H + Cl → HCl?
Does the orientation factor depend on temperature?
Which of the following reactions would you expect to proceed at a faster rate at room temperature? Why? (Hint: Think about which reaction would have the lower activation energy.)2Ce4+ (aq) + Hg 22+ (aq) → 2Ce3+ (aq) + 2Hg2+ (aq)H3O+ (aq) + OH - (aq) → 2H 2O (l)
Account for the relationship between the rate of a reaction and its activation energy.
Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.
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 (kJ•mol -1), for this reaction? What is the value of the pre-exponential factor (sometimes frequency factor), A (s -1), 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. 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?