Adapted from Molecular Driving Forces - Statistical Thermodynamics in Chemistry and Biology by Ken A. Dill and Sarina Bromberg, Garland Science, Taylor & Francis Group, New York and London (2003). Hereafter, we will abbreviate the names of the authors as D&B.(a) One mole of a molecular system can

Question

Adapted from Molecular Driving Forces - Statistical Thermodynamics in Chemistry and Biology by Ken A. Dill and Sarina Bromberg, Garland Science, Taylor & Francis Group, New York and London (2003). Hereafter, we will abbreviate the names of the authors as D&B.

(a) One mole of a molecular system can occupy any one of the four energy states below:

________ [latex]E _{3}=11 kCal / mol[/latex]
[latex]E _{1}=8 kCal / mol[/latex] ________ ________ [latex]E _{2}=8 kCal / mol[/latex]
________ [latex]E _{0}=3 kCal / mol[/latex]

1. Calculate U at T = 300 K.
2. Calculate the probability that a given snapshot of the system will have an energy of 3 kCal/mol at 300 K.
3. If the energy of each state is increased by 2 kCal/mol, what is the probability
that a given snapshot of the system will have an energy of 3 kCal/mol at
300 K?
4. Calculate U as T gets very large.
5. Calculate U as T gets very small.

(b) A four-bead chain can adopt several conformations that may be grouped as shown in the energy ladder below:

The distance between the chain ends is 1 lattice unit in the compact conformation, 3 lattice units in the extended conformation, and [latex]\sqrt{5}[/latex] lattice units in each of the other three chain conformations.

1. Calculate the average end-to-end distance of the chain (in lattice units) as a function of temperature.
2. Calculate the maximum value of the average end-to-end distance of the chain (in lattice units). At which temperature will this be realized?
3. Calculate the temperature at which the average end-to-end distance of the chain is equal to one half of its maximum value.

Accepted Answer

(a1) Because the number of moles, the temperature, and the system volume are constant, we can use the Canonical ensemble to solve this problem. Working on a per mole basis, underbars will not be carried in the solution of this problem. It is most convenient to compute the average energy, [latex]\langle E angle[/latex], or internal energy, U, as follows:

[latex]U=\langle E angle=\frac{\sum_{i=0}^{3} E_{i} e^{-E_{i} / R T}}{Q}[/latex] (1)

where the Canonical partition function, Q, is given by:

[latex]Q=\sum_{i=0}^{3} e^{-E_{i} / R T}[/latex] (2)

In Eqs. (1) and (2), we consider four distinguishable energy states, [latex]E_{0}, E_{1}, E_{2}[/latex], and [latex]E_{3}[/latex] (see the energy ladder in the Problem Statement). At T = 300 K, it follows that:

[latex]\left.R T ight|_{T=300 K }=\left(1.98717 imes 10^{-3} kCal /( molK ) ight)(300 K )=0.5962 kCal / mol[/latex] (3)

Using Eqs. (2) and (3), along with [latex]E_{0}=3 kCal / mol , E_{1}=E_{2}=8 kCal / mol[/latex], and [latex]E_{3}=11 kCal / mol[/latex], we obtain:

[latex]Q=e^{-(3) /(0.5962)}+2 e^{-(8) /(0.5962)}+e^{-(11) /(0.5962)}=0.006527[/latex] (4)

In addition, the numerator in Eq. (1) is given by:

[latex]\sum_{i=0}^{3} E_{i} e^{E_{i} / R T}=\left[(3) e^{-(3) /(0.5962)}+2(8) e^{-(8) /(0.5962)}+(11) e^{-(11) /(0.5962)} ight] kCal / mol[/latex]

[latex]\sum_{i=0}^{3} E_{i} e^{E_{i} / R T}=0.019596 kCal / mol[/latex] (5)

Using Eqs. (5) and (4) in Eq. (1) yields the desired result:

[latex]U=\frac{0.019532 kCal / mol }{0.006506}=3.00229 kCal / mol[/latex] (6)

(a2) The probability that the system will be in the ground state (0) having an energy [latex]E_{0}=3 kCal / mol[/latex] is given by:

[latex]p_{0}=\frac{e^{-E_{0} / R T}}{Q}[/latex] (7)

Using [latex]E_{0}=3 kCal / mol[/latex], along with Eqs. (3) and (4), in Eq. (7) yields:

[latex]p_{0}=\frac{e^{-3 / 0.5962}}{0.006506}=0.99954, ext { or } 99.954 \%, ext { a very high probability! }[/latex] (8)

(a3) If each energy is increased by 2 kCal/mol, then, the possible energy states will be [latex]E_{0}=5 kCal / mol , E_{1}=E_{2}=10 kCal / mol ext {, and } E_{3}=13 kCal / mol[/latex]. Therefore, the lowest possible energy state that the system can attain is now 5 kCal/mol. As a result, the system can never have E = 3 kCal/mol, and hence P (E = 3 kCal/mol) = 0!

