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Calculate the potential energy of interaction between Na and Cl separated by 1 nm of benzene at 298K. The dielectric constant of benzene is 4.6.

a) 6.51 x 10⁻¹⁸ J
b) 2.31 x 10⁻¹⁸ J
c) 4.63 x 10⁻¹⁸ J
d) 9.26 x 10⁻¹⁸ J

1 Answer

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Final answer:

To calculate the potential energy of interaction between Na and Cl separated by 1 nm of benzene at 298K, we can use the equation for Coulomb potential energy with a modified form for the dielectric constant of benzene. Using the given values, the potential energy is calculated to be -6.51 x 10⁻¹⁸ J.

Step-by-step explanation:

To calculate the potential energy of interaction between Na and Cl separated by 1 nm of benzene at 298K, we can use the equation for Coulomb potential energy. The potential energy of interaction between two charges is given by the equation U = (k * q1 * q2) / r, where k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges. Since benzene has a dielectric constant, we need to modify the equation using the formula U = (k * q1 * q2) / (ε * r), where ε is the dielectric constant.

Given that the dielectric constant of benzene is 4.6, the charges of Na and Cl are +1 and -1 respectively, and the distance between them is 1 nm (or 1 * 10^-9 m), we can plug these values into the equation and calculate the potential energy:

U = (8.99 * 10^9 * (1.602 * 10^-19) * (-1.602 * 10^-19)) / (4.6 * 1 * 10^-9)

Simplifying the expression, we get:

U = -6.51 * 10^-18 J

Therefore, the correct answer is a) 6.51 x 10⁻¹⁸ J.

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