Chem. 1A Midterm 2
Version B
November 9, 2016
First initial
of last name
Name:__________________________________________
Print Neatly. You will lose 1 point if I cannot read your name or perm number.
Perm Number:___________________________________
All work must be shown on the exam for partial credit. Points will be taken off for
incorrect or missing units. Calculators are allowed. Cell phones may not be used as
calculators. On fundamental and challenge problems you must show your work in order
to receive credit for the problem. If your cell phone goes off during the exam you will
have your exam removed from you.
Fundamentals
(of 36 possible)
Problem 1
(of 16 possible)
Problem 2
(of 18 possible)
Multiple Choice
(of 30 possible)
Midterm Total
(of 100 possible)
1
Fundamental Questions
Each of these fundamental chemistry questions is worth 6 points. You must show work to get
credit. Little to no partial credit will be awarded. Make sure to include the correct units on your
answers.
1) 6 pts
Balance the following equation in acidic solution
Cu(s) + NO3-(aq)  Cu2+(aq) + NO(g)
CuCu2+
CuCu2+ + 2e-
NO3-NO
NO3- NO +2H2O
NO3- + 4H+ NO +2H2O
NO3- + 4H+ + 3e-  NO +2H2O
3(CuCu2+ + 2e-)= 3Cu3Cu2+ + 6e2(NO3- + 4H+ + 3e-  NO +2H2O-)= 2NO3- + 8H+ + 6e-  2NO +4H2O
3Cu(s) + 2NO3-(aq) + 8H+(aq)  3Cu2+(aq) + 2NO(g) +4H2O(l)
2)
6 pts
What are the (a) molecular, (b) complete ionic, and (c) net ionic reactions when
the following are mixed? If no reaction occurs write no reaction. Do not forget to
include states.
Fe(NO3)3(aq) and (NH4)2SO4
No reaction
CaCl2(aq) and K2SO4(aq)
Molecular: CaCl2(aq) + K2SO4(aq)  2KCl(aq) + CaSO4(s)
Complete Ionic Equation: Ca2+(aq)+2Cl-(aq)+2K+(aq)+SO42-(aq)
2K+(aq)+2Cl-(aq)+CaSO4(s)
2+
2Net Ionic Equation: Ca (aq) + SO4 (aq)  CaSO4(s)
3)
6 pts
Consider the following samples of gases at the same temperature.
Arrange each of these samples in order from lowest to highest.
a. Pressure
ii = vi < i = iv = v = viii < iii = vii
b. Average kinetic energy
All sample have same kinetic energy
c. Density
ii < i = iv =vi < iii = v = viii < vii
d. Root means square velocity
v = vi = vii = viii < i = ii = iii = iv
2
4)
6 pts
Each sketch below shows a flask with some gas and a pool of mercury in it. The gas is at a
pressure of 0.5 atm. A J-shaped tube is connected to the bottom of the flask, and the
mercury can freely flow in or out of this tube. (You can assume that there is so much
more mercury in the pool than can fit into the tube that even if the J-tube is completely
filled, the level of mercury in the pool won't change.)
Notice also that in the left sketch the J-tube is open at its other end, so that air can freely
flow in or out of the tube. On the other hand, in the right sketch the J-tube is closed at its
other end, and you should assume there is no gas between the mercury and the closed
end of the tube. To answer this question, you must decide what the mercury level will be
when the mercury finally stops flowing in or out of the tube and draw the level in on the
picture. You must show calculation to get full credit.
