Answers

Solutions to Study Problems

Experiment V:

Thermodynamics

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> 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)

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