characteristics of chemical equilibrium...5 © 2008 brooks/cole 25 when: meaning of the equilibrium...

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© 2008 Brooks/Cole 1

Chapter 14: Chemical Equilibrium


Characteristics of Chemical Equilibrium


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

Reactants convert to products a A + b B c C + d D

Equilibrium is Dynamic


N2(g) + 3 H2(g) 2 NH3(g)

Equilibrium is Independent of Direction of Approach


Equilibrium and Catalysts


For the 2-butene isomerization:

At equilibrium: rate forward = rate in reverse

An elementary reaction, so: kforward[cis] = kreverse[trans]

C=C CH3

H H

H3C C=C

CH3

H

H

H3C

The Equilibrium Constant

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At equilibrium the concentrations become constant.



or = [trans] [cis]

kforward kreverse

We had: kforward[cis] = kreverse[trans]

“c” for concentration based

Kc = = = 1.65 (at 500 K) kforward kreverse

[trans] [cis]



The Equilibrium Constant For a general reaction:

a A + b B c C + d D

Kc = = [C]c [D]d [A]a [B]b

kforward kreverse

Products raised to stoichiometric

powers…

…divided by reactants raised to their stoichiometric

powers



N2(g) + O2(g) 2 NO(g) Kc = [NO]2

[N2] [O2]

S8(s) + O2(g) SO2(g) 1 8 Kc = [SO2] [O2]


[Solid] is constant throughout a reaction.

•  pure solid concentration =

•  [S8] = dS8 / mol. wt S8 •  d and mol. wt. are constants, so [S8] is constant. •  This constant factor is “absorbed” into Kc.

Equilibria Involving Pure Liquids and Solids

Pure liquids are omitted for the same reason.

density mol. wt

g / L g / mol


Equilibria in Dilute Solutions

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

Kc = [NH4+][OH-]

[NH3] = 1.8 x 10-5 (at 25°C)

(Units are customarily omitted from Kc)

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The Equilibrium Constant What is the equilibrium constant for: SiH4(g) + 2 O2(g) SiO2(s) + 2 H2O(g) ?

Write down Kc for: 2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O(l)

[Na2SO4] [H2SO4][NaOH]2

Omit SiO2 (a solid) Include H2O (a gas)

Omit H2O (aq. solution)

[H2O]2 [SiH4][O2]2


Kc for Related Reactions

Change the stoichiometry… …change Kc

This new Kc is the square root of the original Kc. •  Multiply an equation by a factor… •  … Raise Kc to the power of that factor.

Kc = [NH3]2

[N2][H2]3 N2(g) + 3 H2(g) 2 NH3(g)

3 2 ½ N2(g) + H2(g) NH3(g) Kc =

[NH3] [N2]½[H2]

3 2

reaction divided by 2


Kc for Related Reactions

N2(g) + 3 H2(g) 2 NH3(g)

Kc = [NH3]2

[N2][H2]3

2 NH3(g) N2(g) + 3 H2(g)

Kc = [N2][H2]3 [NH3]2

Reverse a reaction… … Invert Kc:


Kc(step1) x Kc(step2) = [NO]2

[N2][O2] [NO2]2

[NO]2[O2]

The overall Kc is the product of the steps:

If a reaction can be written as a series of steps:

Kc = [NO2]2

[NO]2[O2] 2 NO(g) + O2(g) 2 NO2(g)

Kc = [NO2]2

[N2][O2]2 N2(g) + 2 O2(g) 2 NO2(g)

Kc = [NO]2

[N2][O2] N2(g) + O2(g) 2NO(g)

= [NO2]2

[N2][O2]2

Kc for a Reaction that Combines Reactions


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.:

Ideal gas: PV = nRT

and for gas A: PAV = nART

PA= RT nA V

= [A]RT

P is easily measured, so another K is used…


Kp = PCc PDd PAa PBb

“p” for pressure based

In general, Kp ≠ Kc. Your text shows: Kp = Kc(RT)Δngas

Δngas = (c + d) – (a + b) =(moles of gaseous products) − (moles of gaseous reactants)


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Calculate Kp from Kc for the reaction: N2(g) + 3 H2(g) 2 NH3(g) Kc = 5.8 x 105 at 25°C.

Kp = Kc(RT)Δngas T = 25 + 273 = 298 K Δngas = 2 – (3 + 1) = −2

R must have L/mol units. ([ ] has mol/L units).

Kp = 5.8 x 105 0.0821 L atm mol-1 K-1(298 K) -2

= 9.7 x 102 mol2 L-2 atm-2 mol and L units are assumed to cancel (units omitted from Kc).

Kp = 9.7 x 102 atm-2



If all the equilibrium concentrations are known it’s easy to calculate Kc for a reaction….

Kc = = [C]c [D]d [A]a [B]b

kforward kreverse

For: a A + b B c C + d D

Determining Equilibrium Constants

Kc = [B]/[A]2 = 4.0/(2.0)2 = 1.0

At equilibrium [A] = 2.0 M & [B] = 4.0 M. What is Kc?


