Answer to Question #97096 in General Chemistry for Beverlie

Question

Calculate the electrostatic potential energy for the reaction of gaseous potassium and chloride given the following information: The radius of K is 133 pm and the radius of Cl is 181 pm

Answer 1

Q97096

Solution:

The reaction of gaseous potassium and chloride yields the ionic compound KCl, comprising of $K^+$ and $Cl^-$ ions.

$|q_1|=|q_2|=1e^-=1.6×10^{−19} J$

$r=r_++r_-$

$=133+181$

$=214 pm.$

Electrostatic Potential Energy is;

$U=kq_1q_2/r$

where;

$k=9*10^9 Nm^2/C^2$

$U=9*10^9*(1.6*10^{-19})^2/(214*10^{-12})$

$=10.766*10^{-19} J.$ (Answer)

$=10.766*10^{-19}/(1.6*10^{-19}))eV$

$=6.729eV.$ (Answer)

Thus, Electrostatic Potential Energy is $10.766*10^{-19} J.$ or $6.729eV.$