Le Chatelier’s Principle
8
Le Chatelier’s Principle equilibrium balance forward reaction reverse reaction ges in experimental conditions disturb balance equilibrium shifts counteract disturban concentration pressure (gas phase) temperature
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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
Transcript of Le Chatelier’s Principle
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
