Download Chem. 401 Unit 2 of 4 Review

Chem. 401 Unit 2 of 4 Review
1. If an equation for a reaction is multiplied by n, the value of Kc is altered by:
____raising it to the nth power. _____________________________________________________
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2. A reaction that can proceed in both the forward and reverse directions is said to be:
____reversible ___________________________________________________________________
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3. LeChatlier’s Principle _______________says that a system at equilibrium, when disturbed, will shift
in the direction that relieves the stress of the disturbance.
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4. True or false: The smaller the value of Kc the more reactant favored the reaction.True __________
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5. What is the conjugate base of: subtract one hydrogen ion from the formula!
a) HBr
Br-
b) H2SO4
HSO4-
6. What is the conjugate acid of: add one hydrogen ion from the formula!
a) H2O
H3O+
b) HCO3-
H2CO3
7. Identify the Lewis Acid and Lewis Base in the reactants of:
LA
LB
2+
a) Cu + 4 NH3  Cu(NH3)42+
Metal cations are common Lewis acids!
LA
LB
b) Zn(OH)2 + 2 OH-  Zn(OH)428. For a solution of 7.6 x 10-3 M NaOH, determine
a) [OH-]
= 7.6 x 10-3M
b) [H3O+]
= 1.0 x 10-14 / 7.6 x 10-3 = 1.3 x 10-12M
c) pOH
= log(7.6 x 10-3) = 2.12
d) pH
= 14.00 – pOH = 14.00 – 2.12 = 11.88 (OR = -log(1.31 x 10-12) = 11.88)
9. Which is the stronger acid,
a) the one with Ka = 5.6 x 10-2 or This one. (The larger the Ka the stronger the acid.)
b) the one with Ka = 6.8 x 10-4?
10. For the reaction between carbonic acid and cyanide ion:
a) Write the balanced Rxn:
H2CO3(aq) + CN-(aq) ↔ HCO3-(a) + HCN(aq)
b) Identify the conjugate pairs:
Acid
CB
Base
c) To which side does the equilibrium lie? Ka HCN = 4.0 x 10-10,
Since HCN is weaker than H2CO3, it lies to the product side.
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CA
Ka H2CO3 = 4.3 x 10-7
Chem. 401 Unit 2 of 4 Review
11. Predict whether the following solutions are acidic, basic, or neutral.
a) KNO3
neutral
b) LiCN
basic
c) AlCl3
acidic (metal cations are Lewis acids!)
12. HNO3 is a strong acid. Which of the following is NOT true?
a) HNO3 is essentially 100% dissociated.
b) A 1.0 M solution of HNO3 would have a {H3O+] = 1.0 M
c) The conjugate base, NO3-, is a strong base.
False
13. Write the equilibrium constant expression for the reaction below in terms of Kc.
CaCO3(s) + 2H3O+(aq) ↔ Ca2+ + 2H2O(l) + CO2(g)
Kc =
[Ca 2 + ][CO 2 ]
[H 3 O + ] 2
remember not to include solids or liquids!
14. The following exothermic reaction is at equilibrium:
4NH3 (g) + 3O2 (g) ↔ 6H2O (g) + 2N2 (g)
How will the equilibrium shift if:
a. The volume is increased: Right (to side with more moles of gas)
b. A catalyst is added: No Shift (equilibrium is just reached faster.)
c. The temperature is increased: Left (Heat is a product for exothermic reactions.)
15. For the reaction: A + B ↔ C + D
correct answer is a (true for all reactions at equilibrium)
a) ratef = rater
b) kf = kr
(true if Kc = 1)
c) kf > kr
(true if Kc > 1)
(true if Kc < 1)
d) kf < kr
16. What is the equilibrium constant, Kp, for the reaction below if the following partial pressures were
measured at equilibrium at 25°C: PSO2 = 0.55 atm, PO2 = 0.30 atm, and PSO3 = 0.15 atm.
Also determine Kc.
SO3 (g).
Unbalanced reaction: SO2 (g) + O2 (g)
Balanced reaction: 2 SO2 (g) + O2 (g)
2 SO3 (g).
(PSO3)2
(0.15)2
Kp = --------------- = ------------------- = 0.25
(PSO2)2 PO2
(0.55)2 (0.30)
Kp = Kc(RT)∆n
Kc = Kp =
0.25
= 0.25 = 6.1
(RT)∆n [(0.08206)(298)](2-3) 24.45-1
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Chem. 401 Unit 2 of 4 Review
17. What is the equilibrium constant for a reaction that has ∆G°rxn= 0.450kJ/mol at 25°C?
∆G°rxn= -RTlnKc
lnKc = -∆G°rxn =
-450J/mol
= -0.182
RT
(8.314J/K mol)(298K)
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Kc = e-0.182 = 0.834
18. Consider the following equation:
2 HBr(g) ↔ H2(g) + Br2(g); Kc = 1.8 x 10-9 at some temperature T.
A 4.50 L container has 1.69 mol each of HBr and Br2 and 5.45 x 10-5 mol H2 at temperature T. Is
the system at equilibrium? If not, will it shift towards reactants or products?
[ H 2 ][ Br2 ] (5.45 x 10 −5 / 4.50)(1.69 / 4.50) (1.211 x 10 −5 )(0.3756)
=
=
= 3.22 x 10 −5
Q=
2
2
2
(1.69 / 4.50)
[ HBr ]
(0.3756)
Q = 3.22 x 10-5 > K (1.8 x 10-9) so the system is not at equilibrium.
Reaction will shift to the LEFT.
19. This equation is at equilibrium: CO(g) + H2O (g) ↔ CO2(g) + H2 (g)
a. If a 10.00L vessel has 8.00 mol CO2 and H2, and 1.00 mol CO and H2O gas at 588°K, which
way will the reaction proceed? (Kc = 31.4 at 588°K)
b. What are the concentrations of all species at equilibrium?
The initial concentrations can be found by dividing the moles by the volume.
To determine the direction the reaction will go we need to determine Q ( reaction quotient).
[CO 2 ][[H 2 ] (0.800)(0.800)
=
= 64
Q=
[[CO][H 2 O] (0.100)(0.100)
Since Q > Kc. There are excess products and equilibrium will shift left.
CO(g)
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Kc =
0.100
+x
0.100+x
+
H2O (g)
↔
0.100
+x
0.100+x
CO2 (g)
+
0.800
-x
0.800-x
H2 (g)
0.800
-x
0.800-x
(0.800 − x)
(0.800 − x) 2
= 31.4 = 5.604
= 31.4 This is a perfect square so
2
(0.100 + x)
(0.100 + x)
0.800-x = 0.5604 + 5.604x
(0.800-0.5604) = (5.604x + x)
0.2396 = 6.604x
x = 0.2396 / 6.604 = 0.0363
Final Equilibrium concentrations:
[CO2]=[H2] = 0.100 + x = 0.100 = 0.0363 = 0.1363 ⇒ 0.136M
[CO] = [H2O] = 0.800-x = 0.800 – 0.0363 = 0.7637 ⇒0.764M
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Chem. 401 Unit 2 of 4 Review
20. What are the [H3O+] and pH of a 0.35M solution of NH4NO3? (Ans: 1.4 x 10-5; 4.85)
Start by deciding which ion(s) undergo hydrolysis. (NH4+, nitrate is neutral)
Ka = Kw/Kb(NH3) = 1.0x10-14/1.76x10-5 = 5.68x10-10
Write the hydrolysis reaction and set up an ICE table.
NH4+(aq)
0.35
-x
0.35-x
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C
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Ka = 5.68x10-10 =
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+
↔
H2O(l)
/
/
/
NH3(aq)
0
+x
x
H3O+(aq)
0
+x
x
+
x2
≈ x2 .
(0.35-x) 0.35
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X2 = (0.35)( 5.68x10-10) = 1.99x10-10
x = 1.99 x10 −10 = 1.41x10-5
[H3O+] = 1.4x10-5
pH = -log(1.41x10-5) = 4.85
21. What is the Kb for methylamine, CH3NH2, if a solution prepared by dissolved 0.82 mol of
methylamine in 425 mL of H2O has a pH of 12.46? (Ans: 4.4 x 10-4)
If pH = 12.46. then pOH = 14.00-12.46 = 1.54 and [OH-]= 10-1.54 = 0.0288M
[CH3NH2] = 0.82mol / 0.425L = 1.93M
Since the pH is basic, (and we are asked for Kb), CH3NH2 must react with water as a base by
accepting a proton (hydrogen ion) and forming OH-.
CH3NH2(aq)
1.93
-x
1.93-x
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+
CH3NH3+ (aq)
0
+x
x
↔
H2O(l)
/
/
/
+
OH-(aq)
0
+x
x
From the initial pH we know that x=0.0288
x2
≈ (0.0288)2 . = 0.000832 = 0.0004376 ⇒ 4.4x10-4
(1.93-x) (1.93-0.0288)
1.901
Kb =
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22. A 0.100 M solution of formic acid, HCOOH, is 4.0% dissociated. Calculate Ka for formic acid.
(Ans: 1.7 x 10-4)
4.0% of 0.100 = 0.0040 = x
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HCOOH(aq)
0.100
-x
0.100-x
0.100-0.0040
+
H2O(l)
/
/
/
Ka = (0.0040)2 = 1.7x10-4
0.096
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4
↔
COOH-(aq)
0
+x
x
0.0040
+
H3O+(aq)
0
+x
x
0.0040