CHM 115 QUIZ #5
PRACTICE
Fill in the blanks or circle one answer as appropriate (2.5 points per question).
1.
The equation which relates expansion work (w) done by a system to the change in
the number of moles of gas in a reaction is:
w = -∆ngRT
2.
For a reaction which is endothermic, the final enthalpy of the system (Hf) is
the initial enthalpy (Hi) of the system.
a. >
3.
b. <
c. =
d. ≥
>
e. ≤
The sign of q (positive or negative) for heat lost by the system is
negative .
The sign of w (positive or negative) for work done on the system is positive
4.
What is the change in the energy of a system that has 89 J of work done on it
while losing 100 J of heat.
a. - 189 J
5.
b. + 189 J
c. - 11 J
d. + 11 J
e. none of the above
What is the formation reaction for C8H18 (l)? [C(s) is the stable form of carbon.]
a. 8 C (s) + 9 H2 (g) Æ C8H18 (g)
b. 2 C4H8 (s) + H2 (g) Æ C8H18 (l)
c. 2 C4 (s) + 9 H2 (g) Æ C8H18 (l)
d. 8 C (g) + 9 H2 (g) Æ C8H18 (l)
e. none of the above
Complete the following calculations (Show all of your work!)
(10 points)
6.
0.6549 gram sample of liquid octane, C8H18 (l), a component of gasoline,
undergoes complete combustion in a bomb calorimeter that has a heat capacity of
11.93 kJ/ºC. If the initial temperature of the calorimeter is 20.25ºC and the final
temperature of the calorimeter is 22.87ºC, what is the heat of combustion at
constant volume (qv or ∆E) of octane in the units kJ /mol ? [the molar mass of
octane = 114.23 g/mole]
∆T = 22.87ºC – 20.25ºC = +2.62ºC
0 = qcal + qcomb . . . so . . .
∆E = qcomb = - qcal = - (11.93 kJ/ºC)( +2.62ºC) = - 31.26 kJ
moles consumed = 0.6549 g /(114.23 gmol-1) = 5.733x10-3 mole
∆E = -31.26 kJ / 5.733x10-3 mole = -5,452 kJ/mole
167
(15 points)
7.
(a) Given the values for ∆Hºf for the reactants and products, calculate ∆Hºrxn
for the following reaction in the units specified:
2 C8H18 (l) + 25 O2 (g) Æ 16 CO2 (g) + 18 H2O (l)
∆Hºrxn = ?? kJ /mole C8H18
∆Hºf [C8H18 (l)] = -250.1 kJ / mol
∆Hºf [CO2 (g)] = - 393.5 kJ / mol
∆Hºf [H2O (l)] = - 285.8 kJ / mol
∆H = Σ m ∆Hf (prod) - Σ n ∆H f (react)
Simply divide the reaction two to find the energy change per mole of octane
∆H = 8mole*(-383.5 kJ/mol) + 9mole*(- 285.8 kJ/mol) – 1mole*(-250.1 kJ/mol)
∆H = -5,390.1 kJ / mole octane
Alternatively, do the calculation for the moles as shown and then divide by two.
(b) Explain why the answers to questions 6 and 7(a) do not, and should not,
match one another.
The reaction at constant volume releases different energy per mole of octane due to
the absence of the work term present at constant pressure.
(12.5 points)
8.
Given the reactions below along with their enthalpy changes, use Hess’ Law to
calculate the enthalpy of reaction for the following reaction in the units specified:
2 NOCl (g) Æ 2 NO (g) + Cl2 (g)
∆Hºrxn = ?? kJ/mol NO
(a)
(b)
(c)
2 NOCl (g) + O2 (g) Æ N2O4 (g) + Cl2 (g)
2 NO2 (g) Æ 2 NO (g) + O2 (g)
2 NO2 (g) Æ N2O4 (g)
∆Hºrxn = - 95.52 kJ
∆Hºrxn = + 113.05 kJ
∆Hºrxn = - 58.03 kJ
rxn (a)
rev rxn (c)
rxn (b)
2 NOCl (g) + O2 (g) Æ N2O4 (g) + Cl2 (g)
N2O4 (g) Æ 2 NO2 (g)
2 NO2 (g) Æ 2 NO (g) + O2 (g)
∆Hºrxn = - 95.52 kJ
∆Hºrxn = + 58.03 kJ
∆Hºrxn = + 113.05 kJ
2 NOCl (g) Æ 2 NO (g) + Cl2 (g)
∆Hºrxn = ?? kJ/mol NO
∆Hºrxn =?? kJ/mol NO = (1/2)(-95.52+58.03+113.05)kJ = +75.56/2 = +37.78 kJ/mol NO
168