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(a) In which of the following reactions would you expect the orientation factor to be more important in leading to reaction: \(\mathrm{O}_{3}+\mathrm{O} \longrightarrow 2 \mathrm{O}_{2}\) or \(\mathrm{NO}+\mathrm{NO}_{3} \longrightarrow 2 \mathrm{NO}_{2} ?\) (b) What is related to the orientation factor? Which, smaller or larger ratio of effectively oriented collisions to all possible collisions, would lead to a smaller orientation factor?

Short Answer

Expert verified
The orientation factor is more important in the second reaction \(\mathrm{NO}+\mathrm{NO}_{3}\longrightarrow 2 \mathrm{NO}_{2}\), as both reacting species are linear molecules and must be aligned for the reaction to occur. The orientation factor is related to the effectiveness of collisions and represents the fraction of collisions that are effectively oriented. A smaller ratio of effectively oriented collisions to all possible collisions would lead to a smaller orientation factor, indicating that molecular orientation has a significant impact on the reaction rate.

Step by step solution

01

Part (a): Comparing the importance of the orientation factor

To determine the importance of the orientation factor, we will analyze the reacting species in each reaction. 1. In the first reaction, \(\mathrm{O}_{3}+\mathrm{O}\longrightarrow 2\mathrm{O}_{2}\), the reacting species are O3 and O. O3 is a bent molecule and O is a single atom. 2. In the second reaction, \(\mathrm{NO}+\mathrm{NO}_{3}\longrightarrow 2\mathrm{NO}_{2}\), the reacting species are NO and NO3. Both NO and NO3 are linear molecules. In the first reaction, since O3 is a bent molecule, the orientation factor is less important. The O atom doesn't have any orientation restriction as it is a single atom and can react with any side of the O3 molecule. In the second reaction, both reacting species are linear molecules, and the orientation factor is more important because the reactive sites in the linear molecules must be aligned for the reaction to occur. Therefore, the orientation factor is more important in the second reaction \(\mathrm{NO}+\mathrm{NO}_{3}\longrightarrow 2 \mathrm{NO}_{2}\).
02

Part (b): Relation of the orientation factor and effectiveness of collisions

The orientation factor is related to the effectiveness of collisions between reacting molecules. The orientation factor represents the fraction of collisions that are effectively oriented out of the total possible collisions between the reacting species. It indicates the importance of molecular orientation in leading to a reaction. A smaller ratio of effectively oriented collisions to all possible collisions would lead to a smaller orientation factor. This means that only a small proportion of collisions are effectively oriented, and molecular orientation has a significant impact on the reaction rate. A larger orientation factor indicates that most of the collisions are effectively oriented, and molecular orientation is less critical for the reaction to occur.

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Key Concepts

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

Orientation Factor
In chemical reactions, not all collisions between molecules lead to a reaction. For a reaction to happen, molecules must collide with the right orientation. The orientation factor is a crucial element that tells us how important the alignment of molecules is in making reactions happen.
The orientation factor is particularly important when the molecules involved in a reaction have specific parts that need to line up to react. For example:
  • In reactions involving complex molecules.
  • When the structure of molecules makes certain orientations more reactive.
In simpler terms, the orientation factor measures how picky the molecules are about their alignment during collisions. A high orientation factor means alignment is less crucial, while a lower factor indicates that precise orientation is critical.
Collision Theory
Collision Theory helps us understand how chemical reactions occur. It highlights two main ideas:
  • Molecules must collide to react.
  • The collisions must have enough energy and the proper orientation.
When molecules collide with sufficient energy and proper alignment, they can break bonds and form new ones, resulting in a chemical reaction. Energy plays a vital role in this theory, as only collisions with enough energy can overcome the activation energy barrier, which is the minimum energy required for a reaction to occur.
Think of it like playing billiards. Not every shot sinks the ball in a pocket. The angle and the force applied matter, just as the energy and orientation of colliding molecules matter for reactions to take place.
Reaction Rate
The reaction rate tells us how quickly a reaction proceeds. It is influenced by several factors:
  • Concentration of reactants: Higher concentrations generally increase reaction rates.
  • Temperature: Higher temperatures usually increase rates by providing more energy to the molecules.
  • Orientation: The proper alignment of molecules during collisions is crucial.
  • Presence of a catalyst: Catalysts lower the activation energy needed for a reaction, thus increasing the rate.
When molecules collide more frequently and with the proper orientation and sufficient energy, reactions happen more quickly. This is why temperature and concentration often play large roles in determining how fast reactions happen, but the correct molecular orientation is always essential.
Molecular Orientation
Molecular orientation refers to how molecules position themselves relative to each other during a collision. Some molecules, due to their shape and complexity, require a precise alignment to react effectively.
For example, linear molecules, like NO and NO3 in the given reaction, must align their reactive sites to enable a successful reaction. Bent or asymmetric molecules might have more flexibility, but certain parts still need to be aligned correctly.
Here's why molecular orientation matters:
  • Ensures that reactive sites on molecules meet each other.
  • Determines whether a reaction proceeds by affecting the collision's success.
  • In complex reactions, it can dictate the pathway and outcome of the reaction.
Understanding the importance of molecular orientation helps chemists design more efficient reactions and understand why some reactions might be slow or inefficient.

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Most popular questions from this chapter

The dimerization of \(\mathrm{C}_{2} \mathrm{~F}_{4}\) to \(\mathrm{C}_{4} \mathrm{~F}_{8}\) has a rate constant \(k=0.045 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(450 \mathrm{~K} .\) (a) Based on the unit of \(k\) what is the reaction order in \(\mathrm{C}_{2} \mathrm{~F}_{4} ?(\mathbf{b})\) If the initial concentration of \(\mathrm{C}_{2} \mathrm{~F}_{4}\) is \(0.100 \mathrm{M}\), how long would it take for the concentration to decrease to \(0.020 \mathrm{M}\) at \(450 \mathrm{~K}\) ?

(a) Can an intermediate appear as a reactant in the first step of a reaction mechanism? (b) On a reaction energy profile diagram, is an intermediate represented as a peak or a valley? (c) If a molecule like \(\mathrm{Cl}_{2}\) falls apart in an elementary reaction, what is the molecularity of the reaction?

(a) The activation energy for the reaction \(\mathrm{A}(g) \longrightarrow \mathrm{B}(g)\) is \(100 \mathrm{~kJ} / \mathrm{mol}\). Calculate the fraction of the molecule A that has an energy equal to or greater than the activation energy at \(400 \mathrm{~K} .(\mathbf{b})\) Calculate this fraction for a temperature of \(500 \mathrm{~K}\). What is the ratio of the fraction at \(500 \mathrm{~K}\) to that at \(400 \mathrm{~K}\) ?

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are \(154 \mathrm{~kJ} / \mathrm{mol}\) and \(136 \mathrm{~kJ} / \mathrm{mol}\), respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

A certain enzyme catalyzes a biochemical reaction. In water, without the enzyme, the reaction proceeds with a rate constant of \(6.50 \times 10^{-4} \mathrm{~min}^{-1}\) at \(37^{\circ} \mathrm{C} .\) In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.67 \times 10^{4} \mathrm{~min}^{-1}\) at \(37^{\circ} \mathrm{C}\). Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

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