1
$\begingroup$

If you have 2 ions of equal but opposite charge, will the force between them be larger in a vacuum and smaller in water? Would this be because the relative permittivity of water is greater than 1 (around 80)? Also, what does this mean for the interaction energy? Would an interaction be larger in vacuum? I am trying to figure out whether a reaction is more likely to take place in the two different media between the two ions...

$\endgroup$
5
$\begingroup$

Yes, much larger in vacuum. The interaction is proportional to the reciprocal of the dielectric constant (relative permittivity) $1/\epsilon$ so is less in water. Think of it as an attenuation of the electric field around an ion, the larger $\epsilon$ is the more the field is attenuated and is true whatever the charges on the ions are making them attractive or repulsive. The force at distance r between charges $q_1, q_2$ is

$$F=\frac{q_1q_2}{(4\pi\epsilon_0)\cdot\epsilon r^2}$$

in SI units; $\epsilon_0 $ is the permittivity of free space. The charges are $q=ze$ where e is the charge on the electron and z the ionic valency, $\pm 1, 2 \cdots $ etc.

The interaction energy is

$$U=\frac{q_1q_2}{(4\pi\epsilon_0)\cdot\epsilon r} \qquad \mathrm{Joule}$$

The electric field around charge $q_1$ is

$$E=\frac{q_1}{(4\pi\epsilon_0)\cdot\epsilon r^2}$$

in V/m. The force acting on a second charge $q_2$ is $F=Eq_2$

Its worth plugging in some numbers for the energy for different $\epsilon$ and comparing this to thermal energy $k_BT$.

$\endgroup$
  • $\begingroup$ thank you-very informative. Just to check, if I was to calculate a force in water, would I use 80 in place of ϵ in the "ϵ0ϵ " term? Also, if then I had 2 +1 charges, and I wanted them to react, would it be best in water based on your argument of weakening the field? $\endgroup$ – gamma1 Jun 2 '17 at 21:07
  • $\begingroup$ yes use $80 $ for water as its the relative permittivity no units, $\epsilon_0 = 8.854\cdot 10^{-12} \pu{F\,m^{-1}}$. Yes also if charges are similar high dielectric is best. The rate const is $\ln(k)=\ln(k_0)-U/k_BT$ where $k_0$ is rate const at infinite dielectric const. You might also want to consider increasing the ionic strength of the solution, look up 'primary kinetic salt effect'. $\endgroup$ – porphyrin Jun 2 '17 at 21:31
  • $\begingroup$ @porphyrin It could also be worth mentioning the reason for dielectric constant. Water molecules are polar and orient themselves to become antiparallel with the field. $\endgroup$ – Pritt Balagopal Jun 3 '17 at 4:15
  • 1
    $\begingroup$ @Pritt , both polar and non-polar, i.e all molecules have a dielectric constant. In non-polar ones this is $\epsilon-1 \sim n\alpha$ where n is the number density and $\alpha$ the polarisability. In polar ones additionally there is a term $\epsilon-1 \sim n\mu^2/(k_BT)$ where $\mu$ is the dipole moment. Some partial alignment takes place but this is opposed by thermal motion via the $k_BT$ term. It is too strong a statement to say that molecules become aligned, they are only ever partially so in a liquid . $\endgroup$ – porphyrin Jun 3 '17 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.