chemical equilibrium chapter 15. equilibrium - state in which there are no observable changes with...
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Chemical EquilibriumChapter 15
Equilibrium - state in which there are no observable changes with time
Achieved when:
• rates of the forward and reverse reactions are equal and
• concentrations of the reactants and products remain constant
• dynamic equilibriumdynamic equilibrium Physical equilibrium
H2O (l)
Chemical equilibrium
N2O4 (g)
H2O (g)
2NO2 (g)
colorless red-brown
Fig 15.1 The N2O4(g) ⇌ 2 NO2 (g) equilibrium
colorless red-brown
Since both reactions are elementary, we use kinetics,
• Forward reaction: N2O4 (g) 2 NO2 (g)
• Rate Law: Rate = kf [N2O4]
• Reverse reaction: 2 NO2 (g) N2O4 (g)
• Rate law: Rate = kr [NO2]2
At equilibrium: Ratef = Rater or kf [N2O4] = kr [NO2]2
42
22
r
feq ON
NO
k
kK
N2O4 (g) 2NO2 (g)
Fig 15.2 (a) Achieving chemical equilibrium for:
Start with N2O4
Equilibrium occurs when concentrations no longer change
N2O4 (g) 2NO2 (g)
Fig 15.2 (b) Achieving chemical equilibrium for:
Equilibrium occurs when kf and kr are equal
Start with N2 and H2 or with NH3
Result: same proportions of all three substances at equilibrium
N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g)
e.g., the Haber process to form ammonia
aA + bB cC + dD
KC = [C]c[D]d
[A]a[B]bLaw of Mass Action
Must be caps!
Equilibrium constant
Lies to the right Lies to the left
Table 15.1 Initial concentrations in gas phase at 100 oC
N2O4 (g) 2NO2 (g)
Fig 15.1 Concentrationchanges approaching
equilibrium
constant
N2O4 (g) 2NO2 (g)
Kc = [NO2]2
[N2O4]Kp = NO2
P2
N2O4P
In most cases
Kc Kp
Kp = Kc(RT)Δn
Δn = moles of gaseous products – moles of gaseous reactants
The equilibrium constant may be expressed in terms ofconcentration or in terms of pressure
In the synthesis of ammonia from nitrogen and hydrogen,
Kc = 9.60 at 300 °C. Calculate Kp for this reaction at this temperature
N2 (g) + 3 H2 (g) 2 NH3 (g)
Sample Exercise 15.2 p 634
The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 74°C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g) COCl2 (g)
Kc = [COCl2]
[CO][Cl2]=
0.14(0.012) · (0.054)
= 220
Kp = Kc(RT)n n = 1 – 2 = -1
R = 0.0821 (L·atm)/(mol·K) T = 273 + 74 = 347 K
Kp = 220 ·[0.0821 · 347]-1 = 7.7