chemical equilibrium reactions don’t just stop, they find balance
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Chemical Equilibrium Reactions Don’t Just Stop, They find Balance. Equilibrium is Attained When the Rates of the Forward and Reverse Reactions are the Same. Forward Rate = k f [A]. Reverse Rate = k r [B]. k f [A] = k r [B]. Equilibrium : The Haber Process and Nitrogen Fixation. - PowerPoint PPT PresentationTRANSCRIPT
Chemical EquilibriumChemical Equilibrium Reactions Don’t Just Stop, They find Reactions Don’t Just Stop, They find
BalanceBalance
Equilibrium is Attained When the Rates of the Equilibrium is Attained When the Rates of the Forward and Reverse Reactions are the Same.Forward and Reverse Reactions are the Same.
Forward Rate = kf [A]Reverse Rate = kr [B]
kf [A] = kr [B]
Equilibrium : The Haber Process andEquilibrium : The Haber Process andNitrogen FixationNitrogen Fixation
N2(g) + 3H2(g) 2NH3(g)
Chemistry at work, p. 521
Equilibrium : The Haber Process andEquilibrium : The Haber Process andNitrogen FixationNitrogen Fixation
Note that equilibrium can be reached from either the forwardor reverse direction
jA + kB pR + q S The LAW OF MASS ACTIONLAW OF MASS ACTION allows us to express the relative concentrations of reactants and products at equilibrium in terms of a quantity called the equilibrium constant, ‘K’ such that:
[R]p [S]q
[A]j [B]kK =
The Haber Process and the Law of Mass ActionThe Haber Process and the Law of Mass Action
N2(g) + 3H2(g) 2NH3(g)
[NH3]2
[N2][H2]3K =
The equilibrium constant expression depends only on the stoichiometry of the reaction, not on its mechanism.
Suppose we discover that the equilibrium concen-trations of NO2 and N2O4 are 0.0172 M and 0.00140 M, respectively
[NO2]2
[N2O4]Kc =
[0.0172]2
[0.00140]= = 0.221
N2O4 2 NO2
Heterogeneous EquilibriaHeterogeneous Equilibriaa reaction which may possess reactants orproducts which are in different phases.
CaCO3(s) CaO(s) + CO2(g)
The density of a pure liquid or solid is a constant at any given temperature and changes very little with temperature. Thus the effective concentration of a pure liquid or solid is constant regardless of how much pure liquid or solid is present.
K = [CaO(s)][CO2]
[CaCO3(s)]
Given: CaCO3(s) CaO(s) + CO2(g)
so K = [CO2]
Even though they do not appear in the equilibrium constant expression,pure solids and liquids must be present for equilibrium to be established
The Magnitude of Equilibrium The Magnitude of Equilibrium ConstantsConstants
What does it mean when the constant equals
K = 1 x 1040
orK = 1 x 10-40
In one of their experiments, Haber and co-workers introduced a mixture of hydrogen and nitrogen into a reaction vessel and allowed the system to
attain chemical equilibrium at 472 °C. The equilibrium mixture of gases was analyzed and
found to contain 0.01207 M H2, 0.0402 M N2, and 0.00272 M NH3. From these data, calculate the
equilibrium constant for:
N2(g) + 3H2(g) 2NH3(g) (.00272)2
(.0402)(.01207)3= 104.7
Converting from Kc to Kp
Solutions are Understood in terms of Molarities Gas Pressure is Understood
in Terms of AtmospheresKp = Kc(RT)n, n = (mol. of prod – mol. of react)
Using the value of Kc = 0.105 for the reaction: N2(g) + 3H2(g) 2NH3at 472 °C Convert to Kp
Kp = 0.105 (0.0821 L-atm/mol-K)(745 K)-2
Kp= (0.0821 L-atm/mol-K)(745 K)2
0.105
Kp= 2.81 x 10-5 atm
Le Châteliers principle : If a system at Equilibrium is disturbed
by a change in temperature, pressure or the concentration of a component, the system will shift its equilibrium position so as to
counteract the effect of the disturbance.
Predicting the DirectionPredicting the Direction of of EquilibriumEquilibrium
Effects of Pressure and Volume ChangesIf a system is at equilibrium and the total
pressure is increased by the application of an external pressure by a change in volume, the system will respond by a
shift in equilibrium in the direction that reduces the pressure by shifting to the
side with less moles.
2NO2(g) N2O2(g)
What would happen if 1 atm of argon gas were added to
the following reaction already at equilibrium.
N2 + 3H2 2 NH3
Effects of Temperature ChangesWhen heat is added to a system, the equilibrium shifts in the
direction that absorbs heat. In an endothermic reaction reactants are converted to products, and K increases. In an exothermic reaction, the opposite occurs.
Co(HCo(H22O)O)662+2+
(aq)+ 4Cl-(aq)CoClCoCl44
2-2-(aq)+6H2O(l) H > 0
heating coolingRoom temperature
Predicting the Direction of Equilibrium
Given, N2(g) + 3H2(g) 2NH3(g)
[NH3]2
[N2][H2]3K = and Kc = 0.105
If the equilibrium concentrations were to start at:
[2.00]2
[1.00][2,00]3Q = = 0.500
What must happen in order for the 0.500 value, lets now call it the reaction quotient ‘Q’, to return to Kc= 0.105?
[NH3]2
[N2][H2]3
Q = = 0.500 and Kc = 0.105
Equilibrium will re-emerge if the concentration of NH3 decreases and the concentrations of N2 and H2 increase. In a closed system, this would require that the reaction favor the formation of reactants
N2(g) + 3H2(g) 2NH3(g)
Predicting the Direction of Equilibrium
If Q > K, the reaction will shift to the reactantsIf Q > K, the reaction will shift to the reactants
If Q < K, the reaction will shift to the productsIf Q = K, the reaction is at equilibriumIf Q = K, the reaction is at equilibrium
Kc = [HI]2
[H2] [ I2]
ICEBOX A mixture of 5.00 x 10-3 mol of H2 and 1.00 x 10-2 mol of I2 is placed in a 5.00 L container
at 448°C and allowed to come to equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc
at 448 °C for the reaction: H2(g) + I2(g) 2HI
H2(g) + I2(g) 2HIInitial
Change
Equilib.
1.00 x 10-3 M 2.00 x 10-3 M O M
1.87 x 10-3 M
+-- 1.87 x 10-3 M.935 x 10-3.935 x 10-3
1.065 x 10-3.065 x10-3
= (1.87 x 10-3 )2
(1.065 x 10-3)(.065 x10-3)
Initial
Change
Equilibrium
A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol I2 at 448°C. The value of the equilibrium constant for the following reaction is 50.5. What are the equilibrium concentrations of H2, I2, and HI? H2(g) + I2(g) 2HI
H2(g) + I2(g) 2HI1.000 M 2.000 M 0 M
Initial
Change
Equilibrium
H2(g) + I2(g) 2HI
1.000 M 2.000 M 0 M
- x M - x M + 2x M
(1.000 - x) (2.000 - x) 2x
50.5 = [HI]2
[H2][I]=
(2x)2
(1.000 - x)(2.000 - x)
4x2 = 50.5(x2 - 3.00 x + 2.00)
46.5x2 - 151.5x +101.0 = 0
x =-(-151.1) + (-151.4)2 - 4(46.5)(101.0)
2(46.5)x= 2.323 or 0.935
[H2] = 0.065 M[I2 ] = 1.065 M[HI] = 1.870 M