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)