Download Acids-Bases and Chemical Equilibria II - Homework 5 | CH 302 and more Quizzes Chemistry in PDF only on Docsity!
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001 10.0 points Calculate the equilibrium constant at 25◦C for a reaction for which ∆G^0 = − 1 .62 kcal/mol.
- 8049
- − 15. 4025
- 70123
- 025
- 4025 correct
Explanation: T = 25◦C + 273 = 298 K ∆G^0 = −1620 cal/mol At equilibrium
∆G^0 = −R T ln K −1620 = (− 1 .987 cal/mol · K) × (298.15 K)(ln K) K = 15. 4025
002 10.0 points The standard molar Gibbs free energy of for- mation of NO 2 (g) at 298 K is 51.30 kJ · mol−^1 and that of N 2 O 4 (g) is 97.82 kJ · mol−^1. What is the equilibrium constant at 25◦C for the reaction
2 NO 2 (g) ⇀↽ N 2 O 4 (g)?
2. 1. 02 × 10 −^10
3. 9. 72 × 109
- None of these
- 6.88 correct
8. 7. 01 × 10 −^9
Explanation: ∆G^0 products = 97.82 kJ · mol−^1 ∆G^0 reactants = 51.30 kJ · mol−^1
∆G^0 rxn =
n ∆G^0 products
−
n ∆G^0 reactants = 97. 82 − (2)(51.30)
= (− 4 .78 kJ/mol)
1000 J
kJ
= −4780 J/mol
∆G^0 = −R T ln K K = e−∆G
(^0) /(R T )
= exp
[
−4780 J/mol (8.3145 J/mol · K)(298 K)
]
003 10.0 points The reaction
A + B ⇀↽ C + 2 D
has an equilibrium constant of 3. 7 × 10 −^3. Consider a reaction mixture with [A] = 2. 0 × 10 −^2 M [B] = 1. 7 × 10 −^4 M
[C] = 2. 4 × 10 −^6 M
[D] = 3. 5 × 10 −^3 M
Which of the following statements is defi- nitely true?
- The forward reaction can occur to a greater extent than the reverse reaction until equilibrium is established. correct
- No conclusions about the system can be made without additional information.
- Heat will be evolved.
- The system is at equilibrium.
- The reverse reaction can occur to a
greater extent than the forward reaction until equilibrium is established.
Explanation:
Q =
[C] [D]^2
[A] [B]
(2. 4 × 10 −^6 M) (0.0035 M)^2
(0.02 M) (0.00017 M)
= 8. 64706 × 10 −^6
Since Q < K the foward reaction is favored.
004 10.0 points The reaction
N 2 (g) + 3 H 2 (g) ⇀↽ 2 NH 3 (g) ,
has an equilibrium constant of 4. 0 × 108 at 25 ◦C. What will eventually happen if 44. moles of NH 3 , 0.452 moles of N 2 , and 0. moles of H 2 are put in a 10.0 liter container at 25 ◦C?
- More NH 3 will be formed. correct
- Nothing; the system is at equilibrium.
- More N 2 and H 2 will be formed.
Explanation:
K = 4. 0 × 108 [NH 3 ] =
44 .0 mol 10 L
[N 2 ] =
0 .452 mol 10 L
[H 2 ] =
0 .108 mol 10 L
Q =
[NH 3 ]^2
[N 2 ] [H 2 ]^3
(4.40 M)^2
(0.0452 M) (0.0108 M)^3
= 3. 4 × 108
Since Q < K equilibrium will shift to the right, forming more NH 3.
005 10.0 points Kc = 2. 6 × 108 at 825 K for the reaction
2 H 2 (g) + S 2 (g) ⇀↽ 2 H 2 S(g)
The equilibrium concentration of H 2 is 0. M and that of S 2 is 0.0010 M. What is the equilibrium concentration of H 2 S?
