Question

In molecules of gaseous sodium chloride, the chloride ion has one more electron than proton, and the sodium ion has one more proton than electron. These ions are separated by about \(0.236 \mathrm{nm} .\) How much work would be required to increase the distance between these ions to \(1.00 \mathrm{~cm} ?\)

Short answer

Answer: To calculate the work required, follow the solution steps above and find the difference between the final potential energy and the initial potential energy. Plug in the given values and solve for \(W\).

Step-by-step solution

Convert distances to meters

First, we need to convert the given distances from nanometers (nm) and centimeters (cm) to meters (m), as we will use the standard unit of meter (m) in our calculations. \(0.236 \mathrm{nm} = 0.236 \times 10^{-9} \mathrm{m}\) and \(1.00 \mathrm{cm} = 1.00 \times 10^{-2} \mathrm{m} \)

Determine the charges of sodium and chloride ions

The charge of an electron is \(q_e = -1.6 \times 10^{-19} \mathrm{C}\) and the charge of a proton is \(q_p = 1.6 \times10^{-19} \mathrm{C}\). For the sodium ion, there is one more proton than electron, so its charge \(q_1\) is: \(q_1 = q_p - q_e = 2 \times 1.6 \times 10^{-19} \mathrm{C}\) For the chloride ion, there is one more electron than proton, so its charge \(q_2\) is: \(q_2 = -q_e = -1.6 \times 10^{-19} \mathrm{C}\)

Calculate the initial electrostatic force

To find the initial electrostatic force, we will use Coulomb's law: \(F = \frac{k \times q_1 \times q_2}{r^{2}}\) where \(F\) is the electrostatic force between the two ions, \(k = 8.99 \times 10^{9} \mathrm{Nm^{2}C^{-2}}\) is the electrostatic constant, \(q_1\) and \(q_2\) are the charges of sodium and chloride ions, respectively, and \(r\) is the initial distance between the ions. \(F = \frac{8.99 \times 10^{9} \mathrm{Nm^{2}C^{-2}} \times 2 \times 1.6 \times 10^{-19} \mathrm{C} \times -1.6 \times 10^{-19} \mathrm{C}}{(0.236 \times 10^{-9} \mathrm{m})^2} \)

Calculate the initial potential energy

We can calculate the initial potential energy between the ions using the following formula: \(U_{initial} = \frac{k \times q_1 \times q_2}{r}\) \(U_{initial} = \frac{8.99 \times 10^{9} \mathrm{Nm^{2}C^{-2}} \times 2 \times 1.6 \times 10^{-19} \mathrm{C} \times -1.6 \times 10^{-19} \mathrm{C}}{0.236 \times 10^{-9} \mathrm{m}} \)

Calculate the final potential energy

Now, we will calculate the final potential energy between the ions when they are separated by a distance of \(1.00 \mathrm{cm}\): \(U_{final} = \frac{k \times q_1 \times q_2}{r'}\) \(U_{final} = \frac{8.99 \times 10^{9} \mathrm{Nm^{2}C^{-2}} \times 2 \times 1.6 \times 10^{-19} \mathrm{C} \times -1.6 \times 10^{-19} \mathrm{C}}{1.00 \times 10^{-2} \mathrm{m}}\)

Calculate the work required

Finally, we can calculate the work required to increase the distance between the ions by finding the difference between the final potential energy and the initial potential energy: \(W = U_{final} - U_{initial}\) \(W = \left(\frac{8.99 \times 10^{9} \mathrm{Nm^{2}C^{-2}} \times 2 \times 1.6 \times 10^{-19} \mathrm{C} \times -1.6 \times 10^{-19} \mathrm{C}}{1.00 \times 10^{-2} \mathrm{m}}\right) - \left(\frac{8.99 \times 10^{9} \mathrm{Nm^{2}C^{-2}} \times 2 \times 1.6 \times 10^{-19} \mathrm{C} \times -1.6 \times 10^{-19} \mathrm{C}}{0.236 \times 10^{-9} \mathrm{m}}\right)\) Calculate the value of \(W\) to obtain the work required to increase the distance between the ions of gaseous sodium chloride to \(1.00 \mathrm{cm}\).