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Chapter 5: Problem 18

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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Most popular questions from this chapter

Without doing any calculations, predict the sign of \(\Delta H\) for each of the following reactions: (a) $\mathrm{NaCl}(s) \longrightarrow \mathrm{Na}^{+}(g)+\mathrm{Cl}^{-}(\mathrm{g})$ (b) \(2 \mathrm{H}(g) \longrightarrow \mathrm{H}_{2}(g)\) (c) \(\mathrm{Na}(g) \longrightarrow \mathrm{Na}^{+}(g)+\mathrm{e}^{-}\) (d) \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(l)\)

It is interesting to compare the "fuel value" of a hydrocarbon in a hypothetical world where oxygen is not the combustion agent. The enthalpy of formation of \(\mathrm{CF}_{4}(g)\) is \(-679.9 \mathrm{~kJ} / \mathrm{mol}\). Which of the following two reactions is the more exothermic? $$ \begin{aligned} \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{CH}_{4}(g)+4 \mathrm{~F}_{2}(g) & \longrightarrow \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g) \end{aligned} $$

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ} $$ For this reaction, calculate \(\Delta H\) for the formation of (a) $1.36 \mathrm{~mol}\( of \)\mathrm{O}_{2}\( and \)(\mathbf{b}) 10.4 \mathrm{~g}$ of \(\mathrm{KCl} .(\mathbf{c})\) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2},\) is likely to be feasible under ordinary conditions? Explain your answer.

(a) Write an equation that expresses the first law of thermodynamics in terms of heat and work. (b) Under what conditions will the quantities \(q\) and \(w\) be negative numbers?

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