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)

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)