The Arrhenius Equation illustrates the temperature dependence of the rate constant k.
Temperature is one of the four factors that directly affects the rate of a reaction. It affects the rate of a reaction by impacting the rate constant k.
Concept: The Arrhenius Equation3m
Hey guys! In this new video, we're going to take a look and see what exactly is the effect that temperature has on the rate of a chemical reaction. What we should realize is that temperature has a huge and profound effect on the rate of a reaction because temperature has a direct effect on our rate constant, k.
In fact, we're going to say increasing the temperature or adding or increasing the amount of catalyst will increase our k. If we increase the rate constant k, then that will increase the rate. We're going to say here to show this, we use the Arrhenius equation. The Arrhenius equation is k equals A times e to the negative Ea over Rt. Here we’re going to say A is called our frequency factor. Ea is called our energy of activation or activation energy. Now R is our constant. Because we're using energy, here the constant is 8.314 joules over moles times k. Then temperature is t, but the temperature here is in Kelvins.
Recall, one of the factors for a reaction to occur is that molecules have to collide in order to react. Molecule A has to hit molecule B. We're going to say for a reaction to be successful, the molecules must collide with sufficient energy and with the correct orientation. They have to be going fast enough so that they hit each other with enough force, but also they have to hit the activation site so that they actually stick together.
The Arrhenius Equation shows the dependence of the rate constant on the pre-exponential factor, absolute temperature and activitation energy of the reaction.
In terms of what makes for a fast rate we can relate all these variables together.
Concept: Besides the general Arrhenius equation there are also the two-point form and the linear/graphical form.3m
Now we're going to say here we can convert the Arrhenius equation so that it gets a logarithmic form, also called the graphical form sometimes. If we have a plot or graph of something and we're trying to relate the Arrhenius equation to it, we usually use this graphical form also known as the plot wise approach to the Arrhenius equation.
When do we use this particular form? This is called the two-point form of the Arrhenius equation. We use this form anytime have two k values or we have two temperatures. That’s the time we use this. We’re going to say the Arrhenius equation can be converted so it forms this plot-wise approach. This equation is equal to y equals mx plus b, y is ln k, m is negative Ea over R, 1 over t is x, and ln A is b.
How did we come up with this equation? If we take a look, what you want to do here is you want to do ln in front of both these. But we’re also going to do ln in front of this e. That's how we got the ln k here. When we do ln in front of A and in front of e, it’s going to become plus ln A here. This right here is inverse log. When you do natural log in front of it, it basically cancels them out so then it becomes Ea over Rt. You can actually separate out the t here where it becomes negative Ea over R times 1 over t.
By putting ln in front of both things, that's how we’re able to separate it into this plot-wise approach. We'll see how best to use this later on. Realize that this equation here relates to this graph here. Remember, every time we plot something, it's y versus x. My y is ln k. My x here is 1 over t. Ea over R is m which is my slope. Remember, slope is just the change in y over the change in x. It's in this way that we can find the energy of activation. Later on when we do calculations, we'll see how best to use this type of equation.
When dealing with TWO rate constants or TWO temperatures then we use the Two-Point Form of the Arrhenius Equation.
In order to relate the plot of a graph to the Arrhenius equation then we manipulate it into the plot-wise approach of the equation.
Example: The reaction 2 HCl (g) → H2 (g) + Cl2 (g) has Ea of 1.77 x 104 kJ/mol and a rate constant of 1.32 x 10-1 at 700 K. What is the rate constant at 685 K?7m
Problem: If a first order reaction has a frequency factor of 3.98 x 1013 s-1 and Ea of 160 kJ, then calculate the rate constant at 25oC.4m
Problem: Generally, the slower the rate of the reaction then the ___________ the energy of activation (Ea) and the higher the temperature, the ____________ the value of rate constant, k.2m
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
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
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
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.
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?
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
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.
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
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?
According to collision theory and the Arrhenius concept, which of the following statements about gas-phase chemical reactions are TRUE?
i. Reaction rates are proportional to collision frequency
ii. The number of effective collisions at a given temperature is related to both the kinetic energy and orientation of the reacting molecules or atoms
iii. As the temperature increases, the number of species reaching the transition state of the rate-determining step increases
iv. As the temperature increases, the activation energy of a reaction remains constant.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i), (ii), and (iii)
(d) (ii), (iii), and (iv)
(e) All of the statements are correct
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
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?
d) 2.1×10 36
e) 8.3×10 9
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
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?
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
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.
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 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?
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.
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?