chemical equilibrium chapter 15. equilibrium - state in which there are no observable changes with...

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