1. Given kB=0.0020 kcal/mol-K, find the probability of each of two states: one w

Question

1. Given kB=0.0020 kcal/mol-K, find the probability of each of two states: one with energy 12.32 kcal/mol and the other with an energy of 14.91 kcal/mol at a temperature of 381 K.

2. Find the probability of each of two states at 333 K. One state has an energy of 1.23 kcal/mol and an entropy of 5.51 kcal/mol-K and the other state has an energy of 2.91 kcal/mol-K and an entropy of 5.63 kcal/mol-K.

3. At 363 K, one state is twice as likely as the other. The first (more likely) state has an energy of 9.19 kcal/mol and an entropy of 0.915 kcal/mol-K. The other state has an energy of 2.19 kcal/mol more then the first state. What is the entropy of the second (less likely) state?

4. Given two states, will the lower energy state every be disfavored? Under what circumstances? Include entropy in the discussion.

kB is Boltzman's Constant.

Explanation / Answer

iGiven kB = 0.0020 kcal/mol-K.

Part 1:

For any system, if T > 0, then if the system can exist in more than one energy states, then the relative population of them in those specified energy levels is given by the Boltzmann distribution. At a temperature T, if Na and Nb are population of two states with corresponding energy eigenvalues Ea and Eb then the Boltzmann distribution is given by:

Na/Nb = exp[-(Ea - Eb)/kBT]

Given Ea = 14.91 kcal/mol and Eb = 12.32 kcal/mol and T = 381 K. then;

Na/Nb = exp[-(14.91 - 12.32)/0.0020*381] = 0.033

If Na and Nb are considered as my probabilities, then Na + Nb = 1.

Solving for Na and Nb,we have Na = 0.97 and Nb = 0.03.

Part 2:

When entropy comes into picture, we know that the probabiltily doesn't only depend on the energy. Let's do the math.

From Boltzmann interpretation, we know that S = kBlnW where W is the number of microstates(or the total number of arrangements that the molecule can have by keeping the total energy of the system constant.

From the entropy equation, we have W = exp(S/kB). Now, the probabilities will be weighted along with the number of microstates possible.

Therefore pi = W*exp(-Ei/kBT) = exp(Si/kB)*exp(-Ei/kBT) = exp[-(Ei -TSi)/kBT] = exp(-Gi/kBT)

where G = E - TS is the Gibbs free energy.

Now, T = 333 K , Ea = 1.23 kcal/mol and Eb = 2.91 kcal/mol, Sa = 5.51 kcal/mol-K and Sb = 5.63 kcal/mol-K.

So, from the above developed theory, we have the Boltzmann distribution in the following form:

Na/Nb = exp[-(Ga - Gb)/kBT]
Or, Na/Nb = exp[-((Ea - TSa) - (Eb- TSb))/kBT]

Or, Na/Nb = exp[-((1.23 - 5.51*333) - (2.91-5.63*333))/0.0020*333] = 1.09*10-25 (this value is negligible)

Also here Na + Nb= 1, so, Na = 0 and Nb = 1.   Note that the Gibbs energy for state a is higher than that of b .

Part 3:

Let the unknown entropy be S
From the similar equation, Na/Nb = exp[-((Ea - TSa) - (Eb- TSb))/kBT], we have;

2/1 = exp[-((9.19 - 0.915*363) - (2.19 - 363*S))/0.0020*363]

From here, S is evaluated to be 0.894 kcal/mol-K.

Part 4:

From the previous evaluated solution it is evident that the energy is not the only factor that is governing the likeliness of different states. When entropy is involved, it is the Gibbs free energy that is the cumulative responsible factor to determine the probability of different states. Lesser the Gibbs energy of a state, more is the probability corresponding to that state., despite it being in either higher energy of lower energy.

G1 = E1 -TS1 and G2 = E2 - TS2. Then even if E1 > E2 then also state 2 may be disfavored as compared to 1, if and only if G1 < G2

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