Answered

Here's the 2nd question. Calculate the distance between the electrostatic interaction of potassium and bromide ions which typically results in a molar Coulomb potential energy of (-)970 kJ/mol in nanometers. Charge of ions= 1.602x10^-19 C, permittity=8.854x10^-12C^2/Jm, 6.02x10^23/mol



Answer :

MathPhys

Answer:

0.143 nm

Explanation:

The Coulomb potential energy between two charged particles can be found with:

[tex]\large \text{$ U=\ $} \LARGE \text{$ \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r} $}[/tex]

where

  • U is the electric potential energy
  • ε₀ is the vacuum permittivity (8.854×10⁻¹² C²/J/m)
  • q₁ and q₂ are the charges of the particles
  • r is the distance between the particles

Plug in values and solve for the distance r.

[tex]\large \text{$ U=\ $} \LARGE \text{$ \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r} $}\\\\\large \text{$ 970\frac{kJ}{mol}\times\frac{1\ mol}{6.02\times 10^{23}}\times\frac{1000 J}{kJ} =\ $} \LARGE \text{$ \frac{1}{4\pi \times (8.854\times10^{-12}\ C^2/J/m)}\frac{(1.602\times 10^{-19}\ C)^2}{r} $}\\\\\large \text{$ 1.611\times 10^{-18} J=\ $} \LARGE \text{$ \frac{2.307\times 10^{-28} Jm}{r} $}\\\\\large \text{$ r=1.43\times10^{-10}\ m $}\\\\\large \text{$ r=0.143\ nm $}[/tex]

Rounded to three significant figures, the distance between the ions is 0.143 nanometers.

Other Questions