Characteristics of Chemical Equilibrium

Home , Chemical equilibrium

Chapter 14: Chemical Equilibrium

Equilibrium is Dynamic Equilibrium is Independent of Direction of Approach

Reactants convert to products N2(g) + 3 H2(g) 2 NH3(g) a A + b B c C + d D

Species do not stop forming OR being destroyed Rate of formation = rate of removal Concentrations are constant.

Equilibrium and Catalysts The Equilibrium Constant

For the 2-butene isomerization:

H3C CH3 H3C H C=C C=C H H H CH3 At equilibrium: rate forward = rate in reverse An elementary reaction, so:

kforward[cis] = kreverse[trans]

1 The Equilibrium Constant The Equilibrium Constant

At equilibrium the concentrations become constant. We had: kforward[cis] = kreverse[trans]

kforward [trans] or = kreverse [cis]

kforward [trans] Kc = = = 1.65 (at 500 K) kreverse [cis]

“c” for concentration based

The Equilibrium Constant The Equilibrium Constant For a general reaction:

a A + b B c C + d D [NO]2 N2(g) + O2(g) 2 NO(g) Kc = Products raised to [N2] [O2] stoichiometric powers… k [C]c [D]d forward …divided by reactants Kc = = a b kreverse [A] [B] raised to their stoichiometric [SO ] 1 2 powers 8 S8(s) + O2(g) SO2(g) Kc = [O2]

Equilibria Involving Pure Liquids and Solids Equilibria in Dilute Solutions [Solid] is constant throughout a reaction.

density g / L • pure solid concentration = mol. wt g / mol

• [S8] = dS8 / mol. wt S8 + - NH3(aq) + H2O(l) NH4 (aq) + OH (aq) • d and mol. wt. are constants, so [S8] is constant. • This constant factor is “absorbed” into K . c [NH +][OH-] 4 -5 Kc = = 1.8 x 10 (at 25°C) [NH3] Pure liquids are omitted for the same reason.

(Units are customarily omitted from Kc)

2 The Equilibrium Constant Kc for Related Reactions What is the equilibrium constant for: 2 [NH3] SiH (g) + 2 O (g) SiO (s) + 2 H O(g) ? N2(g) + 3 H2(g) 2 NH3(g) Kc = 3 4 2 2 2 [N2][H2]

2 Omit SiO2 (a solid) [H2O] Change the stoichiometry… … 2 Include H2O (a gas) [SiH4][O2] change Kc

3 [NH3] N2(g) + 2 H2(g) NH3(g) K = 3 Write down Kc for: c 2 [N2] [H2] 2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O(l) reaction divided by 2

[Na SO ] 2 4 This new Kc is the square root of the original Kc. Omit H2O (aq. solution) 2 [H2SO4][NaOH] • Multiply an equation by a factor… • … Raise Kc to the power of that factor.

Kc for Related Reactions Kc for a Reaction that Combines Reactions If a reaction can be written as a series of steps: N2(g) + 3 H2(g) 2 NH3(g) [NO]2 N (g) + O (g) 2NO(g) K = 2 2 2 c [NH ] [N2][O2] K = 3 c [N ][H ]3 2 2 2 [NO2] 2 NO(g) + O2(g) 2 NO2(g) Kc = 2 [NO] [O2] Reverse a reaction… … Invert Kc: 2 [NO2] 2 NH3(g) N2(g) + 3 H2(g) N2(g) + 2 O2(g) 2 NO2(g) Kc = 2 [N2][O2] [N ][H ]3 K = 2 2 The overall Kc is the product of the steps: c [NH ]2 3 2 2 2 [NO] [NO ] [NO2] Kc(step1) x Kc(step2) = 2 2 = 2 [N2][O2] [NO] [O2] [N2][O2]

Equilibrium Constants in Terms of Pressure Equilibrium Constants in Terms of Pressure In a constant-V reaction, partial pressures change as concentrations change.