Open Tube
Close Tube
Open Tube:
Gravity: 1.5 m
760 𝑚𝑚𝐻𝑔
) = 380
1 𝑎𝑡𝑚
760 𝑚𝑚𝐻𝑔
𝑎𝑡𝑚 ( 1 𝑎𝑡𝑚 ) = −760
𝑃𝐵𝑢𝑙𝑏 : 0.5 𝑎𝑡𝑚 (
𝑚𝑚𝐻𝑔 = 0.380 𝑚𝐻𝑔
𝑃𝐸𝑥 : −1
𝑚𝑚𝐻𝑔 = −0.760 𝑚𝐻𝑔
𝑇𝑜𝑡𝑎𝑙 𝑅𝑖𝑠𝑒: 1.5 𝑚 + 0.380 𝑚 + −0.760 𝑚 = 1.12 𝑚
Close Tube:
Gravity: 1.5 m
𝑃𝐵𝑢𝑙𝑏 : 0.5 𝑎𝑡𝑚(7601 𝑚𝑚𝐻𝑔
) = 380 𝑚𝑚𝐻𝑔 = 0.380 𝑚𝐻𝑔
𝑎𝑡𝑚
𝑇𝑜𝑡𝑎𝑙 𝑅𝑖𝑠𝑒: 1.5 𝑚 + 0.380 𝑚 = 1.88 𝑚
5)
6 pts
During a titration a 0.02500 L sample of H2SO4 required 24.16 mL of 0.106 M
NaOH to reach the equivalence point. What is the initial concentration of the
H2SO4?
2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l)
1𝐿
0.106 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
) = 0.00256 𝑚𝑜𝑙
1 𝐿 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝐻2 𝑆𝑂4
𝑁𝑎𝑂𝐻 ( 2 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
) = 0.00128 𝑚𝑜𝑙 𝐻2 𝑆𝑂4
24.16 𝑚𝐿 𝑁𝑎𝑂𝐻 (1000 𝑚𝐿) (
𝑁𝑎𝑂𝐻
0.00256 𝑚𝑜𝑙
𝑛 0.000128 𝑚𝑜𝑙
𝑀= =
= 0.0512 𝑀
𝑉
0.02500 𝐿
6)
6 pts
When the equilibrium is disturbed, in which direction will the reaction proceed?
Circle the correct answer.
2CO(g) + O2(g) ⇌ 2CO2(g)
exothermic
Remove CO2
Reactants
Products
No Change
Reduce the volume
Reactants
Products
No Change
Increase the temperature
Reactants
Products
No Change
Add Ne
Reactants
Products
No Change
Add O2
Reactants
Products
No Change
3
Challenge Problems
Each of the following short answer questions are worth the noted points. Partial credit will be
given. You must show your work to get credit. Make sure to include proper units on your answer.
1a) 6 pts
A mixture of 1.00 g H2 and 8.60 g O2 is introduced into a 1.500 L flask at
25˚C. What is the total gas pressure in the flask?
𝑛𝑡𝑜𝑡 𝑅𝑇 (𝑛𝐻2 + 𝑛𝑂2 )𝑅𝑇
=
𝑉
𝑉
Determine nH 2
𝑃𝑡𝑜𝑡 =
1 𝑚𝑜𝑙 𝐻
𝑛𝐻2 = 1.00 𝑔 (2.0158 𝑔 𝐻2 ) = 0.496 𝑚𝑜𝑙 𝐻2
2
Determine nO2
1 𝑚𝑜𝑙 𝑂
𝑛𝑂2 = 8.60 𝑔 (31.998 𝑔 𝑂2 ) = 0.269 𝑚𝑜𝑙 𝑂2
2
𝑃𝑡𝑜𝑡 =
1b)
10 pts
𝐿∙𝑎𝑡𝑚
(0.496 𝑚𝑜𝑙 + 0.269 𝑚𝑜𝑙)(0.08206 𝑚𝑜𝑙∙𝐾
)(298𝐾)
1.500 𝐿
= 12.5 𝑎𝑡𝑚
When the mixture is ignited, an explosive reaction occurs in which water
is the only product. What is the total gas pressure when the flask is
returned to 25˚C? (The vapor pressure of water at 25˚C is 23.8 mmHg.)
Reaction of Interest
2H2(g) + O2(g)  2H2O(g & l)
This is a limiting reagent problem
H2 (L.R.)
