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Atkins' Physical Chemistry | 11th Edition | ISBN: 9780198769866 | Authors: Atkins, Peter; De Paula, Julio; Keeler, James ISBN: 9780198769866 2042

Solution for problem P14C.1 Chapter 14C

Atkins' Physical Chemistry | 11th Edition

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Atkins' Physical Chemistry | 11th Edition | ISBN: 9780198769866 | Authors: Atkins, Peter; De Paula, Julio; Keeler, James

Atkins' Physical Chemistry | 11th Edition

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Problem P14C.1

A simple pair distribution function has the form

\(g(r)=1+\cos \left(\frac{4 r}{r_{0}}-4\right) \mathrm{e}^{-\left(r / r_{0}-1\right)}\)

for \(r \geq r_{0}\) and g(r)=0 for \(r<r_{0}\). Here the parameter \(r_{0}\) is the separation at which the Lennard-Jones potential energy function (eqn 14B.12) \(V=4 \varepsilon\left\{\left(r_{0} / r\right)^{12}-\left(r_{0} / r\right)^{6}\right\}\) is equal to zero.

(a) Plot the function g(r). Does it resemble the form shown in Fig. 14C.1?

(b) Plot the virial \(v_{2}(r)=r(\mathrm{~d} V / \mathrm{d} r)\).

Text Transcription:

g(r)=1+cos(4r/r_0 -4) e^{-(r/r_0-1)

r geq r_0

r<r_0

r_0

V=4 varepsilon{r_0/r)^12-(r_0/r)^6}

v_2(r)=r(dV/dr)

Step-by-Step Solution:

REDOX REACTIONS Redox reactions refer to the simultaneous oxidation and reduction of reactants in a chemical reaction.  The substance being reduced gains an electron, and thus is oxidizing the other substance the oxidizing agent.  The substance being oxidized loses an electron, and thus is reducing the other substance  reducing agent. o A helpful pneumonic to remember this is OIL RIG: oxidation is losing; reduction is gaining. Balancing redox reactions can be tricky, and depend on whether you are in acidic or basic conditions. The first step is to write the reaction into 2 separate half reactions: Ex. Cu + 2Ag  Cu 2++ 2Ag Step 1: Our first half reaction: Cu  Cu 2+ (it is already obvious that this is what is being oxidized, since it loses 2 electrons)

Step 2 of 3

Chapter 14C, Problem P14C.1 is Solved
Step 3 of 3

Textbook: Atkins' Physical Chemistry
Edition: 11
Author: Atkins, Peter; De Paula, Julio; Keeler, James
ISBN: 9780198769866

This textbook survival guide was created for the textbook: Atkins' Physical Chemistry, edition: 11. The full step-by-step solution to problem: P14C.1 from chapter: 14C was answered by Aimee Notetaker, our top Chemistry solution expert on 04/25/22, 03:45PM. Atkins' Physical Chemistry was written by Aimee Notetaker and is associated to the ISBN: 9780198769866. The answer to “?A simple pair distribution function has the form\(g(r)=1+\cos \left(\frac{4 r}{r_{0}}-4\right) \mathrm{e}^{-\left(r / r_{0}-1\right)}\)for \(r \geq r_{0}\) and g(r)=0 for \(r<r_{0}\). Here the parameter \(r_{0}\) is the separation at which the Lennard-Jones potential energy function (eqn 14B.12) \(V=4 \varepsilon\left\{\left(r_{0} / r\right)^{12}-\left(r_{0} / r\right)^{6}\right\}\) is equal to zero. (a) Plot the function g(r). Does it resemble the form shown in Fig. 14C.1? (b) Plot the virial \(v_{2}(r)=r(\mathrm{~d} V / \mathrm{d} r)\).Text Transcription:g(r)=1+cos(4r/r_0 -4) e^{-(r/r_0-1)r geq r_0r<r_0r_0V=4 varepsilon{r_0/r)^12-(r_0/r)^6}v_2(r)=r(dV/dr)” is broken down into a number of easy to follow steps, and 75 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 327 chapters, and 1120 solutions. Since the solution to P14C.1 from 14C chapter was answered, more than 201 students have viewed the full step-by-step answer.

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