Chemical Equilibrium - Lecture notes 6 PDF

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Dr. Dunlavy's general chemistry II lecture notes on chemical equilibrium...

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Chemical Equilibrium Friday, September 21, 2018

11:52 AM

Characteristics of an Equilibrium System: •

Forward and reverse reactions occurring at the same rate. ○ Dynamic- continually moving/reacting/changing Reaction isn't "over", hasn't "stopped" ○ Constant amounts of everything ○ Some of everything present Mathematically, ratios of amounts give rise to a constant

•

Can be changed by ○ Adding/removing reactants or products ○ Changing temperature or pressure/volume If changed, there is a different amount of everything but still some of everything

H2 (g) + I2 (g)

2HI

H2 I2

HI

C

i

Dynamic Equilibrium

Concentration

Time

Reactant-favored (K1) Reactant Products

Products

Reactants

Equilibrium

Equilibrium

Law of Mass Action (Equilibrium Expressions) [C]c[D]d K= a b [A] [B]

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

[]

Molarity

Kc

Molarity in gases

Kp

Ratio of pressures in atmospheres

K

Aqueous solutions in molarity

practice

c P d P C D Kp = PAa PBb

NH4+(aq) + OH-(aq)

NH3(aq) + H2O (l)

+][OH-] [NH 4 K= [NH3] N2(g) + 3H2(g)

Kc= [NH3]

2NH3(g)

2

[N2][H2]3

(PNH3)2 Kp = (Pn2)(PH2)3

LHS- Left hand side RHS- right hand side If K1 (104 or so) • a lot more RHS present @ equilibrium than LHS • "product favored equilibrium" • forward reaction does happen to a great extent

• • • •

Solids and liquids never appear in equilibrium expression Gases can appear either as partial pressures (atm) or as concentration (M) Aqueous solutions are expressed as molarity K depends on temperature- If T is changed, K changes

•

In general 𝐾 = &$% Water does not appear in equilibrium expressions aA + bB cC + dD

• •

#

%$ aA + bB

cC + dD

Manipulating Eq. Expressions The value for the equilibrium constant depends on how we write the reaction The value of K depends on how we write our reaction

Changing direction 1. S(s) + O2(g)

SO2(g)

SO2(g)

Kc = [SO2]/[O2] = 4/2 = 2

S(s) + O2(g)

Kc' = [O2]/[SO2] = 2/4 = 1/2

Inverse relationship AKA reciprocal relationship 2. N2(g) + 3H2(g)

2NH3(g)

2NH3(g)

N2(g) + 3H2(g)

K = 3 x 108

K' = 1/3 x 10-8

Changing coefficients 1. S(s) + 3/2 O2(g)

2 S(s) + 3 O2(g)

SO3(g)

2 SO3(g)

To change these coefficients, we multiplied everything by an "x factor" in this case, x = 2

(Kc)x = Kc' Combining balanced chemical equations H2CO3 + H2O

H3O+ + HCO3-

K1 = 4.20 x 10-7 = [H3O+][HCO3-]/[H2CO3]

HCO - + H O

H O+ + CO

K

2-

4 80 x 10-1 [H O+][CO 2-]/ [HCO -]

HCO3 + H2O H3O+ + CO32 K2 = 4.80 x 10 1 = [H3O+][CO32 ]/ [HCO3 ] If we add the two reactions together we get: H2CO3 + HCO3- + 2H2O 2H3O+ + HCO3- + CO32Use this information to determine the equilibrium constant for the reaction: H2CO3 + 2H2O

2H3O+ + CO32-

K=?

K= (K1)(K2) = [H3O+][HCO3-]/[H2CO3] x [H3O+][CO32-]/ [HCO3-] = [H3O+]2[CO32-]/[H2CO3]

Changing direction

Inverse

Changing coefficients (Kc)x = Kc' Combining equations K= (K1)(K2)

Reaction Quotient Tells us if we are at equilibrium or not. Symbol = Q ( 𝑅𝐻𝑆) -𝑎𝑐𝑡𝑢𝑎𝑙-𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛𝑠 (𝐿𝐻𝑆)-𝑎𝑐𝑡𝑢𝑎𝑙-𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛𝑠 ( 𝑅𝐻𝑆) -𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚-𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛𝑠 𝐾= (𝐿𝐻𝑆)-𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚-𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛𝑠 𝑄=

If… Q=K

we are at equilibrium

QK

Too much RHS present Numerator is too big The reverse reaction needs to happen to reduce the RHS (increase LHS) so that Q = K

Equilibrium lies to left practice Butane

Isobutane

K = 2.5 If [Iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? Q = .35/.15 = 2.3333 Q...