1. 10 M
- 1.02 M correct
- 0.0010 M
- 0.10 M
Explanation: Kc = 2. 6 × 108 [H 2 ]eq = 0.0020 M [S 2 ]eq = 0.0010 M
2 H 2 (g) + S 2 ⇀↽ 2 H 2 S
Kc =
[H 2 S]^2
[H 2 ]^2 [S 2 ]
[H 2 S] =
Kc [H 2 ]^2 [S 2 ]
=
(2. 6 × 108 ) (0.0020 M)^2 (0.0010 M)
= 1.0 M
006 10.0 points Suppose the reaction
H 2 (g) + I 2 (g) ⇀↽ 2 HI(g)
has an equilibrium constant Kc = 49 and the initial concentration of H 2 and I 2 is 0.5 M and HI is 0.0 M. Which of the following is the correct value for the final concentration of HI(g)?
- 0.250 M
- 0.219 M
- 0.389 M
- 0.599 M
- 0.778 M correct
Explanation: Kc = 49 [H 2 ]ini = 0.5 M [I 2 ]ini = 0.5 M [HI]ini = 0 M
How many moles of H 2 O(g) are present in an equilibrium mixture resulting from the addition of 1.00 mole of H 2 , 2.00 moles of CO 2 , 0.75 moles of H 2 O, and 1.00 mole of CO to a 5.00 liter container at 990◦C?
- 1.4 mol
- 0.60 mol
- 1.7 mol
- 0.80 mol
- 1.1 mol correct
- 1.0 mol
Explanation: Kc = 1. 6
[H 2 ] =
1 .00 mol 5 L
[CO 2 ] =
2 .00 mol 5 L
[H 2 O] =
0 .75 mol 5 L
[CO] =
1 .00 mol 5 L
Q =
[H 2 O][CO]
[H 2 ][CO 2 ]
0 .75 mol 5 L
1 mol 5 L
1 mol 5 L
2 mol 5 L
= 0. 375 < Kc = 1. 6 Therefore equilibrium moves to the right. H 2 (g) + CO 2 (g) ⇀↽ H 2 O(g) + CO(g) 0.2 0.4 0.15 0. −x −x x x
- 2 − x 0. 4 − x 0 .15 + x 0 .2 + x
(0.15 + x)(0.2 + x) (0. 2 − x)(0. 4 − x)
x^2 + 0. 35 x + 0. 03 x^2 − 0. 6 x + 0. 08
x^2 + 0. 35 x + 0.03 = 1. 6 x^2 − 0. 96 x + 0. 128
- 6 x^2 − 1. 31 x + 0.098 = 0
x =
(1.31)^2 − 4(0.6)(0.098)
= 7. 756 × 10 −^2 M
mol H 2 O = 5.00 L
×
0 .15 + 7. 75 × 10 −^2
) (^) mol L = 1.1 mol H 2 O
010 10.0 points
What happens to the concentration of NO(g) when the total pressure on the equilib- rium reaction 3 NO 2 (g) + H 2 O(ℓ) ⇀↽ 2 HNO 3 (aq) + NO(g) is increased (by compression)?
- Unable to determine
- decreases
- remains the same
- increases correct
Explanation: Increasing the total pressure on the sys- tem by decreasing its volume will shift the equilibrium toward the side of the reaction with fewer numbers of moles of gaseous com- ponents. If the total number of moles of gas is the same on the product and reactant sides of the balanced chemical equation, then changing the pressure will have little or no ef- fect on the equilibrium distribution of species present. The amount and concentration of NO will increase.
011 10.0 points Consider the reaction
2 SO 2 (g) + O 2 (g) ⇀↽ 2 SO 3 (g)
where ∆Hrxn = −198 kJ. The amount of SO 2 (g) at equilibrium increases when
- the volume is increased. correct
- the temperature is decreased.
- the pressure is increased.
- more oxygen is added.
- SO 3 is removed.
Explanation:
2 SO 2 (g) + O 2 (g) ⇀↽ 2 SO 2 (g) + heat According to Le Chatelier’s principle, the amount of reactant SO 2 (g) is increased when the equilibrium shifts to the left. This will happen when another reactant (O 2 ) is re- moved. For an exothermic reaction decreasing temperature removes heat and sends equilib- rium to the right. Increasing pressure sends the equilibrium in the direction that has fewer numbers of moles of gas. Increasing the vol- ume is equivalent to decreasing gas pressure.
012 10.0 points For an exothermic reaction, what would hap- pen to the numerical value of Kc, if we in- crease the temperature at constant pressure?
