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A magnesium ion, \(\mathrm{Mg}^{2+}\), with a charge of \(3.2 \times 10^{-19} \mathrm{C}\) and an oxide ion, \(\mathrm{O}^{2-},\) with a charge of \(-3.2 \times 10^{-19} \mathrm{C},\) are separated by a distance of \(0.35 \mathrm{nm}\). How much work would be required to increase the separation of the two ions to an infinite distance?

Short Answer

Expert verified
To find the work required to separate a magnesium ion, \(\mathrm{Mg}^{2+}\), and an oxide ion, \(\mathrm{O}^{2-}\), to an infinite distance, we can use the formula for work done against electrostatic force: \(W = -k \frac{q_1q_2}{r_1}\). Given the charges of the ions and their initial distance, we calculate the work as: \(W = -(9 \times 10^9 \mathrm{Nm^2/C^2}) \frac{(3.2 \times 10^{-19} \mathrm{C})(-3.2 \times 10^{-19} \mathrm{C})}{0.35 \times 10^{-9} \mathrm{m}}\). Upon simplifying, we find the work required is approximately \(2.64 \times 10^{-18} \mathrm{J}\).

Step by step solution

01

Identify the given information

We are given: - Charge of magnesium ion (Mg2+): \(q_1 = 3.2 \times 10^{-19} \mathrm{C}\) - Charge of oxide ion (O2-): \(q_2 = -3.2 \times 10^{-19} \mathrm{C}\) - Separation between ions: \(r_1 = 0.35 \mathrm{nm}\) (it is necessary to convert this value to meters)
02

Convert the distance

Before we can use the given information to find the work, we must first convert the given distance from nanometers to meters: $$r_1 = 0.35 \mathrm{nm} = 0.35 \times 10^{-9} \mathrm{m}$$
03

Write the formula for work done against electrostatic force

The formula for the work done in moving a charged particle against an electrostatic force is given by: $$W = k \frac{q_1q_2}{r_2} - k \frac{q_1q_2}{r_1}$$ where: - \(W\) is the work done, - \(k\) is the electrostatic constant (\(9 \times 10^9 \mathrm{Nm^2/C^2}\)), - \(q_1\) is the charge of the first ion, - \(q_2\) is the charge of the second ion, - \(r_1\) is the initial distance between the ions, and - \(r_2\) is the final distance between the ions (infinite distance).
04

Calculate the work for an infinite separation

As the final separation distance is infinite, the first term of the formula becomes zero, since any value divided by infinity is zero. Thus, we only need to calculate the second term of the formula: $$W = -k \frac{q_1q_2}{r_1}$$ Now, we can plug in the given values and the electrostatic constant: $$W = -(9 \times 10^9 \mathrm{Nm^2/C^2}) \frac{(3.2 \times 10^{-19} \mathrm{C})(-3.2 \times 10^{-19} \mathrm{C})}{0.35 \times 10^{-9} \mathrm{m}}$$
05

Simplify and find the work done

Multiply and divide the given values to find the work done: $$W = 2.64 \times 10^{-18} \mathrm{J}$$ Hence, the amount of work required to separate the magnesium ion and the oxide ion to an infinite distance is approximately \(2.64 \times 10^{-18} \mathrm{J}\).

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Key Concepts

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

Coulomb's Law
Coulomb's law is an essential principle in the study of electrostatics. It quantifies the force between two charged particles. This force is described by the equation: \[ F = k \frac{|q_1q_2|}{r^2} \]where:
  • \( F \) is the magnitude of the electrostatic force between the charges.
  • \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \mathrm{Nm^2/C^2} \).
  • \( q_1 \) and \( q_2 \) are the charges of the particles.
  • \( r \) is the separation distance between the charges.
Coulomb's law illustrates that the force is directly proportional to the product of the charges. It is inversely proportional to the square of their separation distance. This means that as the charges are moved further apart, the force between them decreases. This decrease follows the square of the distance, so the force drops off quickly with increasing distance.
Ion Separation
Ion separation is a significant concept when examining charged particles. It refers to the distance between two ions. When ions are close together, they experience a strong force due to their charges. As explained by Coulomb's law, moving these ions apart reduces the force they exert on each other. In chemical and physical processes, like dissolving salts in water or in crystal structures, ion separation is crucial. It influences properties such as solubility, conductivity, and the stability of compounds. By knowing the initial and final separation distances, one can calculate changes in potential energy as ions move.
Work Done in Electrostatics
In electrostatics, work is done when a force causes an ion to move. The work required to change the separation between charged particles is derived from the energy associated with the forces they exert on each other.The formula for work done, when moving charges in electrostatics, is:\[ W = k \frac{q_1 q_2}{r_2} - k \frac{q_1 q_2}{r_1} \]
  • \( W \) represents the work done.
  • \( q_1 \) and \( q_2 \) are the charges of the ions.
  • \( r_1 \) and \( r_2 \) are the initial and final separation distances.
When ions are moved to an infinite distance apart, as in this exercise, \( r_2 \) approaches infinity. The potential energy at this state drops to zero, and the work done becomes equivalent to the potential energy calculated at the initial separation \( r_1 \). Calculating this work helps understand the energy required for ion separation, which is key in understanding interactions in ionic compounds.

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Most popular questions from this chapter

Consider the following hypothetical reactions: $$ \begin{array}{l} \mathrm{A} \longrightarrow \mathrm{B} \quad \Delta H_{I}=+60 \mathrm{~kJ} \\ \mathrm{~B} \longrightarrow \mathrm{C} \quad \Delta H_{I I}=-90 \mathrm{~kJ} \end{array} $$ (a) Use Hess's law to calculate the enthalpy change for the reaction \(\mathrm{A} \longrightarrow \mathrm{C}\). (b) Construct an enthalpy diagram for substances A, B, and C, and show how Hess's law applies.

The gas-phase reaction shown, between \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), was run in an apparatus designed to maintain a constant pressure. (a) Write a balanced chemical equation for the reaction depicted and predict whether \(w\) is positive, negative, or zero. (b) Using data from Appendix C, determine \(\Delta H\) for the formation of one mole of the product. [Sections 5.3 and 5.7\(]\)

From the enthalpies of reaction $$ \begin{aligned} \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) & \longrightarrow 2 \mathrm{HF}(g) & & \Delta H=-537 \mathrm{~kJ} \\ \mathrm{C}(s)+2 \mathrm{~F}_{2}(g) & \longrightarrow \mathrm{CF}_{4}(g) & & \Delta H=-680 \mathrm{~kJ} \\ 2 \mathrm{C}(s)+2 \mathrm{H}_{2}(g) & \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g) & & \Delta H=+52.3 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction of ethylene with \(\mathrm{F}_{2}\) : $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+6 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g) $$

At the end of 2012, global population was about 7.0 billion people. What mass of glucose in kg would be needed to provide 1500 Cal/person/day of nourishment to the global population for one year? Assume that glucose is metabolized entirely to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) according to the following thermochemical equation: $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ} &=-2803 \mathrm{~kJ} \end{aligned} $$

(a) Use enthalpies of formation given in Appendix C to calculate \(\Delta H\) for the reaction \(\mathrm{Br}_{2}(g) \longrightarrow 2 \operatorname{Br}(g)\), and use this value to estimate the bond enthalpy \(D(\mathrm{Br}-\mathrm{Br})\). (b) How large is the difference between the value calculated in part (a) and the value given in Table 5.4 ?

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