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irreversible reaction or complete reaction reversible reaction → A + B → AB N 2(g) + 3 H 2(g) 2 NH 3(g) at equilibrium ----the rate of forward reaction equal the rate of backward reaction
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Chemical equilibrium
ByBy Dr. Hisham Ezzat Abdellatef Dr. Hisham Ezzat Abdellatef
Professor of Pharmaceutical Analytical Professor of Pharmaceutical Analytical ChemistryChemistry
First Year 2011-2012
Classification of reaction
Homogeneousonly one phase
Heterogeneousmixture is not uniform
e.g: In the gas phase:H2(g) + I2 → 2 HI(g)
In the liquid phase: CH3COCl + CH3OH → CH3COOCH3 + HCIAcetyl Chloride methyl alcohol methylacetate
irreversible reaction or complete reaction
reversible reaction →
A + B → AB N2(g) + 3 H2(g) 2 NH3(g)
at equilibrium ----the rate of forward reaction equal the rate of backward reaction
Factors influencing equilibrium:
1. forward and reverse rates2. partial pressures3. concentrations4. temperature
Law of mass action
A2(g) + B2(g) 2 AB(g)
rate of the forward reaction = Kf. [A2] [B2]rate of the reverse reaction is = Kr [AB]2
At equilibrium ratef = rater
Kf [A2] [B2] = Kr [AB] 2
]][B[A
[AB]K22
2
K which is called the equilibrium constant
aA + bB eE + fF eE +fF aA + bB
ba
fe
c [B] [A][F][E] K
In general, for any reversible reaction
fe
ba\
[F][E][B] [A] K
cK1 \ K
Mechanisms of more than one step 1
2 NO2CI(g) 2 NO2(g) + CI2(g)
22
22
2
Cl][NO][Cl][NO Kc
mechanism consisting of two steps:
1- NO2CI NO2 + Cl
2. NO2CI NO2 + CI2
Cl][NO[Cl] ][NOK
2
21
[Cl] ][NO][Cl ][NOK
2
222
[Cl] ][NO
][Cl ][NOCl][NO[Cl] ][NO= K K = K
2
22
2
221c 2
2
22
2
Cl][NO][Cl][NO
• For reactions involving gasesP α C
using partial pressures instead of concentration.
N2(g) + 3 H2(g) 2 NH3(g) 3H2N2
2NH3
p .PPp K
322
23
c ][H ][N][NHK
3H2N2
2NH3
p .PPp K
(Kc ≠ Kp)
The Relationship Between Kp and KC:
aA + bB eE + fF
bB
aA
fF
eE
p P .PP .P K ba
fe
c [B] [A][F][E]K
Assuming ideal gas PV= nRT
concentration of a gas X RTP
Vn = [X] XX Px is its partial pressure
Px = [X] RT
bB
aA
fF
eE
P .PP .P Kp bbaa
ffee
(RT)[B] (RT)[A](RT)[F] (RT)[E]
Kp = b)(af)(e
ba
fe
(RT)[B] [A][F][E]
Kp = Kc. (RT)n(g)
Example 1:For the reaction N2O4(g) 2 NO2(g)
The concentrations of the substances present in an equilibrium mixture at 25°C are[N2O4] = 4.27 x 10-2 mol/L [NO2] = 1.41 x 10-2 mol/L
what is the value of Kc for this temperature.
Solution:
Kc = mol/l4.66x10mol/l)(4.27X10mol/l)(1.41x10
]O[N][NO 3
2
22
42
2
2
Example 2:At 500 K. 1.0 mol of ONCI(g) is introduced into a one - liter container. At equilibrium the ONCI(g) is 9.0% dissociated:
2 ONCI(g) 2 NO(g) + CI2(g)
Calculate the value of Kc for equilibrium at 500 K.