- Kc would increase.
- Kc would decrease. correct
- Kc would either increase or decrease, de- pending on the number of moles of gas in- volved.
- Kc would either increase or decrease, de- pending on the concentrations.
- Kc would not change.
Explanation: An exothermic reaction gives off heat as products are formed, so at a higher tem- perature the reverse endothermic reaction is favoured, decreasing Kc, which reflects the ratio of the products to reactants.
013 10.0 points Suppose the reaction mixture
N 2 (g) + 3 H 2 (g) ⇀↽ 2 NH 3 (g)
is at equilibrium at a given temperature and pressure. The pressure is then increased at constant temperature by compressing the re- action mixture, and the mixture is then al- lowed to reestablish equilibrium. At the new equilibrium,
- the nitrogen is used up completely.
- there is the same amount of ammonia present as there was originally.
- there is more ammonia present than there was originally. correct
- there is less ammonia present than there was originally.
Explanation: LeChatelier’s Principle states that if a change occurs in a system at equilibruim, the system responds to relieve the stress and reach a new equilibrium. Here, the number of moles of gaseous reactants is greater than the number of moles of products. Increasing the pressure of the above system will result in the reaction proceeding to reduce that pressure increase. The system will shift to the right (the side that has fewer moles of gas), so the pressure will be reduced; thus more ammonia will be produced.
014 10.0 points Consider the system
2 N 2 O 5 (g) ⇀↽ 2 N 2 O 4 (g) + O 2 (g) + heat
at equilibrium at 25◦C. If the temperature were raised would the equilibrium be shifted to produce more N 2 O 5 or more N 2 O 4?
- more N 2 O 5 correct
- There would be no effect.
- more N 2 O 4
Explanation: This is an exothermic reaction and so in- creasing temperature provides more heat to
- At point D, the reaction will move toward the reactants to get to equilibrium.
- None of the statements is false. correct
- At point C, the system is at equilibrium.
Explanation: ∆G◦^ is negative (point E is lower free en- ergy than point A), K > 1. At point B, the reaction will move to the products (Q < K). At point E, the reaction will move to the re- actants (Q > K). At point C, the reaction is at equilibrium (Q = K).
018 10.0 points The following figure represents the progress of a given reaction at 298 K.
b b b b
b G
A
B
C D
E
rxn progress
At point B on this figure, what is the relation- ship of Q to K?
- Cannot be determined
- Q = K
- Q > K
- Q < K correct
Explanation: Point B is on the reactants-heavy side of equilibrium, so Q is less than K. Note also that dG (slope) is negative here which means the reaction would be spontaneous in the for- ward direction. Spontaneous in a forward direction corresponds to Q being less than K.
019 10.0 points Given the hypothetical reaction
X(g) ⇀↽ Y(g) predict what will happen when 1.0 mol Y is placed into an evacuated container.
- Nothing; the products are already formed, so no reaction occurs.
- Q will decrease until Q = K. correct
- ∆G◦^ will decrease until ∆G◦^ = 0.
- Q will increase until Q = K. Explanation: The container begins with products only. The reverse reaction will predominate until equilibrium is reached. The reaction quotient Q starts out larger than K and decreases until Q = K.
020 10.0 points
Consider the reaction: Cgraphite(s) + O 2 (g) ↔ CO 2 (g) ∆G◦= -400 kJ · mol−^1 · K−^1 Which of the following is a possible value of K for this reaction?
- 10 −^70
- 1070 correct
- -0. Explanation: ∆G◦= -400 kJ · mol−^1 · K−^1 is very large and negative, requiring that K be a very large number.
021 10.0 points The hydronium ion is.
- H 2 O+
- OH−
- HO+
- H 3 O+^ correct
Explanation: The hydronium ion H 3 O+^ represents hy- drated protons in aqueous solution.
022 10.0 points A strong acid (or base) is one which
- dissociates completely in aqueous solu- tion. correct
- should only be used when wearing goggles and gloves.
- dissolves glass.
- dissolves metals.
- reacts with a salt to form water.
Explanation: By definition, a strong acid is completely dissociated in solution.