(a4) As [latex]T ightarrow \infty[/latex], the term [latex]e^{-E_{i} / R T}[/latex] tends to 1. It then follows that:

[latex]p_{i}=\frac{e^{-E_{i} / R T}}{Q}=\frac{1}{Q}=\frac{1}{4}[/latex] (9)

Equation (9) indicates that the four available energy states are likely to be equally populated. As a result, the ensemble-averaged energy, U, is equal to the average of [latex]E_{0}, E_{1}, E_{2}, ext { and } E_{3}[/latex]. That is:

[latex]U=\langle E angle=\frac{[3+(2)(8)+11] kCal / mol }{4}=\left(\frac{30}{4} ight) kCal / mol =7.5 kCal / mol[/latex] (10)

(a5) As [latex]T ightarrow 0[/latex], the term [latex]e^{-E_{i} / R T}[/latex] tends to zero. However, it tends to zero more rapidly for the higher-energy states than for the lower-energy states. As a result, the only energy state that is likely to be populated is the ground state (0) having energy [latex]E_{0}=3 kCal / mol[/latex]. In other words, as [latex]T ightarrow 0[/latex], the ensemble-averaged energy is equal to the energy of the ground state. In mathematical terms, one obtains:

[latex]\left.U ight|_{T ightarrow 0}=\left.\langle E angle ight|_{T ightarrow 0}=\lim _{T ightarrow 0} \frac{\sum_{i=0}^{3} E_{i} e^{-E_{i} / R T}}{\sum_{i=0}^{3} e^{-E_{i} / R T}}=\lim _{T ightarrow 0} \frac{E_{0} e^{-E_{0} / R T}}{e^{-E_{0} / R T}}=E_{0}=3 kCal / mol[/latex] (11)

(b1) As can be seen from the energy ladder in the Problem Statement, there are five conformations of the four-bead chain: the conformation with the bead-bead contact is taken to have energy [latex]\varepsilon=0[/latex] (the ground state). The other four conformations have no bead-bead contacts, and all have energy [latex]\varepsilon=\varepsilon_{0}[/latex], where [latex]\varepsilon_{0}[/latex] is a constant. The average end-to-end distance of the four-bead chain is given by:

[latex]\langle d angle=\sum_{i=1}^{5} d_{i} p_{i}[/latex] (12)

where according to the Problem Statement (in lattice units):

[latex]\begin{aligned}&d_{1}=1 \&d_{2}=d_{3}=d_{4}=\sqrt{5} \&d_{5}=3 \&\varepsilon_{1}=0 \&\varepsilon_{2}=\varepsilon_{3}=\varepsilon_{4}=\varepsilon_{5}=\varepsilon_{0}\end{aligned}[/latex] (13)

In addition, the various probabilities are given by:

[latex]\begin{aligned}p_{1} &=\frac{e^{-\varepsilon_{1} / k_{B} T}}{q}=\frac{1}{q}, \quad p_{2}=p_{3}=p_{4}=p_{5}=\frac{e^{-\varepsilon_{0} / k_{B} T}}{q}, ext { and } q=\sum_{i=1}^{5} e^{-\varepsilon_{i} / k_{B} T} \&=1+4 e^{-\varepsilon_{0} / k_{B} T}\end{aligned}[/latex] (14)

Using Eqs. (13) and (14) in Eq. (12) yields:

[latex]\langle d angle=1\left(\frac{1}{q} ight)+\frac{3(\sqrt{5}) e^{-\varepsilon_{0} / k_{B} T}}{q}+\frac{3 e^{-\varepsilon_{0} / k_{B} T}}{q}[/latex] (15)

Using q in Eq. (14) in Eq. (15) yields:

[latex]\langle d angle=\frac{1+9.71 e^{-\varepsilon_{0} / k_{B} T}}{1+4 e^{-\varepsilon_{0} / k_{B} T}}[/latex] (16)

(b2) The maximum value of the end-to-end distance of the chain is obtained when [latex]T ightarrow \infty ext { and } e^{-\varepsilon_{0} / k_{B} T} ightarrow 1[/latex]. In that case (in lattice units):

[latex]\lim _{T ightarrow \infty}\langle d angle=\langle d angle_{\max }=\frac{1+9.71}{1+4}=\frac{10.71}{5} \cong 2.14[/latex] (17)

(b3) We are asked to calculate at what temperature the end-to-end distance of the chain is equal to half of [latex]\langle d angle_{\max }[/latex] given in Eq. (17). We therefore require that [latex]\langle d angle[/latex] in Eq. (16) be equal to [latex]\frac{\langle d angle_{\max }}{2}[/latex]. Specifically:

[latex]\frac{1+9.71 e^{-\varepsilon_{0} / k_{B} T}}{1+4 e^{-\varepsilon_{0} / k_{B} T}}=\frac{\langle d angle_{\max }}{2}=1.07[/latex] (18)

[latex]1+9.71 e^{-\varepsilon_{0} / k_{B} T}=1.07\left(1+4 e^{-\varepsilon_{0} / k_{B} T} ight)[/latex]

[latex]\begin{gathered}(9.71-1.07(4)) e^{-\varepsilon_{0} / k_{B} T}=5.43 e^{-\varepsilon_{0} / k_{B} T}=(1.07-1)=0.07 \e^{-\varepsilon_{0} / k_{B} T}=\frac{0.07}{5.43}=0.01305 \-\frac{\varepsilon_{0}}{k_{B} T}=\ln (0.01305)=-4.339\end{gathered}[/latex]

[latex]T=0.2305\left(\frac{\varepsilon_{0}}{k_{B}} ight)=0.2305 T_{0}[/latex] (19)

Question 24.1: Adapted from Molecular Driving Forces - Statistical Thermody...

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