O2
H2O
I (mol) 0.496
0.269
0
C (mol) -2x = -0.496 -x = -0.248 +2x = +0.496
F (mol) 0
0.021
0.496
0.496-2x=0
x=0.248
𝑃𝑡𝑜𝑡 = 𝑃𝐻2 + 𝑃𝑂2 + 𝑃𝐻2 𝑂
Calculate 𝑃𝐻2
𝑃𝐻2 = 0 𝑎𝑡𝑚
Calculate 𝑃𝑂2
𝐿∙𝑎𝑡𝑚
)(298𝐾)
𝑛𝑂 𝑅𝑇 (0.021 𝑚𝑜𝑙)(0.08206 𝑚𝑜𝑙∙𝐾
𝑃𝑂2 = 2
=
= 0.342 𝑎𝑡𝑚
𝑉
1.500 𝐿
Calculate the 𝑃𝐻2 𝑂
At 25˚C the majority of the water will be in the liquid state.
The pressure of H2O comes from the vapor pressure of water.
1 𝑎𝑡𝑚
𝑃𝐻2 𝑂 = 23.8 𝑚𝑚𝐻𝑔 (760 𝑚𝑚𝐻𝑔) = 0.0313 𝑎𝑡𝑚
Calculate Ptot
𝑃𝑡𝑜𝑡 = 𝑃𝐻2 + 𝑃𝑂2 + 𝑃𝐻2 𝑂 = 0 𝑎𝑡𝑚 + 0.342 𝑎𝑡𝑚 + 0.0313 𝑎𝑡𝑚
= 0.373 𝑎𝑡𝑚
4
2a)
8 pts
A 2.4156 g sample of PCl5 was placed in an empty 2.000-L flask and allowed to
decompose to PCl3 and Cl2 at 250.0°C:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
At equilibrium the total pressure inside the flask was observed to be 358.7 torr.
Calculate the partial pressure of each gas at equilibrium and the value of KP at
250.0°C.
1 𝑎𝑡𝑚
𝑃𝑡𝑜𝑡 = 𝑃𝑃𝐶𝑙5 + 𝑃𝑃𝐶𝑙3 + 𝑃𝐶𝑙2 = 358.7 𝑡𝑜𝑟𝑟(760
) = 0.4720 𝑎𝑡𝑚 (at equilibrium)
𝑡𝑜𝑟𝑟
PCl5
2.4156 g
PCl3
0g
Cl2
0g
Initial (𝑛 = 𝑀 )
0.011601 mol
0 mol
0 mol
Initial (𝑃 =
0.24894atm
0 atm
0 atm
Initial (m)
𝑚
𝑥
𝑛𝑅𝑇
)
𝑉
Change
-x
+x
+x
Equilibrium
0.24894-x
x
x
Therefore
𝑃𝑡𝑜𝑡 = 𝑃𝑃𝐶𝑙5 + 𝑃𝑃𝐶𝑙3 + 𝑃𝐶𝑙2 = 0.24894 − 𝑥 + 𝑥 + 𝑥 = 0.4720
𝑥 = 0.2231
𝑃𝑃𝐶𝑙5 = 0.2489 𝑎𝑡𝑚 − 𝑥 = 0.2489 𝑎𝑡𝑚 − 0.2231 𝑎𝑡𝑚 = 0.0258 𝑎𝑡𝑚
𝑃𝑃𝐶𝑙3 = 𝑃𝐶𝑙2 = 𝑥 = 0.2231 𝑎𝑡𝑚
𝑃𝑃𝐶𝑙3 𝑃𝐶𝑙2 (0.2231)(0.2231)
𝐾𝑃 =
=
= 1.93
𝑃𝑃𝐶𝑙5
0.0258
2b)
10 pts
What are the new equilibrium pressures if 0.250 mole of Cl2 gas is added to the
flask?