At constant T, P is proportional to molar conc.: c d “p” for pressure PC PD Kp = a b Ideal gas: PV = nRT based PA PB

and for gas A: PAV = nART In general, Kp Kc. Your text shows: n ngas P = A RT = [A]RT Kp = Kc(RT) A V ngas = (c + d) – (a + b) P is easily measured, so another K is used… =(moles of gaseous products) (moles of gaseous reactants)

3 Equilibrium Constants in Terms of Pressure Determining Equilibrium Constants

Calculate Kp from Kc for the reaction: If all the equilibrium concentrations are known it’s 5 easy to calculate K for a reaction…. N2(g) + 3 H2(g) 2 NH3(g) Kc = 5.8 x 10 at 25°C. c For: ngas Kp = Kc(RT) R must have L/mol units. a A + b B c C + d D T = 25 + 273 = 298 K ([ ] has mol/L units). c d kforward [C] [D] n = 2 – (3 + 1) = 2 K = = gas c k a b -2 reverse [A] [B] 5 -1 -1 Kp = 5.8 x 10 0.0821 L atm mol K (298 K) = 9.7 x 102 mol2 L-2 atm-2

mol and L units are assumed to cancel (units omitted from Kc). At equilibrium [A] = 2.0 M & [B] = 4.0 M. What is Kc?

K = 9.7 x 102 atm-2 2 2 p Kc = [B]/[A] = 4.0/(2.0) = 1.0

Determining Equilibrium Constants Determining Equilibrium Constants

… In other cases stoichiometry is used to find some concentrations. H2(g) + I2(g) 2 HI(g)

[ ]initial 0.01000 0.00800 0

Concentrations change – use stoichiometry:

[ ]change -x -x 2x

Let HI made Stoichiometry: 1H2 2HI If 1I2 2HI HI made = 2x, H lost = x = 2x 2 I2 lost = x

Determining Equilibrium Constants Determining Equilibrium Constants

Add [ ]initial and [ ]change to get [ ]equilib Calculate [ ]equilib and then Kc

H2(g) + I2(g) 2 HI(g) [H2]equilib = 0.01000 - (0.00240) = 0.00760 M

[ ]equilib 0.01000 - x 0.00800 - x 2x [I2]equilib = 0.00800 - (0.00240) = 0.00560 M

[HI]equilib = 2(0.00240) = 0.00480 M Since [I2] = 0.00560 M at equilibrium [HI]2 (0.00480)2 0.00560 = 0.00800 - x Kc = = [H2][I2] (0.00760)(0.00560) -0.00240 M = - x

x = 0.00240 M Kc = 0.541

4 Meaning of the Equilibrium Constant Predicting the Direction of a Reaction When:

For the reaction: aA + bB cC + dD [C]c [D]d Q = [A]a [B]b

Predicting the Direction of a Reaction Predicting the Direction of a Reaction

Reactants Products

[Products] [Products]equilib Q = Kc= [Reactants] [Reactants]equilib

Kc is a constant (only changes if T changes). So:

Predicting the Direction of a Reaction Calculating Equilibrium Concentrations

2 SO2(g) + O2(g) 2 SO3(g) Kc = 245 at 1000 K Predict the direction of the reaction if SO (0.085 M), 2 If Kc is known, [ ]equilib can be calculated. O2 (0.100 M) and SO3 (0.250 M) are mixed in a reactor at 1000 K.

2 2 [ SO3] (0.250) Q = 2 = 2 = 86.5 [SO2] [O2] (0.085) (0.100)

5 Calculating Equilibrium Concentrations Calculating Equilibrium Concentrations [HI]2 (0.0500 – 2x)2 K = = H2(g) + I2(g) 2 HI(g) c 2 [H2][I2] x [ ]initial 0 0 0.0500 Since K = 76 Change on reaching equilibrium: Amount lost c 2 2 [ ]change +x +x 1I2 2HI -2x 76 x = (0.0500 – 2x)

1H2 2HI Take the square root of both sides:

[ ]equilib +x +x 0.0500 – 2x 8.718 x = ±(0.0500 – 2x)

[HI]2 (0.0500 – 2x)2 Kc = = [H2][I2] (x)(x) Ignore the negative root (x cannot be negative).

