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Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be fastest and which would be slowest? (a) \(E_{a}=45 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=35 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=55 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{~kJ} / \mathrm{mol}\)

Short Answer

Expert verified
The fastest reaction is (b) with an activation energy of \(E_{a}=35 \mathrm{~kJ} / \mathrm{mol}\), and the slowest reaction is (c) with an activation energy of \(E_{a}=55 \mathrm{~kJ} / \mathrm{mol}\).

Step by step solution

01

Recall the concept of activation energy

Activation energy (Ea) is the minimum energy required for a reaction to occur between colliding molecules. It acts as a barrier preventing reactions from happening at lower energies. Generally, the lower the activation energy, the faster the reaction rate, as more of the colliding molecules will possess the necessary energy to overcome the barrier.
02

Determine the effect of activation energy on reaction rate

For this problem, it is assumed that all collision factors are the same, which means we're only considering the activation energies of the reactions when determining their relative speeds. Since a lower activation energy will result in a faster reaction, we can compare the activation energies of the three reactions to determine their relative speeds. From the given data, the activation energies are: (a) \(E_{a}=45 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=35 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=55 \mathrm{~kJ} / \mathrm{mol}\)
03

Identify the fastest and slowest reactions

Based on the activation energies, we can now determine the relative speeds of the reactions: 1. Reaction (b) has the lowest activation energy (\(E_{a}=35 \mathrm{~kJ} / \mathrm{mol}\)), so it will be the fastest reaction. 2. Reaction (c) has the highest activation energy (\(E_{a}=55 \mathrm{~kJ} / \mathrm{mol}\)), so it will be the slowest reaction. So, the fastest reaction is (b) and the slowest reaction is (c).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Activation Energy
When exploring the fascinating world of chemical reactions, one concept stands out as crucial: activation energy, often represented as E_{a}. Imagine activation energy as the energetic 'hill' that reactants must climb before they can transform into products. It's the minimum energy necessary for molecules to collide in a way that leads to a reaction.

Considering our daily life, an analogy might help: lighting a match requires one to strike it against the box; that striking is similar to providing the activation energy necessary for the match to ignite. In chemical reactions, lower activation energy means it's easier for molecules to 'strike' effectively and react, just like it's easier to light a match with a smoother stroke.

  • If a reaction has a high E_{a}, fewer molecules will have enough energy at a given temperature to react, resulting in a slower reaction rate.
  • Conversely, a lower E_{a} allows for more molecules to participate, leading to a faster reaction.
Understanding activation energy is not just important for predicting reaction speeds; it's pivotal for controlling reactions in industrial processes, designing medicines, and even baking the perfect loaf of bread!
Collision Theory
Collision theory provides a lens through which we can view the micro-world of molecules and atoms during chemical reactions. At its core is a simple but powerful idea: molecules must collide to react. However, not just any collision will do; they must collide with the right orientation and sufficient energy—akin to the energy mentioned as activation energy—to break existing bonds and form new ones.

In the microscopic coliseum of a chemical reaction, two key factors govern the success of a gladiatorial match between molecules:
  • Orientation: Molecules must align in just the right way, much like a key fitting a lock. This specific alignment allows for the necessary interaction between atoms that can lead to a reaction.
  • Energy: Molecules must bring enough energy to the collision to overcome their mutual activation energy barrier. This is why temperature can influence reaction rates; higher temperatures mean more energetic collisions.
When we discuss reaction rates, we are really talking about how often successful collisions occur within a reaction mixture. Collision theory helps explain why changing conditions, like temperature or concentration, affects these rates by altering the frequency and energy of collisions.
Reaction Rate
Delving into the kinetics of chemistry, the reaction rate emerges as a central topic, being the speed at which reactants transform into products. It is quantifiable and influenced by various factors, much like how the speed of your car can be affected by the slope of the road or the amount of traffic. In the molecular realm, these 'traffic conditions' are represented by concentrations of reactants, temperature, and the presence of catalysts.

Here's a way to visualize it: consider a crowd moving through a turnstile—the wider the turnstile (representing low activation energy), or the more pushed the crowd is (representing high temperature), the faster the people (molecules) get through (react). A higher concentration of reactants is akin to a larger crowd pushing through the turnstile, potentially accelerating the rate at which they pass through.

When chemists want to control reaction rates, they look at these factors:
  • Temperature: Increasing the temperature usually increases the reaction rate.
  • Concentration: Higher concentrations of reactants can lead to more frequent and effective collisions, speeding up the reaction.
  • Catalysts: These substances reduce the activation energy, making it easier for molecules to collide with enough energy to react without being consumed in the process themselves.
Understanding how these parameters affect reaction rates is vital for everything from developing new drugs to reducing the pollutants in car exhaust.

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Most popular questions from this chapter

A colored dye compound decomposes to give a colorless product. The original dye absorbs at \(608 \mathrm{~nm}\) and has an extinction coefficient of \(4.7 \times 10^{4} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at that wavelength. You perform the decomposition reaction in a 1 -cm cuvette in a spectrometer and obtain the following data: From these data, determine the rate law for the reaction "dye \(\longrightarrow\) product" and determine the rate constant.

Consider the hypothetical reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\). The following two-step mechanism is proposed for the reaction: Step 1: \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X}\) Step 2: A \(+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D}\) \(\mathrm{X}\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

Molecular iodine, \(\mathrm{I}_{2}(\mathrm{~g})\), dissociates into iodine atoms at \(625 \mathrm{~K}\) with a first-order rate constant of \(0.271 \mathrm{~s}^{-1}\). (a) What is the half-life for this reaction? (b) If you start with \(0.050 \mathrm{MI}_{2}\) at this temperature, how much will remain after \(5.12 \mathrm{~s}\) assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?

Many primary amines, \(\mathrm{RNH}_{2}\), where \(\mathrm{R}\) is a carbon- containing fragment such as \(\mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2}\), and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a \(\mathrm{C}\) atom for " \(\mathrm{R}^{w}\) ). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) \(2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (c) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(g)\)

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