Download Name:_____________ Chemistry 112 Third Hour Exam

Name:_____________
Chemistry 112
Third Hour Exam
This test contains 110 points. I will only count 100 points. Thus, over the whole test, you have 10 free
points that will not count against you.
1. The air we breath has a mole fraction of about 0.8 N2 and 0.2 O2. If I have 22.42 L of air at STP...
A. (4 points) How many moles of N2 and O2 do I have?
1 mole of gas occupies 22.42 R at STP. Since I am at STP, and I have 22.42 R volume, I must
have a total of 1 mole of a gas. Sine that is the case
Moles of N2 = 1 mole of gas × .8 (P N2) = .8 mole N2
Moles of O2 = 1 mole of gas × .2 (P O2) = .2 mole O2
B. (4 points) What are the partial pressures of N2 and O2? (assume the total pressure is 760 mmHg)
= Pgas 1 / Ptotal
O gas 1 × Ptotal = Pgas1
if Ptotal = 760 mm Hg, then
O gas
1
PN2 = 760 mm Hg × .8 = 608 mm Hg
PO2 = 760 mm Hg × .2 = 152 mm Hg
2. (8 points)Define and give an example of each of the following
A state function - A property of a system that depends only on the current state of the system,
not on the path used to get to that state. Examples include ) E and )H.
A path function - A property of a system that depends on the path taken between two state of
a system. Examples include work (w) and heat (q).
An extensive variable - A property of the system that depends on the amount of a substance
(extent) within the system. Examples include heat, moles, volume (of a liquid or solid).
An intensive variable - A property of the system that does NOT depend on the amount of a
substance within the system. Examples include temperature and pressure.
2
3. A (4 points) If I have 35 moles of an ideal gas crammed into a 5.5L container at 300K, what should
the pressure in the container be?
PV=nRT
P = nRT/V
= 20 (.08206)300/5.5
=89.5
Units = mole(R@atm/K@mol)K/R = atm
=89.5 atm
In class I gave an equation for the pressure of a non-ideal gas as:
Pobs
nRT
 n
=
− a 
V
V − nb
2
B. (6 points) If we have NH3 gas then a=5.46 atm@L2/mol2 and b =.0371 L/mol
What is the pressure in the 5.5 L container if we fill it with 20 moles of non-ideal NH3 gas at 300K
20(.08206)300
 20 
=
− 5.46 
 55
55
. − 20(.0371)
. 
2
=103.5-72.2
= 31.3
=31.3 atm
4. (10 points) Calculate )E for a system in which q = +100 kJ and w = -37 kJ. In this case is the
system doing work on the surroundings, or are the surroundings working on the system?
)E = q+ w; q=+100 kJ, w=-37 kJ
=100kJ + (-37 kJ)
= 63 kJ
Work =w = -37 kJ
A negative sign represents something that is moving from the surroundings into the system,
hence work is being done by the surroundings on the system.
5. (10 points) How much work is associated with expanding a gas from 4.3 L to 5.7 L at a constant
pressure of 120 atm. Give your answer in BOTH units of L@atm and J
w = -P dV
= -120 × (5.7-4.3)
= -168
Units = atm(R) or L@ atm
To change to J
1 L@atm= 101.3 J
-168 L@atm × 101.3L/1 L@atm
= - 17,020 J or 17.0 kJ
3
4
6. (15 points) I have 10 g of Al at 300oC and 20 g of Fe at 200o C. If I place these pieces of metal in
an adiabatic container where they can exchange heat with each other, but not with the surroundings,
what is the final temperature of the two pieces of metal after they have come to equilibrium and have the
same temperature? (Specific heat capacity are: Al 0.89 J/oC@g, Fe 0.45 J/oC@g) (NOTE: 10 points for
proper set-up 5 for final answer))
In adiabatic system no heat exchanges with surroundings, so all heat remains within the system.
