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Le Chatelier’s Principle. equilibrium. balance. forward reaction. reverse reaction. disturb balance. changes in experimental conditions. equilibrium shifts. counteract disturbance. concentration. (gas phase). pressure. temperature. Concentration. Fe 3+ (aq). FeSCN 2+. - PowerPoint PPT Presentation
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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, K = 10.38
K = [NO2]eq2
[N2O4]eq
Q =
K
[NO2]eq2
[N2O4]i
Q <
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]
K = [NO2]2
[N2O4] Q K>
decrease volume decrease nincrease volume increase n
Δn = 0 no effect of pressure
= (3.0)2
(0.87)
= 10.3 Q = (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)
Kp =
P(1.0 L)=
P = 119 atm
= (119)
/ 22= 242
(3/4.87)
KP 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
Calculationsreaction table ICE table
2HI (g)
Change
Equilibrium
H2 (g) + I2 (g)
Initial
at 453oC, at equilibrium,
calculate K
[HI] (M)[H2] (M) [I2] (M)
0.50 0.50 0.00
- x - x +2x
0.50 – x 0.50 – x 2x
K = [HI]2eq
[I2]eq[H2]eq
= (2x)2
(0.50–x)
= 0.50 – x x = 0.393
(0.786)2
(0.107)2
= = 54.3
[H2] = 0.107 M
(0.50–x)
Calculations
Change
Equilibrium
Initial
[H2] (M) [I2] (M) [HI] (M)
.623 .414 .224
-x +2x- x
.623 - x .414 - x .224 + 2x
2HI (g)H2 (g) + I2 (g)K = 54.3
Q = (.224)2
(.623) (.414)
= .195 < KK = 54.3 = (.224 + 2x)2
(.623 – x)(.414 – x)
50.3x2 - 57.2x+ 13.96 = 0ax2 bx c
x = -b ± b2 – 4ac2a
x = .782x = .355