ANSWERS
1. a. ? temp change = -2000 J[(1 deg/187 J)] = -10.7 deg
final temp = 21.0 - 10.7 = 10.3 deg C
b. sp.ht. is defined as Joules/(gram-deg), so in this case...
sp.ht. = (0.187 kJ/deg)/(74.31 g) = 2.52 J/g-deg
2. a. C10H8 (liq) = C10H8 (solid) (delta)H = -19.3 kJ/mol
b. C4H10 (g) + 6 O2 (g) = 4 CO2 (g) + 5 H2O (liq)
c. Ni(g) ---> Ni1+ (g) + e1-
d. 2 C(solid) + 3 H2 (g) + 1/2 O2 (g) = C2H5OH (liq)
3. a. in heat transfers - one substance gains heat (an endothermic
process, so sign of heat energy is positive) and another
substance loses heat (an exothermic process, so sign of heat
energy is negative). The amounts of heat gained/lost are equal.
i. q gained by water = (4.18 J/g-deg)(32 g)(21.2-19.7 deg) = +187 J
ii. q lost by metal = -187 J
iii. sp.ht. metal = (-187 J)/{(9.238 g)(21.1-74.3 deg)} = 0.381 J/g-deg
b. est.at.wt. = 25/0.381 = 65.6 g/mol
4. a. enthalpy of formation: forming cmpd from its elements in std.sts.
eq. 1 Pb(s) + 1/2 O2 (g) = PbO(s) (delta)Hf = -219 kJ/mol
eq. 2 Pb(s) + O2 (g) = PbO2 (s) (delta)Hf = -277 kJ/mol
can achieve desired equation by reversing eq.1 and adding to eq.2
PbO(s) = Pb(s) + 1/2 O2 (g) (delta)Hf = +219 kJ/mol
Pb(s) + O2 (g) = PbO2 (s) (delta)Hf = -277 kJ/mol
Pb(s) + O2 (g) + PbO(s) = Pb(s) + 1/2 O2 (g) + PbO2 (s) (delta)H = -58 kJ
cancel like substances to achieve the final equation...
PbO(s) + 1/2 O2 (g) = PbO2 (s) (delta)H = -58 kJ
b. (delta)H = -58 kJ/mol
5. q = (ht.cap.)(temp.change) = (sp.ht.)(mass)(temp change)
q(calorimeter) = (9.48 kJ/deg)(17.02-14.37 deg) = 25.12 kJ
q(reaction) = -25.12 kJ; q(reaction)/g = -25.12 kJ/4.37 g = -5.74 kJ/g
? kJ/mol = -5.74 kJ/g[(84 g/1 mol)] = -492 kJ/mol
6. cool steam and warm ice to a common temperature, and then apply any
heat energy that is unaccounted for - to the combined masses
a. cool steam to liquid at 0 deg.C (an exothermic process)
i. cool gas to 100 deg.C
q = (2.03 J/g-deg)(25 g)(100-112 deg) = -609 J
ii. change gas to liquid, at T=100 deg.C (negative sign is applied)
q = 25 g[(1 mol/18 g)(40,700 J/mol)] = -56528 J
iii. cool liquid to 0 deg.C
q = (4.18 J/g-deg)(25 g)(0-100) = - 10450 J
iv. subtotal: heat energy lost = -67587 J
b. i. warm solid to 0 deg.C (an endothermic process)
q = (2.02 J/g-deg)(75 g)(0-(-80)) = +12120 J
ii. change solid to liquid, at T=0 deg.C (positive sign is applied)
q = 75 g[(1 mol/18 g)(6000 J/mol)] = +25,000 J
iii. subtotal: heat energy required = +37,120 J
c. at this point there are (25+75)=100 g liquid water at 0 deg.C, and
(-67,587+37,120)=-30,467 J of heat energy remaining. Apply this
heat energy to the water, i.e., what is the final temperature after
30,467 J of heat energy are added to 100 g water at 0 deg.C?
q = +30,467 J = (4.18 J/g-deg)(100 g)(T-0)
T(final) = 72.9 deg.C
7, 10, 11 solutions not provided
8. a. Fe3+ (aq) + 3 OH1- (aq) = Fe(OH)3 (solid)
b. (delta)H reaction = -84.5 kJ/mol (via thermodynamic tables)
c. ? kJ = 1.375 g Fe(OH)3[(1 mol/106.8 g)(-84.5 kJ/1 mol)] = 1.09 kJ
9. Use information provided in the problem to determine enthalpy
change for the following reaction...
Ba(solid) + S(solid) + 2 O2 (g) = BaSO4 (solid) (delta)Hf = ???
12. moles silver nitrate = (moles silver chromate)*[conversion factor(s)]
(Molarity)(Vol.in L) = (mass/FW)[(2 mol sn/1 mol sc)]
(0.84)(Vol.in L) = (1.00 g/(194 g/mol sc))[2 mol sn/1 mol sc]
Vol.in L = 12.27 mL
13. X + E = name of shape angles
a. 3 + 1 = 4 triangular pyramid less than 109.6 deg
b. 2 + 0 = 2 linear 180 deg
c. 2 + 1 = 3 bent less than 120 deg
d. 3 + 1 = 4 triangular pyramid less than 109.6 deg
e. 4 + 2 = 6 square planar 90 and 180 deg