If this is the case then heat lost by Al cooling must equal heat gained by Fe warming up. That is:
Heat lost by Al = Heat gained by Fe
Heat lost by Al = mass of Al × Specific Heat capacity of Al × temperature lost by Al
= 10 g × 0.89 J/oC@g @ (300- TF)oC
Heat gained by Fe = mass of Fe Al × Specific Heat capacity of Fe × temperature gained by Fe
= 20 g × 0.45 J/oC@g @ (TF-200)oC
5
7. (10 points) Given the following data:
C2H2 (g)+ 5/2 O2(g) 6 2CO2(g) + H2O(l)
C(s) + O2(g) 6 CO2(g)
H2(g) + 1/2O2(g) 6 H2O(l)
)H= -1300 kJ
)H= -394 kJ
)H= -286 kJ
Calculate the )H for the reaction:
2C(s) + H2(g) 6 C2H2(g)
We want an equation that looks like
2C(s) + H2(g) 6 C2H2(g)
The product of this reaction is acetylene C2H2
from our given equations we only have one that contains C2H2, and that is:
C2H2(g) + 5/2 O2(g) 6 2CO2(g) + H2O(l) )H =-1300 kJ
Since we want C2H2(g) as a product we need to reverse this equation
2CO2(g) + H2O(l) 6 C2H2(g) + 5/2 O2(g) ) H =+1300 kJ
One of our given equations starts with C(s) and makes CO2(g) so it looks like we can add it to
the above equation to get rid of the CO2
C(s) + O2(g) 6 CO2(g) )H =-394 kJ
But we need to start with 2 C(s) and make 2 CO2's so let’s double this equation and its )H
2C(s) + 2O2(g) 6 2CO2(g) ) H =-788 kJ
We only have one equation left and it has the H2 on the correct side to cancel out so our last
equation is:
H2(g) + 1/2O2(g) 6 H2O(l)
) H =-286 kJ
Let’s put together our equations in one place:
2CO2(g) + H2O(l) 6 C2H2(g) + 5/2 O2(g)
)H = +1300 kJ
2C(s) + 2O2(g) 6 2CO2(g)
)H = -788 kJ
H2(g) + 1/2O2(g) 6 H2O(l)
)H = -286 kJ
If you add up the equations you get
2C(s) + H2(g) 6 C2H2(g)
and if you add up the )H’s you get
= + 226 kJ
6
8. In one of my demonstration I performed the following reaction:
CaC2 (s)+ 2 H2O(l) 6 Ca(OH)2 (aq) + C2H2(g)
)Hf the reactants and products are:
CaC2(s)
-63 kJ/mol
H2O(l)
-286
Ca(OH)2 (aq) -987
C2H2(g)
227
A. (10 points) What is )H for the net reaction?
)Hrxn = 3 )Hf products - 3 )Hf reactants
=()Hf Ca(OH)2 + )Hf C2H2) - ()Hf CaC2 + 2 )Hf H2O)
= (-987+ 227) - (-63 + 2(-286))
= -987 + 227 + 63 +572
= -125 kJ/mol
B. (10 points) If I did the reaction with .10 mole of CaC2 in 1000 g of water at 25.0oC, what would the
temperature of the water be after the reaction was completed? ( The specific heat capacity of water is
4.18 J/oC@g
From 8A we know heat evolved in this reaction is -125 kJ/mol, or 125,000 J/mol
Problem tells us we are working with .1 mole therefor:
125,000 J/mol × 0.1 mol = 12,500 J
This heat is going to raise the temperature of 1000g of water that is already at 25.0o C.
We can use the equations
Energy = Specific Heat Capacity × amount of material × )T
12,500 J = 4.18 J/oC@g × 1000g × (X-25.0)
Where X will be the new, higher temperature
12,500/(4.18×1000) = X-25
12,500/4,180 = X-25
2.99 = X-25
2.99 + 25 =X
X = 27.99
7
8
9. Yellow Orange light has a wavelength of 660 nm.
A. (5 points) What is the frequency of this light?
c=8<
8 = c/<
= 3x108 m/s / 660x10-9 m
= 4.55 x1014 s-1 Or Hz
B. (5 points) The energy of this light?
E=h<
From 9A < = 4.55 x1014 s-1
E = 6.626x10-34 J/photon ×4.55 x1014 s-1
=3.01x10-19 J
10. (5 points) Which of the following sets of quantum numbers are not allowed for a hydrogen atom?
a.
1p Not allowed
For a 1 principal quantum number, only an s angular momentum quantum number is allowed.
b.
n=1,R=1, mR=0 Not allowed
For a 1 principal quantum number, only an 0 angular momentum quantum number is allowed.
c.
n=8,R=7,mR=-6 Allowed
For a 1 principal quantum number, only an 0,1,2,3,4,5,6,and 7 angular momentum quantum numbers
are allowed.
For a 7 angular momentum quantum number, -7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7 magnetic quantum
numbers are allowed.