Chemistry Pretest 1.3
2014-15
1.
Draw two H2O2 molecules. Label all intermolecular and intramolecular
bonds.
Walter White may have chosen the
pseudonym because of the
Heisenberg Uncertainty Principle
in science. Bonus on test will be
about that principle. Research it if
you find the time.
The blue dots represent
intermolecular bonds.
2.
a.
Draw a reaction profile for an exothermic reaction.
b.
3.
Do the same for an endothermic reaction.
Classify as endothermic or exothermic.
a) If a yellow straw turns blue when placed in ice but reverts to yellow when placed in warm water,
then yellow  blue is __exothermic___
b) A reaction where ∆Hbb> ∆HBF___ endothermic ___
c) A reaction with kJ among the reactants____ endothermic __
d) The electrolysis of water : 2H2O 2H2 + O2__ endothermic ___
e) The condensation of alcohol on a cold glass____ exothermic ___________
f) A reaction in which ∆H = (-)___exothermic ____
4.
Some automobiles and buses are equipped to burn propane gas, C3H8, as a fuel.
The complete combustion of propane is shown by the following chemical equation:
C3 H8 ( g ) + 5 O2 ( g ) →
3 CO2 ( g ) + 4 H2 O( g )
∆ H = − 2044.5
Given the following heats of formation.
H2(g) +
1
O2(g) → H2O(g)
2
C(s) + O2(g) → CO2(g)
∆H = −242.0
kJ
mol
∆H = −393.5
kJ
mol
a) What is the heat of formation of propane?
3 C(s) + 4 H2(g) → C3H8(g)
∆H = ?
kJ
mol
Reverse equation (1):
3 CO2 + 4 H2O(g) → C3H8 + 5 O2
∆H = 2044.5 kJ/mole
4 H2(g) + 2 O2(g) → 4 H2O(g)
∆H = −242.0(4)
Multiply equation (2) by 4:
kJ
mol
Multiple equation (3) by 3:
kJ
∆H = −393.5(3) mol
3C(s) + 3O2(g) → 3CO2(g)
Add them up:
3 C(s) + 4 H2(g) → C3H8(g)
∆H = -104 kJ
b) Express your answer in (a) in kJ/mole of C
∆H = -104 kJ/3mole C = -34.7 kJ/mol C
c) How much heat would be either released or absorbed (specify) if 4.4 g of propane were
formed from carbon and hydrogen?
4.4g /(44 g/mole) = 0.10 mole C3H8(g)
0.10 mole C3H8(g) ( -104 kJ/mole) =-10 kJ (released)
5.
There is a ΔH (heat of vaporization) associated with the evaporation of CCl4(l).
a) Write a simple equation to represent the evaporation of CCl4(l).
CCl4(l) CCl4(g).
b) Use the following thermochemical equations to calculate the heat of vaporization of CCl4(g):
C(s) + 2 Cl2(g) → CCl4(l)
C(s) + 2 Cl2(g) → CCl4(g)
ΔH = -139.5 kJ
ΔH = -103.2 kJ
Reverse 1st equation and add it to the second one:
ΔH = 36.3 kJ (notice that the sig fig rule for addition applies; there was no
multiplication or division, so we preserve decimal places.)
c) Plot the enthalpy versus progress of reaction graph and locate CCl4(l) and CCl4(g) on the graph. Clearly
label both axes.
Draw it like 2(b) with CCl4(l) as a reactant at the bottom of the hill.
6.
Consider the following combustion reaction of ethanol:
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) + 1371 kJ
Bond
H–H
H–O
H–C
C–O
O–O
Energy (kJ/mol)
436
460
413
358
204
Bond
C=O
C–C
C=C
C ≡C
Energy (kJ/mol)
805
347
607
839
Calculate the energy of the double bond between oxygen (O) atoms in the diatomic
molecule.
oxygen gas (O2)
∆Hbb = 3x +5 C-H +1 C-C + 1 C-O + 1 O-H
= 3x + 5(413) + 347 + 358 + 460 = 3x +3230
Products:
∆Hbf = 4(C=O) + 6(O-H) = 4(-805) + 6(-460) = -5980 kJ
∆Hbb + ∆Hbf = ∆H
3x +3230 + -5980 = -1371 kJ ( see original equation)
X = 460. kJ per O2 bond ( an approx. because throughout intermolecular bonds are ignored and
because we assume that all bonds of one type are identical)
7.
Given :
1 H2(g) + ½ O2(g)  1 H2O(g) +
241.8 kJ
What volume of water should a calorimeter contain if you want to burn 4.0 g of hydrogen and have
the temperature increase by only 30.0 oC? Assume that the water is the only thing in the calorimeter
that absorbs heat. Density if water = 1.00 g/ml. Respect sig figs and your grandmother.
4.0 g H2 (mol/2.0g) = 2.0 moles of H2
-241.8 kJ/mol H2 (2.0 moles of H2) = -483.6 kJ
Q = -∆H = 483.6 kJ = 483 600 J
Q = mc ∆T
483 600 J = m 4.19 J/(goC) (30.0 oC)
m = 3847 g
V = 3847 g /(1.00 g/ml) = 3847 ml = 3.8 L
8.
What is the partial pressure of CO2 if its concentration in a 100.0 kPa atmosphere is 396 ppm ?
(September 2014 measurement)? Few people realize that when ppm is expressed for gases, it’s not
𝑛
mg/L, as it is for aqueous solutions. Instead it’s a mole fraction: 𝐴
𝑛𝑇
The reason for this is that 1 L of air does not weigh 1 million mg, but there are a million mg of water in a
1 L of liquid water.
396 ppm (or 396 parts per million) for a gas mixture is 396 moles of CO2 per 1 000 000 moles of dry
air. We could use that as the mole fraction for the formula
𝑛𝐴
𝑃𝐴 = � � 𝑃𝑇
𝑛𝑇
𝑃𝐴 = �
396 𝑚𝑜𝑙𝑒𝑠
𝟏 𝟎𝟎𝟎 𝟎𝟎𝟎 𝑚𝑜𝑙𝑒𝑠
Flashback
9.
� 100.0𝑘𝑃𝑎 = 0.0396 kPa
A 400.0 mL balloon at sea level was submerged to a certain depth and its volume decreased to
84.0 mL.
Under water pressure increases by 100.0 kPa for every 10.0 m of water depth. The initial air
pressure at sea level was 100.0 kPa.
To what depth was the balloon submerged?
P1V1 = P2V2
100(400) = P2(84)
P2 =476.19 kPa due to water and atm. pressure
Change in pressure = 476.19 - 100.0 = 376.19 kPa (due to water alone)
376.19 kPa (10.0m/100.kpa) = 37.6 m of water depth
10.
What am I? I am not a squared quantity but I am a quantity directly proportional to the kinetic
energies of molecules. The slope of that relationship is 1.5R, where R = gas constant.
Kelvin (absolute) temperature; because of Ek = 1.5 RT