Mg(OH) 2 (s)
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Transcript of Mg(OH) 2 (s)
Mg(OH)2 (s) Mg2+ + 2 OH-
K = [OH-]eq [Mg2+]eq
add H+ reacts with OH-
Q = [OH-]0 [Mg2+]0
Le Chatelier’s Principle
equilibrium balance
forward reaction reverse reaction
changes in experimental conditions disturb balance
equilibrium shifts counteract disturbance
concentration
pressure (gas phase)
temperature
Concentration
Fe3+ (aq) + SCN- (aq) FeSCN2+
add Fe(NO3)3 add reactant
add NaSCN add reactant
add C2O42- remove Fe2+
K = [FeSCN2+]
[Fe3+] [SCN-]
at equilibrium change
Q = [FeSCN2+]
[Fe3+] [SCN-]
Q K<
Q K>
ratef = kf [Fe3+] [SCN-]
Pressure
N2O4 (g) NO2 (g)2
add N2O4
at 25oC, Kc = 10.38
Kc = [NO2]eq2
[N2O4]eq
Qc =
Kc
[NO2]eq2
[N2O4]i
Qc <
increase P by adding reactant or product
Pressure
N2O4 (g) NO2 (g)2
decrease volume [N2O4] = mol N2O4
V
increase [N2O4]
[NO2] = mol NO2
V
increase [NO2]
Kc = [NO2]2
[N2O4] Qc Kc>
decrease volume decrease nincrease volume increase n
Δn = 0 no effect of pressure
= (3.0)2
(0.87)
= 10.3 Qc =(6.0)2 = 11.9
(1.74)
Pressure
N2O4 (g) NO2 (g)2 add inert gas
1.00 M Ar
2PNO
increase P = 3.0 M
2 4 PN O
= 0.87 M
= 3.0 mol/L
= 0.87 mol/L
at 298 K PV = nRT
P(1.0 L) = (3.87)(.08206)(298)
P = 95 atm
2PNO
[NO2][N2O4]
=(3/3.87)
= (.87/3.87)
K =
P(1.0 L)=
P = 119 atm
= (119)
/ 22= 242
(3/4.87)
K unchanged
(95) =73 atm
(95) =22atm
(73)2
(4.87)(.08206)(298)
=73atm
Temperature
treat heat reactant product
raising T adding heat as reactant
lowering T removing heat as product
endothermic exothermic
N2O4 (g) NO2 (g)2
ΔH > 0 ΔH < 0
ΔH > 0
ΔH < 0
ΔH = 58.0 kJ
changes K
heat + +heatheat