Name:_____________
(4 points)
Chemistry 114
First Hour Exam
Remember- Show all work for partial credit. Each problem is worth 12 points.
1.
A (10 points) How much work to I do if I blow up 12 one liter balloons in
Spearfish when the atmospheric pressure is .89 atm?
B(2 points) If 1 Calorie = 4,180 J, is blowing up balloons a significant way to burn
Calories for my diet?
-1,081 J x (1 Calorie/4,180 J) = .25 Calorie - not a significant way to burn
Calories
2. Consider the reaction:
2Fe2O3(s) 64Fe(s) + 3O2(g)
ÄHrxn = 824.2 kJ/mol
If 6.0g of O2 gas are formed using the above reaction and the reaction is performed in 1
kg of water...
A. (2 points)Will the water get warmer or colder?
+ÄH reaction will get colder
B. (5 points) How much heat will be evolved or consumed in the above reaction?
C. (5 points) If the specific heat capacity of water is 4.18J/g@K, and the water is
initially at 60o C, what is the final temperature of the water?
Assume Adiabatic: Heat lost by water + heat gained by reaction = 0
Heat lost by water = 1000g×4.18J/g@K×(Tf-60) (ÄT in K = ÄT in oC)
Heat gained by reaction = 51500 J
51500J + 1000g×4.18J/g@K×(Tf-60)=0
1000g×4.18J/g@K×(Tf-60)= -51500J
Tf-60=-51500/(1000×4.18) = -12.32
Tf = -12.32 + 60 = 47.6oC
Note: on some copies of this test the heat capacity of water was given in units of J/mol@K. This
give a very different answer, and I graded those tests accordingly.
2
3. Estimate ÄHrxn for the reaction 2C2H2(g) + 5 O2(g) 64CO2(g) +2 H2O(g)
Bonds Energies
Bond
C-C
C=C
C/C
C-H
Energy (kJ/mol)
347
615
821
347
Bond
C=H
O-O
O=O
C-O
Energy (kJ/mol)
590
142
512
351
Bond
C=O
O-H
O=H
Energy (kJ/mol)
730
464
670
For bond energies ÄHrxn= Bonds broken in reactants - bonds made in products
C2H2 : H-C/C-H, O2 : O=O, CO2: O=C=O, H2O: H-O-H
ÄHrxn = [2(H-C/C-H) + 5(O=O)] -[4(O=C=O) + 2(H-O-H)
=[2(347+821+347) + 5(512)]-[4(730+730) + 2(464+464)]
=[2(1515) + 2560)]-[4(1460)+2(928)]
=1330+2560-5840-1856
= -3806 kJ
Note: the equation in the original test was not balanced correctly and had only 3 O2
molecules as reactants. If you did the problem with 3 O2 you get -3130 kJ, and I gave
you full credit because I gave you the wrong equation.
4. What is the difference between...
A. The system and the surroundings
The system is usually the chemical reaction we are examining, and the surroundings
are usually the solvent in which the reaction occurs.
B. An extensive property and an intensive property
An extensive property depends on the amount of material, and intesve property is
independent of the amount.
C. An exothermic reaction and an endothermic reaction
An exothermic reaction releases heat energy, and endothermic reaction absorbs heat
energy.
D. State function and an energy transfer function
A state function is calculated from the final and initial states of a system. An enegy
transfer function is calculated from the path taken between the states.
E. Adiabatic and non-adiabatic conditions
Adiabatic - no heat is lost or gained, so ÄHsys+ÄHsurr =0. Non-adiabatic means that
some heat can be lost or gained.
F. A constant P calorimeter and a constant V calorimeter
A constant P calorimeter measures heat change at a constant pressure, it is a
‘coffee-cup’ type calorimeter and is used to measure ÄH.
while a constant V calorimeter measures heat change while the volume is constant, it is
a ‘bomb’ calorimeter and it is used to measure ÄU.
3
5.Two 20.0g ices cubes @ -10o C are placed in 245 g of water @ 25.0 oC. Assuming
adiabatic conditions, what is the final temperature of the system when the ice cubes
melt?
Molar Heat Capacity of H2O(s) = 37.0 J/mol@K
Molar Heat Capacity of H2O(l)= 75.3 J/mol@K
Enthalpy of fusion for H2O = 6.01 kJ/mol
Ice cubes = 2x20g = 40 g; 40/18 = 2.22 mole
Warm water = 245 g; 245/18 = 13.6 mole
Again assume adiabatic so heat gained by ice cubes + heat lost by warm water =0
Heat gained by ice cubes = ice @ -10 6Ice at 0+ ice6liquid@0 + water@0 warming to Tf
Ice @ -10 60 = 2.22(37.0J/mol)[0-(-10)] = 821.4J
Ice melting = 2.22 ×6.01 kj = 13.342 kJ = 13,342J
Changing to final temp = 2.22(75.3)(Tf-0) =167.17(Tf-0) = (167.17 ×Tf)
Total for ice cubes warming = 821.4J + 13342J + (167.17J×Tf)
Warm water cooling = 13.6 ×75.3×(Tf-25) = 1024Tf - 25602
Total 0 = 821.4J + 13342J + (167.17J×Tf) +1024J×Tf - 25602J
+25602J-821.4J-13342J = 1191.2J×Tf
11438.6J = 1191.2J×Tf
Tf =11438.6J/1191.2J = 9.6oC
6A. (8 points) In chapter 15 we discussed several different molecular forces. Name
these forces, rank them from strongest to weakest, briefly explain the interaction that
gives rise to the force
Strongest
Electrostatic - Charge-charge attraction or repulsion
Hydrogen bonding - A strong dipole-dipole that occurs when H is bonded
to N, O, or F
Dipole-Dipole - similar to charge-charge interaction, but it occurs between
the partial charges that exist in a molecule that has a dipole.
London forces- as electrons move around in their orbitals, a molecule or atom
can have a very temporary dipole due to an unequal distribution of electrons. This
temporary dipole can then induce a temporary dipole in an adjacent molecule and the
two molecules can be attracted briefly in a dipole-dipole type interaction that breaks
down almost instantly as the electrons continue in their orbitals. This is the weakest
and mot short-ranged of the intramolecular forces we discussed
6B. (4 points) Which of the above forces would you see in interactions between
molecules in:
NaCl(s) Electrostatic
F2 (g) London
HF(l) Hydrogen bonding, dipole, London
H2S(l) Dipole, London
CaCO3(s) Electrostatic
CH3-CH3(l) Dipole
H2CO(l) Dipole, London
He(g) Dipole
7
.
I
4
have an atomic solid which occupies a simple cubic lattice. The distance between
atoms in this lattice is 3.6Å. If I direct 1.54Å X-rays at this substance, at what angle will
the X-rays be reflected off the crystal surface . (Assume a first order reflection where
n=1.)
8. Explain the difference between ionic bonding, covalent bonding, and metalliccovalent bonding. Give an example of a compound or element that uses each kind of
bonding in its crystal structure.
Ionic bonding occurs between oppositely charged ions. It is a purely electrostatic
interaction between an anion and a cation. An example could be any ionic compound,
or specifically NaCl.
Covalent bonding occurs when two non-metals share a pair of electrons. The shared
pair of electrons form an orbital that is localized between the atoms and holds the
atoms in a chemical bond. Example: CH4
Metallic covalent bond occurs between metals. In this bond outermost shell electrons
are shared between atoms, but the bonding is not localized to a particular space
between two atoms, so the electrons can move anywhere in the lattice of metal ions.
Example: Iron