Question

1. A state has an energy  = 76.5×10−21 J and a degeneracy of 15. What is the free energyof the state at 25◦C and at 150◦C? Express your answer in kJ/mol.

2. A molecule has a bond that can rotate between the cis and trans states, both of whichhave zero entropy. It is found that the trans state is three times more common than thecis state when the temperature is 250 K. If the molecule is heated to 450 K, what is thenew ratio of trans to cis?

3. A system consists of two independent subsystems, A and B. Subsystem A has 1.3 × 106
allowed microstates and subsystem B has 4.6 × 107 allowed microstates. What is theentropy of the combined system?

4. Macrostate X consists of 5 microstates, each of which have an energy of x = 1.5×10−21
Joules. Macrostate Y consists of 150 microstates, each of which have an energy y =7.0 × 10−21 J. Calculate probability that the system is in state X if the temperature is25◦C.

5. A molecular system at room temperature (25◦C) is found with energy EA 95% of thetime and with energy EB 5% of the time. It is known that state A consists of an ensembleof 3 microstates while there are only 2 microstates in state B. Determine the differencein energies between the two states (EA − EB).

6. Chemical reactions are often described using a three state modelReactants → Transition State → ProductsIn most cases the energy of the transition state is much higher than the energy of the
reactant state. This means that the reaction cannot proceed until there is a randomthermal fluctuation large enough to ‘kick’ the reactant molecule(s) up to the transitionstate energy. Say we have a reaction in which the transition state is 9.0 × 10−20 J abovethe reactant state.
(a) Calculate the ratio of the probability the system is in the transition state to the probability that it is in the reactant state PTS/PR at 37◦C.

(b) Use your answer above to estimate the time it takes for the reaction to occurspontaneously at 37◦C. Assume that the system samples a new microstate everynanosecond (10−9 s).

(c) One way to speed up the reaction is to heat the system. If the temperature isincreased to 1000 K, how long does it take for the reaction to occur?

(d) In biological systems it is not feasible to accelerate reactions by heating them to1000 K. Instead, the reactions rely on enzymes that catalyze reactions by loweringthe energy of the transition state. If catalyzed reaction has a transition state energy2.0 × 10−20 J, what is the approximate reaction time? Use a temperature 37◦C.
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(e) Finally, estimate the uncatalyzed reaction time if the transition state is not a uniquemicrostate, but actually an ensemble of 100 microstates. Use a temperature 37◦C.

7. A salt ion is dissolved in a large volume of pure water. The wall of the container has asingle defect that can bind the ion with an energy bound = −2.1 × 10−19 J.

(a) If the ion is only bound to the impurity site 2% of the time at room temperature(25◦C), what is the free energy of the ion in the unbound state? Assume that thefree energy of the bound state is equal to the binding energy (the state is non-degenerate).

(b) Calorimetry experiments reveal that the energy of the ion in the unbound state issolv = −5.1 × 10−20 J. Is the change in energy bigger or smaller than kBT at roomtemperature? [i.e. calculate (solv − bound)/kBT]

(c) What is the entropy of the unbound state?

(d) How many microstates are in the unbound ensemble?

Answer 1

1. Free energy of the state = G - G₀ = -nRTln (q/N) .....[N = no of particles ; here N = N_A as we have to                                                                                                            determine G for 1 mole]

q = ge^{- ε/kT} = 15* e^{(76.5×10−21 J)/(1.38*10-23 J.K-1)(298K)}

= 1.25*10^-07

Free energy = - 1*8.314*298 ln (1.25*10^-07)/(6.023*10²³) = 175 kJ/mol

Do the same for T = 423 K

2. Cis <------> Trans ; K= [trans]/[cis]   ; at 250 Kelvin K = 3/1 = 3

You have to calculate the equilibrium at the new temperature (450 K).

We have the value of K at 250K. Let's see what we can get from this.

The relationship between dG and K is: dG = -RT ln K ...... (1)

and

dG = dH - TdS ....(2)

Both of the molecules have zero entropy; so form (2) we get dG = dH....( as dS=0)

At 250 K, dG = dH = -RT ln K = -(8.314 J/K.mol)(250K) ln 3 = -2283.5 J/mol

Now, dG = dH = -RT ln K

Integrating this equation from T₁ to T₂ we get:   ln K₂/K₁ = (dH/R)[T₂ - T₁/T₁T₂] .......(dH of reaction is assumed to be constant over the temperature range]

ln K₂/K₁ = (dH/R)[T₂ - T₁/T₁T₂]

or, ln K₂/3 = (-2283.5 J.mol^-1/ 8.314 J.K^-1.mol^-1) [ 450-250/450*250]K^-1

or, K₂ = 1.84

So the new ratio of trans to cis = 1.84

3. we know , S = klnW where W = no of microstates

For 1 mole of substance, S =RlnW

For system A: S_A = RlnW_A

For system B: S_B = RlnW_B

Total entropy,S = S_A + S_B = RlnW_A + RlnW_B = R ln W_AW_B

S = (8.314 J/K.mol) ln (1.3 × 10 ⁶ * 4.6 × 10⁷)

= 263.7 J/K.mol

4. probability that the system is in state X

= ge^{- ε/kT} / ∑ ge^{- ε/kT}

= 5* e^{- (1.5×10−21 J)/ (1.38*10-23 J.K-1)(298K)} / [5* e^{- (1.5×10−21 J)/ (1.38*10-23 J.K-1)(298K)}] + [150* e^{- (7×10−21 J)/ (1.38*10-23 J.K-1)(298K)}]

= 3.47/3.47+27.34

= 0.11

=11%

5. We can write,

3e^{- Ea/kT} / 2e^{- Eb/kT} = 95/5

or, e^{(Eb- Ea)/kT} = (95/5) * (2/3)

or, (Eb- Ea) = kT ln 12.67

or, Eb- Ea = 1.04*10^-20 J ......(put the value of k and T)

So, Ea - Eb = - 1.04*10^-20 J