Solution: [ONCI(g) ] = 1 mol/L since 9.0% dissociated,Number of moles dissociated =
at equilibrium, [ONCI] = 1.0 mol/L - 0.09 mol/L = 0.91 mol/Lamounts of CI2 :
2 ONCI 2NO + CI2 2 x x
0.09 mol 0.045 mol
ONCI mol 0.09mol x1.0100 9 = [X]
Therefore, at Kc = 22
2
[ONCl]][Cl[NO]
= mol/l 4.4x10
mol/l) (0.9l)(0.045mol/mol/l) (0.09 4
2
2
2ONCI 2NO + CI2
at start 1.0 mol/L ----- ------ Change - 0.09 mol/L + 0.09 + 0.045at equilibrium 0.91 0.09 0.045
Example 3:
For the reaction2 SO3(g) 2 SO2(g) + O2(g) at 1100 K
Kc is 0.0271 mol/L. what is Kp at same temperature.
Solution:n = 3-2 =1 Kp = Kc (RT)+1 = 0.0271 mol/L x (0.0821 L. atm / K. mol) (1100 K) = 2.45 atm
Example 4:What is Kc for the reaction?
N2(g) + 3 H2(g) 2 NH3(g)
At 500°C if Kp is 1.5 x 10-5 / atm-2 at this temperature.
Solution:Kp = Kc (RT)n
Kp = Kc (RT)-2
Kc = Kp (RT)2 = x [0.0821.atm/K.mol x 773K]2 = (1.5 x I0-5/ atm2) (4.03 x 103 L2. atm2 / mol2) = 6.04 x 10-2 L2 / mol2
n = 2 - 4 = -2T = 273 + 500 = 773 K
Try
At 127oC, K = 2.6 x 10-5 mol2/L2 for the reaction2NH3(g) N2(g) + 2H2(g)
Calculate Kp at this temperature
Reaction quotient (Q).
Predicting the Direction of a Relation:• For the reaction
PCl2(g) PCl3(g) + Cl2(g) at 250oC
• Suppose that a mixture of 1.00 mol of PCI5(g)/ 0.05 mol of PCI3(g)/ and 0.03 mol of CI2(g) is placed in 1.0 L container. Is this an equilibrium system, or will a net reaction occur in one direction or the other?
Kc = l0.0415mol/][PCl
]][Cl[PCl
5
23
1- Q < Kc from left to right (the forward direction) to approach equilibrium.
2- Q = Kc The system is in equilibrium.
3- Q > Kc from right to left (the reversible direction) to approach equilibrium.
Q = mol/l 0.015 0.1
03.005.0][PCl
]][Cl[PCl
5
23 x
Q (0.015 mol/L) < kc (0.0415 mol/L). The system is not at equilibrium. The reaction will proceed from left to right.
Example 4:• For the reaction 2 SO2(g) + O2(g) 2 SO3(g)
at 827°C, kc is 36.9 L / mol. If 0.05 mol of SO2(g), 0.03 mol of O2(g), and 0.125 mol of SO3(g) are mixed in a 1.0 L container, in what direction will the reaction proceed?
• Solution:
Since Q (208 L/mol) > kc (36.9 L/mol), the reaction will proceed from right to left (SO3 will dissociate).
L/moL 208][0.03mol/lmol/l] [0.05
l][0.125mol/][O][SO
][SOQ 2
2
22
2
23
Heterogeneous Equilibria:
• The concentration of a pure solid or a pure liquid is constant and do not appear in the expression for the equilibrium constant.
• For exampleCaCO3(S) CaO(S) + CO2(g)
Kc = [CO2]
3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g)
Example 5:
Kc for the HI equilibrium at 425°C is 54.5:
H2(g) + I2(g) 2 HI(g)
A quantity of HI(g) is placed in a 1.01 container and allowed to come to equilibrium at 425°C What are the concentrations of H2(g) and I2(g) in equilibrium with 0.5 mol/L of HI(g)
Solution:at equilibrium [H2] = [I2] [H2] = [I2] = x [HI] = 0.5 mol/L
x = 0.068 mol/L The equilibrium concentration is: [HI] = 0.5 mol/ L [H2] = [I2] = 0.068 mol/ L
54.5][I ][H
[HI]K 422
2
c
54.5Xmol/l] [0.5
2
2
222
2 /Lmol 0.0045654.5(0.5)X
Example 6:
For the reaction
H2(g) + CO2(g) H2O(g) + CO(g)
kc is 0.771 at 750°C. If 0.01 mol of H2 and 0.01 mol of CO2 are mixed in 1 liter container at 750°C, what are the concentrations of all substances present at equilibrium?
