CHM 2046 Module 10
Chapter 16 Sample Multiple Choice test #2 Answers
1. The
correct answer is: (e)
the partial pressure of NH3
and BF3 will remain the same.
Explanation:
For this system the equilibrium expression is
K = [NH3][BF3]. Addition of more solid will have no
effect as H3 NBF3(s) does not appear in the equilibrium
expression.
Solution Reference:
Page 750, The Equilibrium Constant
2. The correct answer is: (c)
Explanation:
The equilibrium expression is a constant when
the concentrations of the reaction products, each raised to a power equal to
its stoichiometric coefficient in the chemical
reaction, appear in the numerator, and the concentrations of the reactants,
each raised to the appropriate power, appear in the denominator.
Solution Reference:
Page 750, The Equilibrium Constant
3. The correct answer is: (d)
2.63.
Explanation:
K = (1/0.145)1/2 = 2.63
Solution Reference:
Page 753, Manipulating Equilibrium Expressions
4. The correct answer is: (e)
1.2 x 10-20.
Explanation:
Solution Reference:
Page 753, Manipulating Equilibrium Expressions
Reverse the 2nd equation which makes the
Sum the equations, cancel common ions/molecules on both
sides, and the K for the total equation is Ktotal
= K1 x 1/
5. The correct answer is: (e)
O3(g) + NO(g) <--> NO2(g)
+ O2(g)
Explanation:
Kc = Kp when Dngas = 0
(Total molecules of products - Total molecules of reactamts)
Solution Reference:
Page 752, Expressing Concentrations: Kc and Kp
6. The correct answer is: (a)
> 0.27.
Explanation:
Dngas = 1 (2 molecules of products - 1 molecule of reactant = 1)
Solution Reference:
Page 752, Expressing Concentrations: Kc and Kp
7. The correct answer is: (d)
2 and 3 only
Solution Reference:
Page 758, The Reaction Quotient
8. The correct answer is: (a)
1 < 2 < 3.
Solution Reference:
Page 757, Meaning of the Equilibrium Constant
9. The correct answer is: (c)
[Fe3+][
Explanation:
Fe(OH)3(s) <--> Fe3+(aq) + 3OH-(aq)
Solution Reference:
Page 751, Writing Equilibrium Constant
Expressions
10. The correct answer is: (d)
BaSO3; K = 8.0 x 10-7
Explanation:
Which salt has the next to largest equilibrium
constant?
Solution Reference:
Page 757, The Meaning of the Equilibrium
Constant
11. The correct answer is: (b)
8.00.
Explanation:
Solution Reference:
Page 762, Calculating an Equilibrium Constant
12. The correct answer is: (b)
0.10 M.
Explanation:
Solution Reference:
Page 766, Using Equilibrium Constants in
Calculations
13. The correct answer is: (b)
49.0.
Explanation:
Solution Reference:
Page 750, The Equilibrium Constant
14. The correct answer is: (e)
the equilibrium constant for the reaction
above increases with an increase in temperature.
Explanation:
For
25oC (assuming a 1 Liter container) :
Kc
= (0.02)2/0.10
= 0.0004/0.10 = 0.004
For
90oC (assuming 1 Liter container):
Kc
= (0.12)2/0.050 = 0.0144/0.050 = 0.288
Therefore
the Kc increases with temperature
N2O4(g) + heat <--> 2NO2(g)
Solution Reference:
Page 770, Effect of Temperature Changes on Equilibria
15. The correct answer is: (a)
1; increase; decrease
Explanation:
Need more products so that Q can approach 1.6. The reaction is endothermic
because DH is positive.
Solution Reference:
Page 758, Reaction Quotient and Page 770,
Effect of Temperature Changes on Equilibrium
16. Question Instructions: This
question refers to the following equilibrium at a given temperature:
4NH3(g) + 3O2(g) <--> 2N2(g) + 6H2O(g)
The
correct answer is: (e) (I have a problem with
this calculation below as it is a 5 L container, change it to a 1 Liter
container then the below is correct, in #18 the author accounts for the five
liters)
0.30 - 2/3 y.
The
answer should be (0.30-2/3y)/5.0 or 0.060-0.132y, which is not a choice using
the 5.0 liter container
Explanation:
Solution Reference:
Page 766, Using Equilibrium Constants in
Calculations
Question Instructions: This
question refers to the following equilibrium at a given temperature, using the
concentrations from #16:
4NH3(g) + 3O2(g) <--> 2N2(g)
+ 6H2O(g)
17. The correct answer is: (c)
0.08 - 0.10 y.
Explanation:
Solution Reference:
Page 766, Using
Equilibrium Constants in Calculations
18. The correct answer is: (e)
O3(g) + NO(g) <--> NO2(g)
+ O2(g)
Explanation:
Dngas = # mol gaseous products - # mol gaseous reactants. When Dngas =
zero, a volume change doesn't effect the concentrations of the products any
differently than the reactants, so there is no shift in equilibrium.
Solution Reference:
Page 770, Disturbing a Chemical Equilibrium
19. The correct answer is: (b)
PF3Cl2(g) <--> PF3(g)
+ Cl2(g)
Explanation:
(I misread the book yesterday: Volume and
pressure changes affect equilibria in which the
number of moles of gas on the two sides is different. (No effect –if the same)
The system will shift to the side occupying the smaller volume to accommodate
an increase in pressure or a reduction in volume. The reverse is opposite
instead of the same as I told you in class last night See Table 16.2 page 781,
then read pages 785-786.)
The
equilibrium will shift to the side that has the most molecules.
Dngas = 2 - 1 = +1
Solution Reference:
Page 770, Disturbing a Chemical Equilibrium
20. The correct answer is: (c)
I
disagree with the author here. If you reduce the volume of the bulb, you will
increase the concentration of the oxygen gas, therefore reducing the mass of
the HgO. Answers b,c,d,e all shift the equilibrium to the right, choice
a is the only shift to the left. I still think the answer is a.
reducing the volume of the bulb.
Explanation:
For this equilibrium, the equilibrium
expression is K = [O2]. Reducing the volume of the bulb would
increase the concentration of the oxygen.
Solution Reference:
Page 770, Disturbing a Chemical Equilibria
21. The correct answer is: (a)
0.25.
Explanation:
Solution Reference:
Page 750, The Equilibrium Constant
22. The correct answer is: (c)
(This
question or the answers do not make sense. Last night I told you the
equilibrium shifts to the right, or product-favored. The product is HI, so it
will increase. So (d) has to be the correct answer, not (c):
The concentrations of H2 and HI
will decrease as the system is approaching equilibrium.
Explanation:
The reaction quotient equals 1.0 so more
product will be produced.
Solution Reference:
Page 758, The Reaction Quotient
23. The correct answer is: (d)
(I
gave you the wrong answer last night—see page 781, Table 16.2.)
increase both the rate and the equilibrium constant.
Solution Reference:
Page 770, Disturbing a Chemical Equilibrium
24. The correct answer is: (c)
3 only
Solution Reference:
Page 770, Disturbing a Chemical Equilibrium
25.The correct answer is: (e)
low temperature and high pressure.
Solution Reference:
Page 770, Disturbing a Chemical Equilibrium