023 10.0 points HCN is classified as a weak acid in water. This means that it produces
- a relatively small fraction of the maximum number of possible hydronium ions. correct
- 100% of the maximum number of possible hydronium ions.
- no hydronium ions.
- a relatively large fraction of the maximum number of possible hydronium ions.
Explanation: In water only a small percentage of a weak acid hydrolyzes, producing H 3 O +^ ions.
024 10.0 points An acid is strong if it
- is very concentrated.
- causes metals to corrode.
- makes acid-base indicators change color.
- ionizes completely in water. correct Explanation: In water a weak acid is partially hydrolyzed and a strong acid is completely hydrolyzed. In both cases H 3 O+^ ions are produced.
025 10.0 points Which of the following substances is NOT a weak acid?
- HCl correct
- HSO 3
- HCN
- H 2 CO 3
- HF
- H 3 PO 4 Explanation: HCl is not a weak acid.
026 10.0 points Which of the following substances is a weak acid?
- HClO 3
- HBr
- HCN correct
- HI
- HNO 3
- HCl
- H 2 SO 4 Explanation: HCN is a weak acid.
027 10.0 points
- A base must contain a hydroxide group. correct
Explanation: According to the Bronsted-Lowry defini- tion of bases, a base must be a proton accep- tor. It does not need to contain a hydroxide group.
032 10.0 points Which statement is true for the following re- action?
CCl 3 COOH + H 2 O ↔ CCl 3 CO− 2 + H 3 O+
- H 3 O+^ is the conjugate base of CCl 3 CO− 2.
- H 2 O is the conjugate acid of CCl 3 CO− 2.
- H 2 O is the conjugate base of CCl 3 CO− 2.
- CCl 3 COOH is the conjugate acid of CCl 3 CO− 2. correct
- Cl 3 COOH is the conjugate base of CCl 3 CO− 2.
- H 3 O+^ is the conjugate acid of CCl 3 CO− 2.
Explanation:
033 10.0 points According to Bronsted-Lowry Theory an acid is
- a proton donor. correct
- amphoteric.
- a proton acceptor.
- a soluble ionic hydroxide.
Explanation: The Bronsted-Lowry theory of acids and bases defines an acid as a substance which is able to donate a proton.
034 10.0 points Which is NOT a conjugate base-acid pair?
- HSO− 4 : SO^24 − correct
- CN−^ : HCN
- F−^ : HF
- H 2 O : H 3 O+
- OH−^ : H 2 O
Explanation: In the Bronsted-Lowry sense, a base is a proton acceptor and an acid is a proton donor. HSO− 4 is a proton donor and an acid. There- fore this is a conjugate acid-base pair, not a conjugate base-acid pair.
035 10.0 points The conjugate base of H 2 SO 4 is:
- SO^24 −
- HSO− 4 correct
- H 2 O
- H 3 O+
- H 2 SO 3
- OH−
- H 3 SO+ 4
Explanation: Remove one H+^ ion from a species to find its conjugate base.
036 10.0 points A given weak acid HZ has a Ka = 4 × 10 −^6. What is the H 3 O+^ concentration of a solution of HZ that has a concentration of 0.5 mol/L?
Correct answer: 0.00141 mol/L.
Explanation:
Ka = 4 × 10 −^6 [H 3 O+] =? The expression for Ka here is
Ka =
[H+] [Z−]
[HZ]
Substituting in values for the concentrations and Ka, we have
4 × 10 −^6 =
(x) (x) (0. 5 − x)
As HZ has a small Ka, we can assume that x is very small compared to 0.5, so we can simplify this equation to
4 × 10 −^6 =
(x)^2 (0.5) 2 × 10 −^6 = x^2 x = 0. 00141
Remembering that we called [H+] x in the equation, [H+] is 0.00141 mol/L.
037 10.0 points Assume that five weak acids, identified only by numbers (1, 2, 3, 4, and 5), have the following ionization constants.
Ionization Acid Constant Ka value 1 1. 0 × 10 −^3 2 3. 0 × 10 −^5 3 2. 6 × 10 −^7 4 4. 0 × 10 −^9 5 7. 3 × 10 −^11
The anion of which acid is the strongest base?
- 5 correct
- 3
- 2
- 1
- 4
Explanation:
