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Title: Practice problems on kinetics and equilibrium
Description: Practice problems on kinetics and equilibrium
Description: Practice problems on kinetics and equilibrium
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DATA Tables provided for Quiz #6, Final Exam
Common Polyatomic Ions
Chemistry 107 SEC 507
Fall 2020
Solubility Rules
usually
soluble
usually
insoluble
Chemistry 107 SEC 507
Ion
Exceptions
Li+, Na+, K+, Rb+, Cs+
none
NH4+
none
Cl-, Br-, I-
Ag+, Hg22+, Pb2+
F-
Mg2+, Ca2+, Sr2+, Ba2+, Pb2+
NO3-
none
ClO4-
none
C2H3O2-
none
SO42-
Ag+, Hg22+, Pb2+, Ca2+, Sr2+, Ba2+
OH-
Gp
...
IA, NH4+, Ca2+, Sr2+, Ba2+
CO32-
Gp
...
IA, NH4+
PO43-
Gp
...
Rate = (9-2)/600
0
...
011 M-1 min–1
Rate = k[NH4NCO]2, k = –0
...
011 min–1
Rate = k[NH4NCO], k = 91 min–1
Chemistry 107 SEC 507
Rate= deltaY/time
Fall 2020
Half Life
t=[A]0/2k
t=0
...
What happens when NH3 is removed from a system that is at equilibrium?
a)
b)
c)
d)
e)
The equilibrium constant stays the same, the reaction shifts to the right to reestablish equilibrium
...
The equilibrium constant stays the same, the reaction does not shift to either the right or left
...
The equilibrium constant decreases
...
What happens when the temperature is increased after the system has
reached equilibrium?
a)
b)
c)
d)
e)
The equilibrium constant stays the same, the reaction shifts to the right to reestablish equilibrium
...
The equilibrium constant stays the same, the reaction does not shift to either the right or left
...
The equilibrium constant decreases
...
What happens when the volume of the container is increased once the system has
reached equilibrium?
a)
b)
c)
d)
e)
The equilibrium constant stays the same, the reaction shifts to the right to reestablish equilibrium
...
The equilibrium constant stays the same, the reaction does not shift to either the right or left
...
The equilibrium constant decreases
...
8
N2(g) + 3H2(g) 2NH3(g)
Equilibrium and Gibbs Free Energy
Chapter 12
...
0 M), hydrogen (3
...
50
M)
...
3 kJ * 1000 = -33300 J
3H2 + N2
2NH3
Q= [NH3]/([H2]^3*[N2])=[0
...
0 M]^3*[1
...
056
T (K) = 25 C + 273
...
15 K
R= 8
...
314 J/mol*K*298
...
056)=
= -33300 J -7144
...
4 kJ
SI = 4
...
314 J/(mol*K) * (298
...
25925*10^-3)= - 44
...
8
∆G = ∆Go + RTlnQ
Q defines our
reaction anytime
∆Go = -RTlnK
The equilibrium point
K = e- ∆Go/RT
Calculate the equilibrium constant at 25oC for the following
reaction:
N2(g) + 3H2(g) 2NH3(g)
∆Go = -33
...
314 J/mol*K; T = 25 C + 273
...
15 K
K= e- 33300 J/8
...
15 K = e- 13
...
83 * 10^5 K = [P]/[R] = [NH3]/[H2]^3*[N2]
K = e- ∆Go/RT = e^-(-33300J/(298
...
314 J/mol*K))
= e^-(-13
...
319=6*10^5
Chapter 12: Part 5
Acid/Base and Solubility Equilibria
Acid Base Equilibria
Chapter 12
...
Stronger acids produce a higher concentration of H3O+
...
HCl, HBr, HI, H2SO4, HNO3, HClO4
HA(aq) + H2O(l)
→
H3O+(aq) + A-(aq)
Weak Acids – Do not dissociate to a great degree
...
7
Acid Base Equilibria
Chapter 12
...
Strong bases produce more OH-
...
LiOH, NaOH, KOH,
BOH(aq)
→
B+(aq) + OH-(aq)
Weak Bases – Produce very little OH-
...
7
Acid Base Equilibria
Chapter 12
...
001 M
(equilibrium)
...
999
[0
...
999]
[H3O+] M
0
+x
0
...
001
HCl pka = -8
...
6
The solubility product, Ksp, is an indicator of the solubility of a salt
...
1 x 10-10
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)
[Ba+2] [SO4−]
Ksp=
Ksp = 5
...
82 M solution
of benzoic acid, HC7H5O2, (Ka = 6
...
82
-x
0
...
39e-3 ± 5%
4
...
20e-2 ± 5%
7
...
55e3 ± 5%
[H3O+] M
0
+x
x
6
...
3*10^-3
[A-] M
0
+x
x
[x] [x]
[0
Title: Practice problems on kinetics and equilibrium
Description: Practice problems on kinetics and equilibrium
Description: Practice problems on kinetics and equilibrium