• Solution:If x mol of H2 reacts with x mol of CO2 out of the
total amount supplied, x mol H2O and x mol CO will be produced. Hence
H2(g) + CO2(g) H2O(g) + CO(g)
At start
0.01 mol/L
0.01 mol/L
----
-----
Change
- x
- x
+ x
+ x
at equilibrium
0.01 -x
0.01 -x
X
X
0.771][CO ][H
[CO] O][HK22
2c
2
2
c X) - (0.01XK square root 878.0
X - 0.01X
X = 0.0878 – 0.878 XX= 0.00468 mol/l
At equilibrium, therefore
[H2] = [CO2] = 0.01 mol/L - 0.00468 mol/L = 0.0053 mol/L [H2O] = [CO] = 0.00468 mol/L
Example 7: •For the reaction
C(s) + CO2(g) 2 CO(g)
Kp is 167.5 atm at 1000°C. What is the partial pressure of CO(g) in an equilibrium system in which the partial pressure ofCO2(g) is 0.1atm?
•Solution:
•PCO2= 16.8•PCO= 4.10 atm
atm 167.5PPK
CO2
2CO
p
atm 167.50.1P2
CO
Example 8:
Kp for the equilibrium:
FeO(s) + CO(g) Fe(s) + CO2(g)
at 1000°C is 0.403. If CO(g) at a pressure of 1.0 atm, and excess FeO(s) are placed in a container at 1000°C, what are the pressures of CO(g) and CO2(g) when equilibrium is attained?
FeO(s) + CO(g) Fe(s) + CO2(g)
At start
Change
1.0 atm
- x
---
+ x
At equilibrium 1.0 - x atm x
403.0PPK
CO
CO2p
403.0X - atm 1.0
atm X
Solution: Let x equal the partial pressure of CO2 when equilibrium is attained
X = PCO2 = 0.287 atm1.0 – X = Pco = .713
• If a change is made to an equilibrium, the equilibrium shifts in the direction that consumes the change – Case 1: Changing the amounts of reactants /
products.– Case 2: Changing the volume by changing
pressure. – Case 3: Changing the temperature.
• If the concentration of substance is increased, the equilibrium will shift in a way that will decrease the concentration of the substance that was added.
e.g: H2(g) + I2(9) 2 HI(g)•
Increase H2 or I2 → shift to to formation of HI• Removal of H2 or I2 ← Reaction shift to
decomposition of HI.
• Increasing the pressure causes a shift in the direction that will decrease the number of moles of gas.
2 SO2(g) + O2(g) 2 SO3(g)3 moles 2 moles
• When the pressure on an equilibrium mixture is increased (or the volume of the system decreased), the position of equilibrium shifts to the right., and vice versa.
• For reactions in which n = 0, pressure changes have no effect on the position equilibrium.
e.g: N2(g) + O2(g) 2 NO(g)
For the reactionN2(g) + 3 H2(g) 2 NH3(g) H = - 92.4 KJ
Since H is -ve, the reaction to the right evolves heat
N2(g) + 3 H2(g) 2 NH3(g) + 92.4 KJ
The highest yields of NH3 will be obtained at the lowest temperatures and high pressures.
Also consider the reactionCO2(g) + H2(g) CO(g) + H2O(g) H = + 41.2KJ
Since H is + ve , we can write the equation
41.2 KJ + CO2(g) + H2(g) —— CO(g) + H2O(g)
Increasing the temperature always favors the endothermic change, and decreasing the temperature always favors the exothermic change.
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