Equilibrium
Added (n)
Added(𝑃 =
𝑛𝑅𝑇
)
𝑉
PCl5
0.0258 atm
0 mol
PCl3
0.2231 atm
0 mol
Cl2
0.2231 atm
0.250 mol
0 atm
0 atm
5.36 atm
Initial
0.0258 atm
0.2231 atm
5.58 atm
Change
+x
-x
-x
Equilibrium
0.0258+x
0.2231-x
5.58-x
𝑃𝑃𝐶𝑙3 𝑃𝐶𝑙2 (0.2231 − 𝑥)(5.58 − 𝑥)
𝐾𝑃 =
=
= 1.93
𝑃𝑃𝐶𝑙5
0.0258 + 𝑥
𝑥 2 − 5.80𝑥 + 1.24
= 1.93
0.0258 + 𝑥
2
𝑥 − 5.80𝑥 + 1.24 = 0.0498 + 1.93𝑥
𝑥 2 − 7.73𝑥 + 1.19 = 0
−𝑏 ± √𝑏 2 − 4𝑎𝑐 7.73 ± √(−7.63)2 − 4(1)(1.19)
𝑥=
=
= 7.57 𝑎𝑛𝑑 0.156
2𝑎
2(1)
Since the pressure cannot be negative x=0.156
Using guess and check x = 0.1578 with significant figures x = 0.158
Partial pressure at equilibrium
𝑃𝑃𝐶𝑙5 = 0.0258 𝑎𝑡𝑚 + 𝑥 = 0.0258 𝑎𝑡𝑚 + 0.156 𝑎𝑡𝑚 = 0.182 𝑎𝑡𝑚
𝑃𝑃𝐶𝑙3 = 0.2231 𝑎𝑡𝑚 − 𝑥 = 0.2231 𝑎𝑡𝑚 − 0.156 𝑎𝑡𝑚 = 0.067 𝑎𝑡𝑚
𝑃𝐶𝑙2 = 5.58 𝑎𝑡𝑚 − 𝑥 = 5.58 𝑎𝑡𝑚 − 0.156 𝑎𝑡𝑚 = 5.42 𝑎𝑡𝑚
5
Multiple Choice Questions
On the ParScore form you need to fill in your answers, perm number, test version,
and name. Failure to do any of these things will result in the loss of 1 point. Your
perm number is placed and bubbled in under the “ID number.” Do not skip boxes
or put in a hyphen; unused boxes should be left blank. Bubble in your test version
(B) under the “test form.” Note: Your ParScore form will not be returned to you,
therefore, for your records, you may want to mark your answers on this sheet.
Each multiple choice question is worth 5 points.
1. How is the observed pressure of a gas related to the ideal pressure?
A) The relationship depends on the gas.
B) The observed pressure is less than the ideal pressure.
C) The observed pressure is greater than the ideal pressure.
D) They are equal.
E) None of the above
2. Consider the equation 2A(g) ⇌ 2B(g) + C(g). At a particular temperature, K = 1.6 x 104. If
you mixed 5.0 mol B, 0.10 mol C, and 0.0010 mol A in a 1-L container, in which direction
would the reaction initially proceed?
A) To the left.
B) To the right.
C) The above mixture is the equilibrium mixture.
D) We cannot tell from the information given.
3. Given the following two reactions what is the equilibrium constant of
2A(aq) ⇌ D(aq) + E(aq)?
A(aq) + B(aq) ⇌ D(aq) + C(aq)
K1 = 5.0
2C(aq) + D(aq) ⇌ 2B(aq) + E(aq)
K2 = 2.0
A) 10.
B) 2.5
C) 0.40
D) 20.
E) None of the above
4. The following experiment was carried out using a newly synthesized chlorofluorocarbon.
Exactly 50 mL of the gas effused through a porous barrier in 157 s. The same volume of
argon effused in 76 s under the same conditions. Which compound is the
chlorofluorocarbon?
A) C2Cl5F
B) C2Cl4F2
C) C2Cl2F4
D) C2ClF5
E) C2Cl3F3
6
5. The equilibrium constant K for the following reaction at 900. C is 0.0028. What is KP at
this temperature? CS2(g) +4H2(g) ⇌ CH4(g) + 2H2S(g)
A) 0.28
B) 2.7 101
C) 2.9 10-3
D) 3.3 106
E) None of the above
6. Which of the following reactions does not involve oxidation-reduction?
A) Mg + 2HI  MgI2 + H2
B) MnO2 + 4HI  I2 + 2H2O + MnI2
C) 2Na + 2H2O  2NaOH + H2
D) LiOH + HCl  H2O + LiCl
E) All of the above are oxidation-reduction reactions
Solutions: B, A, E, C, E, D
7