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What is the electrostatic potential energy (Eel) of the ionic bond between a Ca2+ ion and a S2– ion in joules? Before you do the calculation, predict whether the result you get will be less than (more negative) or greater than (less negative) than the Eel value for KCl. The ionic radii of Ca2+ and of S2– are 100 pm and 184 pm, respectively.

Respuesta :

nuhulawal20

Answer: the electrostatic potential energy is -3.26 × 10⁻¹⁸ J

Explanation:

Given the data in the question;

I think the value of electrostatic potential energy of calcium sulfide CaS formed will be less(more negative) than KCl because the Ca2+ and S2- ions has more charge that is +2 and -2 respectively when compared to Kcl which has +1 and -1 charge.so it will be more negative.

radius = r[tex]_{ca+2}[/tex] + r[tex]_{s-2}[/tex] = 100 + 184 = 284 pm = 2.84 × 10⁻¹⁰ m

we know that; k = 2.31 × 10⁻²⁸ J

q1 = 2  ( charge on Ca⁺² )

q2 = -2( charge on S⁻²)

so

Ep = k × q1q2/r

so we substitute

Ep = 2.31 × 10⁻²⁸ × (+2×-2) / 2.84 × 10⁻¹⁰

Ep = (2.31 × 10⁻²⁸ × -4) / 2.84 × 10⁻¹⁰

Ep = -9.24 × 10⁻²⁸ / 2.84 × 10⁻¹⁰

Ep = -3.26 × 10⁻¹⁸ J

Therefore, the electrostatic potential energy is -3.26 × 10⁻¹⁸ J

mickymike92

Based on the data provided, the electrostatic potential energy is -3.26 × 10⁻¹⁸ J

What is electrostatic potential energy?

The electrostatic potential energy is the work done in moving the charge from infinity to that point against the electric field.

The electrostatic potential energy between any two charges q1, q2 separated by distance r is derived from Coulomb’s law and mathematically given as:

  • E = k × [q1q2/r2]

where

  • k is a constant = 2.31 × 10⁻²⁸ J
  • q1 and q2 are the charges
  • r is the distance of separation

From the data provided:

ionic radii of Ca2+ r1 = 100 pm

ionic radii of S2– = 184 pm.

Chargeon Ca2+ = +2

Charge on S2_ = -2

From the data above, it most likely that the value of electrostatic potential energy of calcium sulfide CaS formed will be less(more negative) than KCl because the Ca2+ and S2- ions has a greater charge of +2 and -2 respectively compared to Kcl which has +1 and -1 charge.

Calculating the electrostatic potential energy, Ep:

radius = r1 + r2 = 100 + 184 = 284 pm

r = 2.84 × 10⁻¹⁰ m

k = 2.31 × 10⁻²⁸ J

q1 = 2  ( charge on Ca⁺² )

q2 = -2( charge on S⁻²)

Then:

Ep = k × q1q2/r

Substituting the values above in the equation:

Ep = 2.31 × 10⁻²⁸ × (+2 ×-2) / 2.84 × 10⁻¹⁰

Ep = (2.31 × 10⁻²⁸ × -4) / 2.84 × 10⁻¹⁰

Ep = -9.24 × 10⁻²⁸ / 2.84 × 10⁻¹⁰

Ep = -3.26 × 10⁻¹⁸ J

Therefore, the electrostatic potential energy is -3.26 × 10⁻¹⁸ J.

Learn more about electrostatic potential energy at: https://brainly.com/question/24648428

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