Chemistry 121. Name _______________________________
Mid-Term Examination #2 [32 Marks] Friday, 26th March, 1999.
1. What is the role of the "Second Law of Thermodynamics" in chemistry?
The Second Law of Thermodynamics states that "in any spontaneous process the entropy of the Universe must increase." This means that for a chemical process to be spontaneous, the entropy of the Universe must increase as it takes place. Calculating the change in the entropy of the Universe for a process can be used to find whether it will be spontaneous or not under the given conditions.
2. The reaction between ethyne and oxygen is used in oxyacetylene welding:
2 C2H2(g) + 5 O2(g) Ž 4 CO2(g) + 2 H2O(g) DHº = -2510 kJ.
a. Is the reaction exothermic or endothermic? Explain your answer.
The enthalpy of the system is decreasing. Since energy cannot be created or destroyed, this must be lost to the surroundings. The process is exothermic.
b. Is the sign of the entropy change for the process +, -, or neither
(ie DSº Ÿ 0)? Explain your answer.
The entropy change will be negative, since the final state of the system is more ordered than the initial.
3. PbI2 dissolves in water: PbI2(s) ¾ Pb++(aq) + 2 I-(aq).
The K value for this equilibrium at 298 K is 8.3 x 10 -9. What is [I-] at equilibrium?
K = a(Pb2+) * a(I-)2 / a(PbI2) = (x)*(2x)2 / 1 = 8.3 x 10-9
So x = 1.28 x 10-3
And [I-] = 2.55 x 10-3 M.
4. Consider the equilibrium and the given data:
By measuring the total pressure in the flask at two different temperatures, you are to calculate the missing entries in the above table for COCl2(g).
a. Why is DHºf(Cl2) = 0 kJ/mol?
By definition, the enthalpy of formation is the change in enthalpy when 1 mole of the compound is made from its elements in their standard states. So converting Cl2(g)into Cl2(g) involves no change.
b. Why is Sº298(Cl2) > Sº298(CO)?
Cl2 is a more complex molecule in that it contains many more electrons, which must be ordered about the two nuclei than does CO.
c. 100 kPa of COCl2 is placed in a 4.00 L flask at 300 K and heated to 800 K. What is the pressure of COCl2 at 800 K?
P1 / T1 = P2 / T2. So P2 = 267 kPa.
d. Define K and Kp for this reaction.
K = a(CO)*a(Cl2) / a(COCl2)
Kp = P(CO)*P(Cl2) / P(COCl2) kPa.
e. At this temperature COCl2 dissociates according to the equation given. At equilibrium, the TOTAL pressure in the flask is found to be 381.8 kPa. Calculate Kp and then K800 at this temperature.
|E||267 - x||x||x|
Kp = x2 / (267 - x)
Need Kp and x! Two unknowns, need another equation:
Total pressure at equilibrium = 381.8 kPa = 267 -x +x +x = 267 + x
So x = 115.1
And Kp = 87.22 kPa.
From which K = 0.8722.
f. The reaction flask is heated to 900 K. If the COCl2 were not dissociated, what would the pressure of it be at this temperature?
P1 / T1 = P2 / T2. So P2 = 300 kPa.
g. In fact the COCl2 does dissociate and the TOTAL pressure at this temperature is found to be 515.8 kPa. Calculate Kp at this temperature and K900.
|E||300 - x||x||x|
Kp = x2 / (300 - x)
Need Kp and x! Two unknowns, need another equation:
Total pressure at equilibrium = 515.8 kPa = 300 -x +x +x = 300 + x
So x = 215.8
And Kp = 553 kPa.
From which K = 5.53.
h. From the two values of K at different temperatures (K800 and K900), calculate DHº and DSº for the reaction. (Note: If you do not have calculated values for K800 and K900 use K800 = 1.00 and K900 = 6.00 in this part. These are not the calculated answers!)
R = 8.314 J. mol-1. K-1.
ln K = DS/R - DH/RT
ln (K1 / K2) = (DH/R)(1/T2 - 1/T1)
from which DH = ln (K1 / K2)*R / (1/T2 - 1/T1) = 110,600 J
then using ln (5.53) = DS/R - 110,600/RT gives DS = 137 J/K
i. Now calculate the missing table entries for COCl2, DHºf and Sº298.
DHº = SmiDHºf(i) = (-1)(X) + (+1)(-110.5) + (+1)(0) = 110.6 kJ (remember UNITS)
from which DHºf(COCl2) = X = -221 kJ
DSº = SmiSº298(i) = (-1)(X) + (+1)(198) + (+1)(223) = 137 J/K
from which Sº298(COCl2) = X = 284 J/K
5. Predict whether BF3 will react with water or not under standard conditions:
BF3(g) + 3 H2O Ž H3BO3(s) + 3 HF(l)
DHºf -1137 -286 -1094 -271 kJ/mol
Sº298 254 70 89 174 J/mol.K
This reaction will be spontaneous of DGº298 < 0.
DHº = SmiDHºf(i) = (-1)(-1137) + (-3)(-286) + (+1)(-1094) + (+3)(-271) = 88 kJ
DSº = SmiSº298(i) = (-1)(254) + (-3)(70) + (+1)(89) + (+3)(174) = 147 J/K
from which DGº298 =DHº - TDSº = 88 - 298*147/1000 = 44.2 kJ (watch UNITS).
Since DGº298 is positive, the reaction is not spontaneous under standard conditions at 298 K.
6. Consider the process: N2O4(g) ¾ 2 NO2(g). (endothermic)
a. At a certain temperature, P(N2O4) = 30 kPa, and P(NO2) = 120 kPa. at equilibrium. If the volume of the container is doubled at constant temperature, calculate the new equilibrium partial pressures.
|I (after the volume change)||15||60|
|E||15 - x||60 + 2x|
KP = P2(NO2) / P(N2O4) kPa = (60 + 2x)2 / (15 - x)
At this point there are two unknowns KP and x. However since we know that before the volume change the system was at equilibrium, we can calculate KP:
Kp = (120)2 / 30 kPa = 480 kPa.
Now the quadratic can be solved giving x = 4.88
and P(NO2) = 69.8 kPa and P(N2O4) = 10.1 kPa
b. Complete the following table for this equilibrium. (Use UP, DOWN, NONE)
|Change||Affect on P(N2O4)||Affect on K.|
|Addition of more N2O4 at constant volume and temperature.||UP||NONE|
|Addition of a catalyst||NONE||NONE|
|Heating the equilibrium mixture||DOWN||UP|