w = -PDV or -DngasRT
DSsurr = -DHsyst/T
qsurr =
mCDt or
CsetupDt
qsyst = -qsurr
DS°rxn = SS°products - SS°reactants
DH°rxn = SDH°fproducts - SDH°freactants
DG°rxn = SDG°fproducts - SDG°freactants
DSuniverse = DSsyst + DSsurr
DG°rxn = DH°rxn -TDS°rxn
DGrxn = DG°rxn +RT lnQ
DG°rxn = -RT
lnKeq
ln(K2/K1) = (DH°/R) x (1/T1 - 1/T2)
A number of the questions on the exam were the
same as those done in class.
(Use bond energies to find DH° for
H-H + H2C=CH2 ---> H3C-CH3)
Coffee Cup calorimeter question done in
class.
similar K1 at T1 and K2 at T2. Determine
DG°, DH°, DS° etc.
1. Use Hess’s Law to find DH° for the
reaction
C2H2 (g) + 2H2 (g) <--->
C2H6(g)
from the following combustion data:
C2H2 (g) + 5/2 O2 (g) <---> 2CO2 (g) + H2O
(l)
DH° = -1320 kJ
H2 (g) + ½ O2 (g) <---> H2O (l)
DH° = - 285 kJ
C2H6(g) + 7/2 O2 (g) <---> 2CO2 (g) + 3H2O
(l)
DH° =- 1560 kJ
Answer: -330.
kJ
2. When a 0.300 g sample of solid camphor (C10H16O)
was burned in a bomb
calorimeter, the temperature
of the calorimeter and its contents increased by
1.54 °C. When an electric heater delivered 1500 W (1 W = 1 J/s)for 10 s to
the
calorimeter, its temperature changed from
25.30 °C to 27.30 °C.
Calculate:
a) the heat
capacity of the calorimeter,
Answer: 7.50 kJ/oC
b) the change in the
internal energy, and
Answer: -11.55 kJ or -5861 kJ/mol
c) the DH° of combustion per mole of
camphor.
Answer:
-5870 kJ/mol
3. For the reaction 2 SO2 (g) + O2 (g) <--->
2SO3 (g), Kc = 280 at 1000 K.
a) What is the
value of DG° at 1000 K?
Answer: -10.2 kJ/mol
b) If 0.40 mol SO2, 0.18 mol O2, and 0.72 mol SO3
are mixed in a 2.50 L flask
at 1000 K, in what
direction will a net reaction occur?
Answer: Forward direction, DG = -15.2 kJ or
Qc<Kc or
Qp<Kp
(note need to use Qp and
Kp)
4. Calcium oxide (burnt lime) may be prepared by
heating calcium carbonate
(limestone) in a lime kiln, in the following reaction:
CaCO3 (s) <---> CaO
(s) + CO2 (g)
The value of DH° for this
reaction is +178.3 kJ; DS° is +0.1606 kJ/K. Find
the
temperature at which this reaction becomes
spontaneous if the partial pressure of CO2
is 0.10
atm.
Answer: T = 992 K or 719
oC (0.10 atm is Kp at Temp T)
7. A 100.0 g sample of a
substance with a density of 0.8200 g/cm³at 20.00 °C is
heated to 150.00 °C under a constant pressure of 1
atm. The sample absorbed 1.500 kJ of heat
during this process and its volume increased by 10.00 %. Find:
a) q Answer:
1.500 kJ b) w Answer: -1.2 J
c) DU, and Answer: 1.499 kJ
d) DH. Answer: 1.500 kJ