Thermodynamics Review Questions
1. At 298K, _G = –141.6 kJ, _H = –198.4 kJ, and _S = –187.9 J/K for the reaction
2SO2(g) + O2(g) _ 2SO3(g). Will the reaction be spontaneous at 900.ºC?
2. In the context of the last problem, at what temperature will the reaction be both
spontaneous and nonspontaneous?
3. A reaction is nonspontaneous at room temperature but is spontaneous at – 40ºC.
What is true about the signs and relative magnitudes of _H, _S, and –T_S?
4. Which species has greater entropy?
a. 1 mol SO2(g) or 1 mol SO3(g)?
b. 1 mol CO2(s) or 1 mol CO2(g)?
c. 3 mol O2(g) or 2 mol O3(g)?
d. 1 mol KBr(s) or 1 mol KBr(aq)?
e. Green tea at 25ºC or lemon tea at 307K?
f. 1 mol CF4 or 1 mol CCl4?
5. Which species has greater entropy? Explain.
a. PCl3(g) or PCl5(g)?
b. CaF2(s) or BaCl2(s)?
c. Br2(g) or Br2(l)?
d. HCl(g) or HCl(aq)?
Answers
1. _G = _H – T_S = –198.4 kJ – (1173K)( –.1879 kJ/K) = –198.4 kJ + 220.4 kJ =
22.0 kJ.
Because _G is positive, the reaction is nonspontaneous at 900.ºC.
2. The reaction is spontaneous and nonspontaneous when _G = 0. Because _H and
_S are known, the temperature can be calculated using the equation _G = _H –
T_S.
_G = 0 = _H – T_S = –198.4 kJ – (T)( –.1879 kJ/K) = –198.4 kJ + (.1879
kJ/K)(T) = 0;
(.1879 kJ/K)(T) = 198.4 kJ;
T = 1056K
3. At a lower temperature, _G becomes negative. Looking at the equation _G = _H
– T_S, the entire T_S term must be decreasing if T is decreasing – T_S
theoretically reaches 0 at 0K, yet _G is still negative at this temperature. Thus,
_H must be negative. Because –T_S is the opposite sign of _H (otherwise, _G
could not change), _S must have the same sign – negative.
4. Which species has greater entropy?
a. 1 mol SO2(g) or 1 mol SO3(g)?
b. 1 mol CO2(s) or 1 mol CO2(g)?
c. 3 mol O2(g) or 2 mol O3(g)?
d. 1 mol KBr(s) or 1 mol KBr(aq)?
e. Green tea at 25ºC or lemon tea at 307K?
f. 1 mol CF4 or 1 mol CCl4? (For similar compounds, entropy increases with
molar mass.)
5.
a. PCl5(g) has higher entropy because its molar mass and molecular
complexity are greater.
b. BaCl2(s) has greater entropy because its molar mass is greater.
c. Br2(g) has greater entropy because gases have more molecular freedom
than liquids.
d. HCl(g) has greater entropy because gases, when dissolved, have lower
entropies: water molecules restrict the freedom of gas molecules.
Colligative Properties
1. Calculate the molal boiling point elevation of 11.00 g. aqueous solution with
0.0230 mol. of methanol (CH3OH). (Kb for water=0.512°C/m)
2. Calculate the mass of NaCl needed to depress the freezing point of a 400.0 g.
aqueous solution by 2.34°C (Kf for water=1.84°C/m; d of NaCl=2.16 g/cm3)
3. By how much is the vapor pressure of a 100.00 g. aqueous solution containing
35.00 g. of KI at 25°C lowered?
4. What mass of solute is needed in order to lower the vapor pressure of a 45.00 g.
aqueous sol-n of KI at 30°C by 5.00 mmHg
5. Calculate the molality of an aqueous solution of NaCl if it has a freezing point
depression of 1.00°C and a Kb of 1.84°C/m
Answers
32.06 g. CH3OH
= 0.737g. CH3OH
1 mol CH3OH
Mass of solvent=Mass of sol-n – Mass of solute=10.26g. sol-n
1.
0.0230 mol CH3OH ×
10.6 g. sol-n ×
1kg.
= 0.0106kg.
1000g.
m=
0.0230mol CH3OH
= 2.17m
0.0106 kg. sol - n
∆T=Kb*m
∆T=2.17m*0.512/m=1.11ºC
2.
∆T=2.34 ºC
m * Kf =2.34 ºC
1.84ºC/m * m=2.34 ºC
mol solute
58.44 g.
m=
mol solute ×
= mass of solute
kg solvent
1mol.
mol solute=mass of solute/58.44 g.
let mass of solute=x
then g. of solvent= 400.0-x
1kg.
400.0 − x
kg.
and kg. of solvent=(400.0-x)g. ×
=
1000g.
1000
x g.
17.11x
58.44 g./mol.
=1.27m
=1.27m
400.0 - x
400 - x
kg.
1000
17.11x=508-1.27x
18.38x=508
x=27.64 g.
3.
∆P=xsolute*Pº
Pº=23.76 mmHg
1mol. KI
=0.2098 mol KI
166.8g. KI
1mol H2O
mol solvent=65.00 g. H2O ×
=3.610 mol. H2O
18.02 g. H2O
mol sol-n=mol solute + mol solvent=3.820 mol sol-n
mol solute=35.00g. KI ×
xsolute=0.2098 mol. KI/3.820 mol sol-n=0.05492
∆P=23.76mmHg*0.05492=1.305mmHg
m=1.27m
4.
∆P=Pº*xsolute
Pº=31.8mmHg
∆P=5.00mmHg
5.00mmHg=31.8mmHg* xsolute
xsolute=0.157
mol solute*166.0g/mol=mass of solute
mass of solute
mol solute=
166.0 g/mol
45.00g. - mass of solute
mol solvent=
18.02 g/mol
mass of solute 45.00g. - mass of solute 7470.g. − 147.98 mass of solute
mol sol-n=
+
=
166.0g/mol
18.02 g/mol
166.0 × 18.02 g/mol
mass of solute
166.0g/mol
0.157=
7470.g. - 147.98mass of solute
18.02 × 166.0 g/mol
mass of solute
7470.g. - 147.98mass of solute
18.02g/mol
7470.g. - 147.98mass of solute
mass of solute=0.157
18.02
115 mass of solute=7470.g.-147.98mass of solute
0.157=
263mass of solute=7470g.
Mass of solute=28.4 g.
5.
∆T=Kb*m
1.00ºC=1.86 ºC/m*m
m=0.538m
Acid – Base
Key terms
Acid-Base Equilibria
Hydronium Ion=H3O+
Neutralization-When acid and base react
pH Scale- scale measuring acidity based on a 0 to 14 scale where
numbers incresing from 7 to 14 are more basic and number decresing from 7 to
0 are more acidic
Polyprotic Acids- Acid with more than one ionizable proton
% ionization=(# of moles ionized/ total number of moles brought into
solution per liter) * 100
Properties of Acids1-conduct electricity
2-react with metals to create H+
3-taste sour
4-turn litmus paper red
5-neutralizes bases
Titration- An expirimental way of determining an unknown consentration of an
acid (or base) by neutralizing it with the base (or acid) with the known
consentration