Solutions to Study Problems

Experiment V:

Thermodynamics

 

1. The vapor pressure of ethyl alcohol (ethanol or etOH) is 100.0mmHg at 34.9oC and 400.0mmHg at 63.5oC. Calculate the average heat of vaporization of etOH across this range.

Solution:

Pvap(EtOH)1=100.0mmHg at 34.9 oC

Pvap(EtOH)2 = 400.0mmHg at 63.5 oC

34.9oC + 273.15K = 308.05K = T1

P1 = 100.0 mmHg (1atm/760mmHg) = 0.13158 atm

P2 = 0.52632 atm

(Do I really think I am justified keeping this many significant figures? No, but I'll round off at the end.)

Let's now use the Clasius-Clapeyron equation as provided by Oxtoby.

ln (P2 / P1) = (DHvap / R)[(1 / T1) - (1 / T2)]

ln (0.52632 / 0.13158) = (DHvap / 8.314 J/mol K)[(1 / 308.05K) - (1 / 336.65K)

DHvap =41792 J / mol = 41.8 kJ / mol

(Note: Using the Clasius-Clapeyron equation is the same thing as just finding the slope of a line plot of ln P vs 1/T using these two points and multiplying it by R like you did for the experimental write-up of this lab)

2. The vapor pressure o f water at 363K is 526mmHg, and the average heat of vaporization across the temperature range from 363K to 373K is 40.8 kJ/mol. Calculate the vapor pressure of water at 373K.

Solution

Here again, the Clasius-Clapeyron equation seems useful.

ln (P2 / P1) = (DHvap / R) [(1 / T1) - (1 / T2)]

ln (P2 / 0.692105 atm) = (40800 J mol -1 / 8.314 J mol-1 K-1)[(1 / 363 K) - (1 / 373 K)]

P2 /0.69211 atm = e .362435

P373 K = 756 mmHg

3. Write the equilibrium constant for the following process:

A(liquid)<-----> A(gas)

Why is the concentration of A(liquid) not included in the equilibrium constant?

Solution:

When you are dealing with the equilibrium of reactions involving gases, you measure concentrations in terms of partial pressures.

Kp = PA(g) [A]l remains constant and P represents partial pressure.

Liquids and solids exert negligible pressures. A (liquid) is not included in the equilibrium constant because [A]l, is constant. Look at a more familiar case of this.

H2O + H2O <----> H3O+ + OH- (auto-ionization of water)

Keq = [H3O+][OH-] / [H2O]2

Keq [H2O]2 = Kw = [H30+][OH-]

This is another example where a fairly constant concentration of a liquid (about 55M of water) was absorbed into an equilibrium constant and the constant renamed. This absorbing of a constant into Keq is fairly common.

4. For the following reaction, DHoand DS o are 40kJ/mol and 80kJ/mol K respectively.

A(g) + B(g) <----> 3C(g)

a. Is this reaction spontaneous at 25 o C and 1 atm? Why? (Pay attention to units!!)

b. At what temperature would this reaction become spontaneous?

c. What's the equilibrium constant for this reaction at 25 o C and 1 atm? (R=8.314J/mol K)

Solution

(a) DG o =DH o - T DS o

DG o = 4 x 104 J / mol - (298 K)(80 J/ mol K)

DG o = 16160 J / mol

Since DGo> 0, the reaction is not spontaneous at 25 oC.

(b) DG o =DH o - T DS o

0 > 4 x 104 J / mol - T(80 J / mol K)

(DG o < 0 for a reaction to be spontaneous)

500 K < T

(c) ln Keq = (DH o / RT) + (DS o / R) (eqn 5.7 in the manual)

ln Keq = (-4 x 104 J / mol) / [(8.314 J / mol K)(298 K)] + (80 J / mol K) / (8.314 J / mol K)

ln Keq = -6.5225

Keq = .00147

5. Circle the sign of ĘS for the following processes?

a) The evaporation of water ĘS > 0 ĘS < 0

b) The solidification of water ĘS > 0 ĘS < 0

c) The chemistry department blows up ĘS > 0 ĘS < 0

d) H2(g) + O2(g) ----->H2O(l) ĘS > 0 ĘS < 0

e) Photosynthesis ĘS > 0 ĘS < 0

Solution:

a)Remember, entropy is a measure of disorder. Gases are in a more chaotic, less ordered state than liquids, ĘS > 0

b)Solids are in a more ordered state than liquids, ĘS < 0

c)CHAOS! ĘS > 0

d)Again, liquid states are more ordered than the gas state, ĘS < 0.

e)CO2+ H20 + light energy ---> CH20 (carbohydrate) + O2 This is the overall equation for photosynthesis; as a result of this process small simple molecules (water and carbon dioxide) are converted into large, complex molecules (carbohydrates). Therefore, there is an increase in order as a result of this reaction, ĘS < 0

6. The change in enthalpy for the following reaction is 170.1 kJ/mol, and the change in entropy is 160 J/mol K. Calculate the lowest temperature at which this reaction occurs spontaneously. Assume that DH o and DS o are independent of temperature.

CaCO3(s) <-------> CaO(s) + CO2(g)

Solution:

DG o =DH o - T DS o

0 > 170100 J mol-1 - T (160 J mol-1K-1)

1063 K < T (This should tell you that calcium carbonate will not decompose spontaneously at 25 oC and 1 atm)