… 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

Let HI made = 2x

2x

Stoichiometry: 1H2 ≡ 2HI If HI made = 2x, H2 lost = x

-x

1I2 ≡ 2HI I2 lost = x

-x



H2(g) + I2(g) 2 HI(g) [ ]equilib 0.01000 - x 0.00800 - x 2x

Add [ ]initial and [ ]change to get [ ]equilib

Since [I2] = 0.00560 M at equilibrium

0.00560 = 0.00800 - x -0.00240 M = - x x = 0.00240 M



Calculate [ ]equilib and then Kc

Kc = = [HI]2

[H2][I2] (0.00480)2

(0.00760)(0.00560)

Kc = 0.541

[H2]equilib = 0.01000 - (0.00240) = 0.00760 M [I2]equilib = 0.00800 - (0.00240) = 0.00560 M [HI]equilib = 2(0.00240) = 0.00480 M


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When:

Meaning of the Equilibrium Constant


Predicting the Direction of a Reaction

Q = [C]c [D]d [A]a [B]b

For the reaction: aA + bB cC + dD


Reactants Products

Q = [Products]

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

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





2 SO2(g) + O2(g) 2 SO3(g) Kc = 245 at 1000 K Predict the direction of the reaction if SO2 (0.085 M), O2 (0.100 M) and SO3 (0.250 M) are mixed in a reactor at 1000 K.

Q = [ SO3]2

[SO2]2[O2] (0.250)2

(0.085)2(0.100) = = 86.5



If Kc is known, [ ]equilib can be calculated.

Calculating Equilibrium Concentrations

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H2(g) + I2(g) 2 HI(g) [ ]initial 0 0 0.0500

Change on reaching equilibrium: [ ]change +x +x -2x

Amount lost

1H2 ≡ 2HI

1I2 ≡ 2HI

[ ]equilib +x +x 0.0500 – 2x

Kc = = [HI]2

[H2][I2] (0.0500 – 2x)2

(x)(x)



Kc = = [HI]2

[H2][I2] (0.0500 – 2x)2

x2

Since Kc = 76

76 x2 = (0.0500 – 2x)2 Take the square root of both sides:

8.718 x = ±(0.0500 – 2x)

Ignore the negative root (x cannot be negative).



So:

[H2]equilib = x = 0.00467 M [I2]equilib = x = 0.00467 M [HI]equilib = 0.0500 – 2x = 0.0407 M

It’s a good idea to check your work:

Kc = = = 76. [HI]2

[H2][I2] (0.0407 M)2

(0.00467 M)(0.00467 M)

Matches Kc given in the problem



Consider:

A(g) B(g) + C(g) Kc = 2.500

A is added to an empty container. [A]initial = 0.1000 M. What are [A], [B] and [C] at equilibrium?



A B + C [ ]Initial 0.100 0 0 [ ]change -x x x [ ]equilib (0.100 – x) x x

Kc = = [B][C]

[A] (x)(x)

(0.1000 – x)

x2 (0.1000 – x) 2.500 =

x2 + 2.500 x – 0.2500 = 0 A quadratic

equation



ax2 + bx + c = 0

-b ± b2 - 4ac 2a has two roots (solutions): x =

x = +0.0963 M or -2.596 M

Here:

x =


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Ignore the negative root (negative concentration!) Then x = 0.0963 M, and:

[ A ] = 0.1000 - x = 0.0037 M [ B ] = x = 0.0963 M [ C ] = x = 0.0963 M

Most of the A is converted to products.

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



Le Chatelier’s Principle

“If a system is at equilibrium and the conditions are changed so that it is no longer at equilibrium, the system will react to reach a new equilibrium in a way that partially counteracts the change”

•  A system at equilibrium resists change. •  If “pushed”, it “pushes back”


Changing Concentrations

a A + b B c C + d D

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 Concentrations

C=C CH3

H H

H3C C=C

CH3

H

H

H3C


Changing V or P in Gaseous Equilibria

Kc=Q=

At equilibrium

[ NO2 ]2 [ N2O4]

Consider: N2O4(g) 2 NO2(g)

Q= = = = ½Kc

After V change, Q < Kc equilibrium shifts to products

(½[ NO2 ])2 ½[ N2O4]

¼[ NO2 ]2 ½[ N2O4]

½[ NO2 ]2 [ N2O4]


Consistent with Le Chatelier

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.

P doubled. Minimize the change by: • Removing molecules (decrease P). •  Shift toward reactants - convert 2 molecules into 1.


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

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

Kc=Q=

At equilibrium

[HI]2 [H2][I2]

Q = = = = Kc

After V change, Q = Kc still at equilibrium!

(½[HI])2 ½[H2]½[I2]

¼[HI]2 ¼[H2][I2]

[HI]2 [H2][I2]

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



PA= RT nA V

= [A]RT



N2(g) + O2(g) 2 NO(g) ΔH° = 180.5 kJ

Changing Temperature

T (°C) Kc 298 4.5 x 10-31 900 6.3 x 10-10 2300 8.2 x 10-3

Kc increases as T increases.

Explained by LeChatelier…


T increased? Minimize by removing heat.  Forward reaction is endothermic  Equilibrium shifts forward, removing heat.

T reduced? Add heat.  Reverse reaction is exothermic  Equilibrium shifts in reverse, releasing heat

N2(g) + O2(g) 2 NO(g) ΔH° = 180.5 kJ





The Haber-Bosch Process

Production of NH3 for fertilizer from atmospheric N2

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…

N2(g) + 3 H2(g) 2 NH3(g)

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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.

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

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



characteristics of chemical equilibrium...5 © 2008 brooks/cole 25 when: meaning of the equilibrium...

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