School of Chemistry and Physics, University of KwaZulu-Natal, Westville
Campus, Durban
General Principles of Chemistry – CHEM110
Tutorial 10 – 2nd May and 4th May 2012
1. A 2.250 g-1 sample of quinone (C6H4O2) is burned in a bomb calorimeter
whose total heat capacity is 7.854 kJ/°C. The temperature of the
calorimeter increases from 23.62 °C to 30.91 °C.
A.
What is the heat of combustion per gram of quinone?
Need the following equation:
q = mCΔT
=
.
× .
℃
qcalorimeter = - qreaction
×(
.
−
.
)=
.
/
Thus –qreaction = - 25.4 kJ/g
Also ΔH~–qreaction
B. What is the heat of combustion per mole of quinone?
Molar Mass = 108.0964 g/mol
Answer = -2750 kJ/mol
.
×
.
.
=
.
2. Using the bond enthalpies tabulated in Table 7.4 in the textbook, estimate
ΔH for each of the following gas-phase reactions:
Equation Needed:
Hrxn = (bond enthalpies of bonds broken) – (bond enthalpies of bonds
formed)
a.
Broken: 1 C=C = 614, 1 O-O = 146, total = 760
Formed: 2 O-C = (2 x 358), 1 C-C = 348, total = 1064
760 – 1064 = -304 kJ/mol
Ans = - 304 kJ
b.
Broken: 1 C=C = 614 and 1 H-C = 413; Total = 1027
Formed: 2 C-C = (2 X 348), and 1 H-C = 413; Total = 1109
1027 – 1109 = – 82
c.
Broken: 6 N-Cl = (6 x 200); Total = 1200
Formed: 1 N≡N = 941, and 3 Cl-Cl = (3 x 242); Total = 1667
1200 – 1667 = - 467
Ans = -467 kJ
3. The hydrocarbons acetylene (C2H2) and benzene (C6H6) have the same
empirical formula. Benzene is an 'aromatic' hydrocarbon, one that is
unusually stable because of its structure
A. By using the data in Appendix C in the textbook (226.77 kJ/mol and 49
kJ/mol), determine the standard enthalpy change for the reaction:
.
Use:
H =  n Hf(products) -  m Hf(reactants)
H = [49] – [3 x (226.77)] = -631.31 kJ/mol
B. Which has greater enthalpy, 3 mol of acetylene gas or 1 mol of liquid
benzene?
3 mol acetylene = 3 x 226.77 = 680.31
1 mol benzene = 1 x 49 = 49
Ans = 3 mol acetylene
C. Determine the enthalpy of combustion in kJ g-1 for acetylene.
H =  n Hf(products) -  m Hf(reactants)
Combustion of acetylene
C2H2(g) + 2.5O2(g)  2CO2(g) + H2O(l)
H = [2(-393.5) + (-285.83)] – [(226.77) + (2.5(0)]
H = -1299.6 kJ mol-1
H = - 1299.6 kJ mol-1 ×
Ans = -50 kJ g-1
=−
.
.
D. Determine the enthalpy of combustion in in kJ g-1 for benzene.
H =  n Hf(products) -  m Hf(reactants)
Combustion of benzene
C6H6(l) + 7.5O2(g)  6CO2(g) + 3H2O(l)
H = [6(-393.5) + 3(-285.83)] – [(49) + (7.5(0)]
H = - 2695.83 kJ mol-1
H = - 2695.83 kJ mol-1 ×
.
=−
.
Ans = -35 kJ g-1
Ans = 40 kJ g-1 When water is produced as a gas
Ans = 41 kJ g-1 When benzene is a gas, and water is produced as a gas
Above methods do not round off the answer until the end…
HOWEVER (sample exercise 4.11, in the assigned textbook, shows…)
H = [6(-393.5) + 3(-285.83)] – [(49) + (7.5(0)]
H = (-2361 – 857.4 – 49) kJ
H = -3267
H = - 3267 kJ mol-1 ×
.
=−−
.
Ans = 42 kJ g-1
Rounding off at each stage of the calculation
4. What is ΔHrxn for the following chemical reaction?
Express your answer numerically, in kJ.
Element Standard Heat of Formation (kJ/mol)
Use:
H =  n Hf(products) -  m Hf(reactants)
H = [(-1150) + (-241.8)] – [(-393.5) +2(-424.70)]
Ans = -149 kJ
5. Given the data
use Hess's law to calculate
for the reaction
Express your answer using four significant figures.
The first equation given remains as written since the product ‘NO’ is on the
right and it is on the right in the final equation:
( )+
( )→
( )
∆
= +
.
Reverse the second equation given because ‘NO’ is a reactant and we want
it as a product (as shown in the final equation given):
( )→
( )+ ( )
Note the sign on the enthalpy changes
∆
= +
.
The third equation remains as written:
( )→
( )+ ( )
∆
= −
.
When you cancel terms on opposite sides of the arrow, note there is a N2
left over, which is not in the final equation, thus must get rid of it, the only
way to do that is re-write the first equation x 2, and remember to multiply
the ΔH by 2
( )+
( )→
( )
∆ = +
.
( )→
( )+ ( )
∆ = +
.
( )→
( )+ ( )
∆ = −
.
Final equation:
( )+
( )→
( )
∆
= +
.
However we want it per mol of N2O, thus divide through by 2:
( )+
Ans = 155.7 kJ
( )→
( )
∆
= +
.