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For the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \longrightarrow \mathrm{AB}(g),\) how many unique collisions between \(A\) and \(B\) are possible if 1.01 mol of \(\mathrm{A}(g)\) and \(2.12 \mathrm{~mol}\) of \(\mathrm{B}(g)\) are present in the vessel?

Short Answer

Expert verified
About 7.76 x 10^{47} unique collisions are possible.

Step by step solution

01

- Calculate the Number of Molecules for A

Use Avogadro's number (ablaA = 6.022 x 10^23 molecules/mol) to calculate the number of molecules of ablaA in 1.01 mol. ablaN(A) = 1.01 mol x 6.022 x 10^{23} molecules/mol ablaN(A) ≈ 6.082 x 10^{23} molecules.
02

- Calculate the Number of Molecules for B

Similarly, use Avogadro's number to calculate the number of molecules of ablaB in 2.12 mol. ablaN(B) = 2.12 mol x 6.022 x 10^{23} molecules/mol ablaN(B) ≈ 1.276 × 10^{24} molecules.
03

- Calculate the Number of Unique Collisions

Each molecule of ablaA can collide with any molecule of ablaB. Thus, the total number of unique collisions is given by ablaN(collisions) = ablaN(A) x ablaN(B). ablaN(collisions) ≈ 6.082 x 10^{23} x 1.276 x 10^{24} ablaN(collisions) ≈ 7.76 x 10^{47} unique collisions.

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Key Concepts

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

Molecular Collisions
In a chemical reaction, molecules must collide for the reaction to occur. Not every collision leads to a reaction, but for those that do, the orientation and energy of the molecules are crucial. Molecules A and B must collide properly to form the product AB, and this concept is key to understanding many reactions.

To calculate unique collisions, consider the number of each type of molecule present. Each molecule of A can potentially collide with every molecule of B. Thus, by knowing the number of molecules of A and B, we can determine the total unique collisions possible.
First, the number of molecules is calculated using Avogadro's number for the given moles. Then multiply the two results to get the total unique collisions during the reaction.
Avogadro's Number
Avogadro's number is a fundamental constant in chemistry, defined as the number of atoms, ions, or molecules in one mole of a substance. This number is approximately \(6.022 \times 10^{23} \) particles per mole. It allows chemists to count entities at the atomic scale by relating amounts to macroscopic measurements.

In the exercise, here's how Avogadro's number is used:
1. Calculate the number of molecules of A:
\[ N(A) = 1.01 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \]
\[ N(A) \backsimeq 6.082 \times 10^{23} \text{ molecules} \]
2. Calculate the number of molecules of B:
\[ N(B) = 2.12 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \]
\[ N(B) \backsimeq 1.276 \times 10^{24} \text{ molecules} \]
After finding the number of molecules using Avogadro's number, you can use these values to calculate the unique collisions between the molecules.
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. It provides the quantitative relationships derived from the balanced chemical equations.

For the reaction representing the formation of the product AB from reactants A and B, stoichiometry helps in understanding how many molecules of A and B react to form AB. In our example, 1 mole of A reacts with 1 mole of B:
\[ \text{A}(g) + \text{B}(g) \rightarrow \text{AB}(g) \]
Knowing the moles of A and B, stoichiometry ensures the calculation of the unique collisions according to the balanced equation.
From the given moles of A and B, we derive the number of molecules using Avogadro's number. Then, using these quantities, we understand how they interact in a stoichiometric ratio to form the product AB. This method demonstrates the integral role of stoichiometry in chemical reactions and quantitative predictions.

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Most popular questions from this chapter

The mathematics of the first-order rate law can be applied to any situation in which a quantity decreases by a constant fraction per unit of time (or unit of any other variable). (a) As light moves through a solution, its intensity decreases per unit distance traveled in the solution. Show that \(\ln \left(\frac{\text { intensity of light leaving the solution }}{\text { intensity of light entering the solution }}\right)\) \(=-\) fraction of light removed per unit of length \(\times\) distance traveled in solution (b) The value of your savings declines under conditions of constant inflation. Show that \(\ln \left(\frac{\text { value remaining }}{\text { initial value }}\right)\) \(=-\) fraction lost per unit of time \(\times\) savings time interval

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16.103 Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]\). The long-accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) In the 1960 s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism: (1) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) [fast] (2) \(\mathrm{H}_{2}(g)+\mathrm{I}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{I}(g)\) [fast] (3) \(\mathrm{H}_{2} \mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) \quad[\) slow \(]\) Show that this mechanism is consistent with the rate law.

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