CHEM. 161
1.
THERMODYNAMICS
Problem Set #1
q = (1.888*104 cal)*(4.184 J/1 cal) = 7.900*104 J
mass coffee cup = 250.mL*(1g/mL) = 250 g
q = Cs*m*T
7.900*104 J = (4.184 J/g-K)*(250 g)*T
T = (7.900*104 J)/[ (4.184 J/g-K)*(250 g)] = (7.900*104)/[1046 K] = 75.5 K
the change in temperature is 75.5o,
T = Tf – Ti
Ti = Tf – T = 95.6 – 75.5 = 20.1oC
K = 273.15 + oC = 273.15 + 21.1 = 293.2 K
_20.1_oC _293.2_K
2. Determine Horxn, formation CH4 from graphite) & hydrogen gas from given steps.
overall equation to find: C(s) + 2 H2(g) ----- CH4(g)
rearrange, balance equations, & calculate Hof:
C(s) + O2(g) ----- CO2(g)
CO2(g) + 2 H2O(l) ----- CH4(g) + 2 O2(g)
2 H2(g) + O2(g) ----- 2 H2O(l)
C(s) + 2 H2(g) ----- CH4(g)
Hof = -393.5 kJ
Hof = 890.3 kJ
Hof = -571.6 kJ
-74.8 kJ
Hof = nHof products - mHof reactants
C(s) + O2(g) ----- CO2(g)
Hof =[Hof CO2(g)] – [(Hof C(s))+(Hof O2(g))] = [-393.5] – [0 + 0] = -393.5 kJ
CO2(g) + 2 H2O(l) ----- CH4(g) + 2 O2(g)
Hof =[(Hof CH4(g))+(Hof 2 O2(g))] – [(Hof CO2(g))+(Hof 2 H2O(l))] =
[-74.8 + (2*0)] – [-393.5 + (2*-285.8)] = 890.3 kJ
2 H2(g) + O2(s) ----- 2 H2O(l)
Hof =[Hof 2 H2O(l))] – [(Hof 2 H2(g))+(Hof O2(g))] = [2*-285.8] – [(2*0) + 0] = -571.6 kJ
3. Formula weight aluminum sulfate. Moles in 515.17 grams. Grams in 3.627 moles
Al2(SO4)3 2 Al = 2 * 27.0 = 54.0
3 S = 3 * 32.1 = 96.3
12 O = 12 * 16.0 = 192.0
342.3g/mol
515.17 g*(1 mol/342.3 g) = 1.680 mol
3.627 mol *(342.3 g/mol) = 1242 g
4. Sulfur dioxide formed sulfur & oxygen. Grams pdt forms with 452.74 g S & 12.56 mols oxygen.
eqn: S(g) + O2(g) --- SO2(g)
total: S0(g) + O20(g) --- S+4O2-2(g)
net: S0(g) + O20(g) --- S+4O2-2(g)
red: O (0 -2) ox: S (0 - +4)
 1 mol SO 
2   12.56mol SO
O : 12.56 mol O 2 * 
2
 1 mol O 
2 

oxygen limiting reagent; (12.56 mol SO2)*(64.1 g/mol) = 805 g SO2
 1 mol   1 mol SO 2 
  14.1mol SO
 * 
S : 452.74 g S * 
2

 32.1 g   1 mol S 