Calculating Equilibrium Concentrations Calculating Equilibrium Concentrations

So:

[H2]equilib = x = 0.00467 M Consider:

[I2]equilib = x = 0.00467 M

[HI]equilib = 0.0500 – 2x = 0.0407 M A(g) B(g) + C(g) Kc = 2.500

Matches Kc given It’s a good idea to check your work: in the problem A is added to an empty container. [A]initial = 0.1000 M. What are [A], [B] and [C] at equilibrium? [HI]2 (0.0407 M)2 Kc = = = 76. [H2][I2] (0.00467 M)(0.00467 M)

Calculating Equilibrium Concentrations Calculating Equilibrium Concentrations A B + C ax2 + bx + c = 0

[ ]Initial 0.100 0 0 2 [ ] -x x x -b ± b - 4ac change has two roots (solutions): x = 2a [ ]equilib (0.100 – x) x x Here:

[B][C] (x)(x) K = = c [A] (0.1000 – x) x2 2.500 = x = (0.1000 – x) A quadratic x2 + 2.500 x – 0.2500 = 0 equation x = +0.0963 M or -2.596 M

6 Calculating Equilibrium Concentrations Le Chatelier’s Principle

Ignore the negative root (negative concentration!) Then x = 0.0963 M, and: “If a system is at equilibrium and the conditions are changed so that it is no longer at equilibrium, the [ A ] = 0.1000 - x = 0.0037 M system will react to reach a new equilibrium in a way [ B ] = x = 0.0963 M that partially counteracts the change” [ C ] = x = 0.0963 M • A system at equilibrium resists change. Most of the A is converted to products. • If “pushed”, it “pushes back”

Check: (0.0963)(0.0963)/0.0037 = 2.5

Changing Concentrations Changing Concentrations

a A + b B c C + d D H3C CH3 H3C H C=C C=C H H H CH3 Add A (or B) and the system adjusts to remove it. more product is created. the reaction shifts forward.

Remove A (or B) the system adjusts to add more. more reactant is created. the reaction shifts backward.

Changing V or P in Gaseous Equilibria Changing V or P in Gaseous Equilibria Consistent with Le Chatelier

N2O4(g) 2 NO2(g) Consider: N2O4(g) 2 NO2(g) V doubled = lower concentration. Minimize change by: • Making more molecules (increase concentration). • Shifting toward products - convert 1 molecule into 2.

[ NO ]2 ([ NO ])2 [ NO ]2 [ NO ]2 P doubled. Minimize the change by: K =Q= 2 Q= 2 = 2 = 2 c [ N O ] [ N O ] [ N O ] [ N O ] • Removing molecules (decrease P). 2 4 = K 2 4 2 4 2 4 c • Shift toward reactants - convert 2 molecules into 1. At equilibrium After V change, Q < Kc equilibrium shifts to products © 2008 Brooks/Cole 41 © 2008 Brooks/Cole 42

7 Changing V or P in Gaseous Equilibria Changing V or P in Gaseous Equilibria

P and V changes have no effect if ngas = 0:

H2(g) + I2(g) 2 HI(g)

If V is doubled, the system cannot adjust – 2 gas molecules each side.

[HI]2 ([HI])2 [HI]2 [HI]2 Kc=Q= Q = [H ] [I ] = = = Kc [H2][I2] 2 2 [H2][I2] [H2][I2] nA PA= RT = [A]RT After V change, Q = K still at equilibrium! V At equilibrium c

Changing Temperature Changing Temperature

N2(g) + O2(g) 2 NO(g) H° = 180.5 kJ

T increased? Minimize by removing heat. N2(g) + O2(g) 2 NO(g) H° = 180.5 kJ Forward reaction is endothermic Equilibrium shifts forward, removing heat. T (°C) K c K increases as T increases. 298 4.5 x 10-31 c T reduced? Add heat. 900 6.3 x 10-10 Reverse reaction is exothermic Explained by LeChatelier… 2300 8.2 x 10-3 Equilibrium shifts in reverse, releasing heat

Changing Temperature The Haber-Bosch Process

Production of NH3 for fertilizer from atmospheric N2

N2(g) + 3 H2(g) 2 NH3(g) Process developed by Fritz Haber (chemist) and Carl Bosch (engineer). Perfected in 1914.

Germany could not import guano from S. America and needed to make explosives…

8 The Haber-Bosch Process The Haber-Bosch Process

N2(g) + 3 H2(g) 2 NH3(g) H° = -92.2 kJ

It is: • Exothermic – product favored at low-T, but… • … very slow at room-T, and must be heated. • Carried out at high-P (200 atm), to favor product. • Uses a catalyst to speed the reaction.

Run at 450°C; NH3 is continually liquefied and removed.